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# 1983 AIME Problems/Problem 14
## Problem
In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$, one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$.
$[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("Q",Q,NW); label("P",P,1.5*dir(80)); label("R",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]$
## Solution
### Solution 1
First, notice that if we reflect $R$ over $P$ we get $Q$. Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$, we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ with radius $6$) that intersects circle $A$ at $Q$. The rest is just finding lengths:
Since $P$ is the midpoint of segment $BC$, $AP$ is a median of triangle $ABC$. Because we know that $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \sqrt{56}$. So now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then
$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.$
Doing routine algebra on the above equation, we find that $x^2=\frac{65}{2}$, so $PQ^2 = 4x^2 = \boxed{130}.$
### Solution 2 (easiest)
$[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label("A",a,S); label("B",b,S); label("M",m,NE); label("N",n,NE); label("P",p,N); label("Q",q,NW); label("R",r,E); label("12",(14,0),SW); label("6",(23,0),S); [/asy]$
Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$.
Since $\frac{AR}{MR}=\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$.
Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$.
Subtracting, $8y^2=28\Rightarrow y^2=\frac72\Rightarrow x^2=\frac{65}2\Rightarrow QP^2=4x^2=\boxed{130}$.
### Solution 3
Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\circ}$. By the Law of Cosines, $\angle APB=\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\cos^{-1}(x/16)$ and $\cos^{-1}(x/12)$. So we have
$\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).$
Taking the $\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$.
### Solution 4 (quickest)
Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle.
The line segment consisting of R and the first intersection of the larger circle has length 10. The length of the diameter of the larger circle be16.
Through power of a point, $$x \cdot 2x = 10 \cdot 26.$$ $$x^2 = 130.$$
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# How To Find A Circle Knowing The Diameter
## Video: How To Find A Circle Knowing The Diameter
A circle is a plane figure whose points are equally distant from its center, and the diameter of a circle is a segment passing through this center and connecting the two most distant points of the circle. It is the diameter that often becomes the value that allows you to solve most problems in geometry by finding a circle.
## Instructions
### Step 1
For example, to find the circumference of a circle, it is enough to determine the known diameter in the form of initial data. Specify that you know the diameter of the circle, equal to N, and draw a circle in accordance with this data. Since the diameter connects two points of the circle and passes through the center, therefore, the radius of the circle will always be equal to the value of the half diameter, that is, r = N / 2.
### Step 2
Use the mathematical constant π to find the length or any other value. It represents the ratio of the value of the circumference to the value of the length of the diameter of the circle and in geometric calculations is taken equal to π ≈ 3, 14.
### Step 3
To determine the circumference, take the standard formula L = π * D and plug in the diameter value D = N. As a result, the diameter multiplied by 3.14 will show the approximate circumference.
### Step 4
In the case when you need to determine not only the circumference of a circle, but also its area, also use the value of the constant π. Only this time, use a different formula, according to which the area of a circle is defined as the length of the radius, erected by the square, and multiplied by the number π. Accordingly, the formula looks like this: S = π * (r ^ 2).
### Step 5
Since in the initial data it is determined that the radius is r = N / 2, therefore, the formula for the area of a circle is modified: S = π * (r ^ 2) = π * ((N / 2) ^ 2). As a result, if you plug in a known diameter into the formula, you get the area you are looking for.
### Step 6
Do not forget to check in which units of measurement you need to determine the length or area of the circle. If the original data specifies that the diameter is measured in millimeters, the area of the circle should also be measured in millimeters. For other units - cm2 or m2, calculations are made in the same way.
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Unveiling the Mysteries: Exploring the Intricacies of Higher-Order Sets
Getting Started
In set theory, a discipline that forms the foundation of mathematics, sets are fundamental objects that allow us to classify and organize elements into distinct groups. Sets can contain any number of elements, including other sets. When a set itself consists of multiple sets, an intriguing question arises: What is a set of sets called? In this article, we will explore this concept and delve into the fascinating world of sets within sets.
Understanding sets
Before we can grasp the notion of a set of sets, it is crucial to have a solid understanding of sets themselves. In set theory, a set is a well-defined collection of distinct objects, called elements or members, which can be anything from numbers to letters or even other sets. For example, let’s look at the following sets:
Set A:
Set B:
Set C:
Here, Set A and Set B consist of individual elements, while Set C contains two sets, namely Set A and Set B. This brings us to the concept of a set of sets.
The Power Set
A set that contains all possible subsets of a given set is called a power set. In other words, the power set of a set is the collection of all possible combinations of its elements, including the empty set and the set itself. If we denote a set as S, then the power set of S is denoted as P(S).
For example, let’s consider the set S = . The power set of S, denoted P(S), would be , , , , }. Here, the power set contains four subsets: the empty set , the sets and , and the set .
Note that the power set of any set always has 2^n elements, where n is the number of elements in the original set. Thus, if a set contains ‘n’ elements, its power set will contain 2^n subsets.
Nesting of sets
In some cases, sets can be nested, forming a hierarchical structure. When a set contains other sets as its elements, it is called a nested set or a set of sets. This nesting can occur at any level, allowing for complex and intricate relationships between sets.
For example, let’s consider the following sets:
Set X:
Set Y:
Set Z:
Here, Set Z is a set of sets because it contains Set X and Set Y as elements. In this nested structure, Set Z is a parent set, while Set X and Set Y are considered child sets.
Alternative terminology
While there is no universally accepted term for a set of sets, it is worth mentioning some alternative terminologies that are used in different contexts. One common term is a “family of sets,” which refers to a collection of sets. This term emphasizes the collective nature of the sets within the group.
Another term sometimes used is a “collection of sets”. This terminology is more general and does not necessarily imply a hierarchical relationship between the sets in question. It is often used to describe a group of sets that share certain properties or characteristics.
Conclusion
In set theory, a power set is a fascinating concept that allows sets to be nested within each other. While there is no specific, universally accepted term to describe a set of sets, the power set provides a way to represent all possible subsets of a given set. The nesting of sets opens up a world of possibilities for organizing and categorizing elements, allowing for intricate relationships and structures in mathematics and beyond. By understanding the fundamentals of sets and their nesting, we can gain deeper insights into the underlying structure of mathematical concepts and scientific phenomena.
FAQs
What is a set of sets called?
A set of sets is called a “power set” or “set of subsets”.
How is a power set defined?
A power set is defined as the collection of all possible subsets of a given set.
What is the cardinality of a power set?
The cardinality of a power set is equal to 2 raised to the power of the cardinality of the original set. In other words, if a set has n elements, its power set will have 2^n elements.
Can a set be an element of its own power set?
Yes, a set can be an element of its own power set. The power set of any set includes both the set itself and the empty set.
Can a power set be empty?
Yes, a power set can be empty if the original set is also empty. In that case, the power set will contain only the empty set.
How do you represent a power set?
A power set is often denoted using the notation P(S), where S is the original set. Alternatively, you can use the notation 2^S to represent the power set, emphasizing the relationship between the cardinality of the set and its power set.
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### Remainder theorem
Introduction
How do we find the remainder when we divide by ?
One way is to use long division
From the long division, we get a remainder of 13
Note that:
is known as the divisor
is known as the quotient
is known as the dividend
In most cases we are only interested in the remainder, there is an easier way of obtaining the remainder without using long division
The easier way is to use the Remainder Theorem
Once again we want to find the remainder when is divided by
How do we know what value to sub in ?
If we are dividing by x - 2, we let x - 2 =0 and get x = 2. So we sub in 2
If we are dividing by x + 2, we let x + 2 =0 and get x = -2. So we sub in -2
If we are dividing by x - a, we let x - a =0 and get x = a. So we sub in a
Question 1
Given that leaves a remainder of 6 when divided by and has a factor of , find the value of a and b.
Next the question says that x + 2 is a factor
Factor means that the remainder is zero
Hence we can apply the remainder theorem
By solving the 2 simultaneous equations we can determine the values of a and b
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Theory:
Rational Expression
An expression is called a rational expression if it can be written in the form $$\frac{p(x)}{q(x)}$$ where $$p(x)$$ and $$q(x)$$ are polynomials and $$q(x) \neq 0$$.
A rational expression is the ratio of two polynomials.
Example:
The following are few examples of rational expressions.
1. $$\frac{x + y}{x - y}$$, where $$x \neq y$$
2. $$\frac{3ab}{5a^2b^3}$$
3. $$\frac{x^2 + 3x}{x^5}$$
Reduction of Rational Expression
A rational expression $$\frac{p(x)}{q(x)}$$ is said to be in its lowest form if the greatest common divisor of $$p(x)$$ and $$q(x)$$ is $$1$$.
That is $$GCD \left(p(x), q(x) \right)$$ = $$1$$.
Working rule to reduce a rational expression to its lowest form:
Step 1: Simplify or factorise the numerator $$p(x)$$ and the denominator $$q(x)$$.
Step 2: Cancel out the common factors in the numerator and the denominator.
Step 3: The final expression obtained after the above two steps is the rational expression in its lowest form.
Example:
Reduce the expression $$\frac{x^2 + 5x + 6}{x + 2}$$.
Solution:
Step 1: Factorise the numerator $$x^2 + 5x + 6$$ by splitting the middle term.
$$x^2 + 5x + 6$$ $$=$$ $$x^2 + 2x + 3x + 6$$
$$=$$ $$x (x + 2) + 3 (x + 2)$$
$$=$$ $$(x + 2)(x + 3)$$
Step 2: Rewrite the expression and cancel out the common factors.
$$\frac{x^2 + 5x + 6}{x + 2}$$ $$=$$ $$\frac{(x + 2)(x + 3)}{x + 2}$$
$$=$$ $\frac{\overline{)\left(x+2\right)}\left(x+3\right)}{\overline{)\left(x+2\right)}}$
$$=$$ $$x + 3$$
Step 3: Write the rational expression in its lowest form.
$$\frac{x^2 + 5x + 6}{x + 2}$$ $$=$$ $$x + 3$$
Therefore, the rational expression $$\frac{x^2 + 5x + 6}{x + 2}$$ is reduced to $$x + 3$$.
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# دوره GRE Test- Practice & Study Guide ، فصل 25 : GRE Quantitative Reasoning- Probability and Statistics
## دربارهی این فصل:
Looking for practice in probability and statistics for the GRE? This chapter offers engaging and fun video lessons to help prepare you for these questions on the GRE test. Test your knowledge with lesson quizzes and a chapter exam to help gauge your understanding of probability and statistics concepts.
## Statistics Are on the Rise
When you decide to move to a new home, whether it is in the same city or across the country, one of the first things that you might do is look at the crime statistics for that area. Are they on the rise, or are they declining? You might also look at the schools. How do they stack up against other schools? Are their scores on the way down, or are they rising year after year?
All of this information, and information on almost any topic you can think of, can be found using statistics. Statistics is the study of the collection and analysis of data.
## Relative and Cumulative Frequency
In mathematics, frequency refers to the number of times a particular event occurs. There are two types of frequency: relative and cumulative. Cumulative frequency is the total number of times a specific event occurs within the time frame given. Relative frequency is the number of times a specific event occurs divided by the total number of events that occur. Let’s use an example:
Your soccer team ended the season with a record of 15 wins and 3 losses. The cumulative frequency of your wins is 15 because that event occurred 15 times. The relative frequency of wins is 15 divided by 18, or 83%, because, out of the 18 total games (or events), your team won 15.
Relative frequency is a good way to predict how often an event might occur in the future. If you know that your soccer team has won 83% of the time in the past, you can reasonably assume that you will win 83% of the time in the future.
## Frequency Table
One of the best ways to tally and organize frequency data is to use a frequency table. A frequency table is a table that lists items and shows the number of times they occur. Below you see an example of a frequency table that describes the different ways students travel to get to school. You can also include a column that gives the relative frequency of each event.
## Percent Increase
Let’s go back to our example of moving to a new town for a minute. If you were to read that the area you wanted to move to had a 2% increase in crime in the last year, you might think twice about moving there. At the very least, you would do some extra research.
What if you discovered in your additional research that the test scores at the local elementary school increased by 23% in the last year? That statistic might be more important to you than the small increase in crime in the same area. The percent increase represents the relative change between the old value and the new value. In order to calculate percent increase, you must have collected data about the same event, just at a different time.
## Calculating Percent Increase
To determine the percent increase between two sets of data, you can use a frequency table and the following formula:
Percent Increase = Frequency 2 - Frequency 1 / Frequency 1 100.
Let’s try an example: Data for crime in a certain area was recorded over two years. The following table shows the occurrences of three different types of crime over a two-year period.
To calculate the percent increase, take each row individually and plug the numbers into the equation.
Percent increase of robbery = ((37 - 33) / 33) 100 = (4 / 33) 100 = 0.12 100 = 12%
Percent increase of murder = ((8 - 2) / 2) 100 = (6 / 2) 100 = 3 100 = 300%
Percent increase of assault = ((16 - 15) / 15) 100 = (1 / 15) 100 = 0.07 100 = 7%
Then, you can complete the table.
Here we have another example: 100 people were asked how often they ate dinner at home in a typical week. Then, they all took a class on how to cook healthy meals at home and after six months were asked again how often they ate at home. What was the percent increase for eating at home 5, 6 and 7 nights a week? Take a minute to try this one on your own.
## Table corresponding to example problem above
And here you see the answers:
The percent increase for eating at home 5 nights a week was 347%, 6 nights a week - 350%, and 7 nights a week - 125%.
## Lesson Summary
Determining the frequency and percent increase of events can be very helpful when trying to interpret certain sets of data. The cumulative frequency of a certain event is the number of times that event occurs in a certain time frame. Relative frequency refers to the percentage a certain event is of the total amount of events that occur. You can determine the percent increase of an event from time to time by using the formula:
Percent Increase = Frequency 2 - Frequency 1 / Frequency 1 100
## این مجموعه تلوزیونی شامل 15 فصل زیر است:
In this lesson, we will examine two of the most widely used types of graphs- bar graphs and pie charts. These two graphs can provide the reader with a comparison of the different data that is displayed.
Measures of central tendency can provide valuable information about a set of data. In this lesson, explore how to calculate the mean, median, mode and range of any given data set.
Simple, compound, and complementary events are different types of probabilities. Each of these probabilities are calculated in a slightly different fashion. In this lesson, we will look at some real world examples of these different forms of probability.
To calculate the probability of a combination, you will need to consider the number of favorable outcomes over the number of total outcomes. Combinations are used to calculate events where order does not matter. In this lesson, we will explore the connection between these two essential topics.
In this lesson, you will learn how to calculate the probability of a permutation by analyzing a real-world example in which the order of the events does matter. We'll also review what a factorial is. We will then go over some examples for practice.
Sometimes probabilities need to be calculated when more than one event occurs. These types of compound events are called independent and dependent events. Through this lesson, we will look at some real-world examples of how to calculate these probabilities.
While the definition of factorial isn't complicated, it's easy to make them trickier by throwing a lot of them together and adding in some fractions. Test your skills here with some algebraic examples that make you use factorials without many numbers.
Maybe it's because I'm a math teacher, but when I watched the Olympics I found myself thinking about how many different ways the swimmers could have finished the race. In this video, you'll learn the answer to this question, why it's important and how it lead to the invention of the mathematical operation called the factorial.
Combinations are an arrangement of objects where order does not matter. In this lesson, the coach of the Wildcats basketball team uses combinations to help his team prepare for the upcoming season.
A permutation is a method used to calculate the total outcomes of a situation where order is important. In this lesson, John will use permutations to help him organize the cards in his poker hand and order a pizza.
Occasionally when calculating independent events, it is only important that the event happens once. This is referred to as the 'At Least One' Rule. To calculate this type of problem, we will use the process of complementary events to find the probability of our event occurring at least once.
Statistics is the study and interpretation of a set of data. One area of statistics is the study of probability. This lesson will describe how to determine the either/or probability of overlapping and non-overlapping events.
In this lesson, we will examine the meaning and process of calculating the standard deviation of a data set. Standard deviation can help to determine if the data set is a normal distribution.
In statistics, one way to describe and analyze data is by using frequency tables. This lesson will discuss relative and cumulative frequencies and how to calculate percent increase using these two methods.
Conditional probability, just like it sounds, is a probability that happens on the condition of a previous event occurring. To calculate conditional probabilities, we must first consider the effects of the previous event on the current event.
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Maths Square and Square Roots part 17 (Repeated Subtraction Method) CBSE Class 8 Mathematics VIII | Summary and Q&A
15.2K views
January 26, 2017
by
LearnoHub - Class 11, 12
Maths Square and Square Roots part 17 (Repeated Subtraction Method) CBSE Class 8 Mathematics VIII
TL;DR
This video explains how to find square roots using the repeated subtraction method and discusses its limitations.
Key Insights
• ⬛ The repeated subtraction method is a simple option for finding square roots but is not favorable for large numbers.
• #️⃣ The method involves subtracting odd numbers from the given number until the result becomes 0.
• ❎ Square roots can be found using this method, but it only works for perfect square numbers.
• 🫚 The repeated subtraction method is based on the relationship between square roots and squaring.
• 🌥️ This method requires multiple steps, making it less efficient for larger numbers.
• ⬛ Alternative methods are needed to find the square root of large numbers.
• 🫚 The repeated subtraction method can be useful for understanding the concept of square roots.
Transcript
hello friends this video on question square root 517 is brought to you by example.com no more your thumb exam so that now next question that comes to your mind is how do we find square root of a number that it seems to be a big challenge that finding squares makes it easier because we only have to multiply even if we go by the actual multiplication... Read More
Q: What is the repeated subtraction method for finding square roots?
The repeated subtraction method involves subtracting odd numbers sequentially from the given number until the result becomes 0. It is used to find the square root of perfect square numbers.
Q: Is the repeated subtraction method suitable for finding the square root of large numbers?
No, the repeated subtraction method is not suitable for finding the square root of large numbers. It requires multiple steps, making it inefficient and time-consuming.
Q: How does the repeated subtraction method relate to square roots?
The repeated subtraction method is based on the fact that the square root is the inverse function of squaring. It uses the concept of adding or subtracting odd numbers to find square roots or square numbers, respectively.
Q: What are the limitations of the repeated subtraction method?
The repeated subtraction method is not suitable for large numbers due to the large number of steps involved. It becomes increasingly time-consuming as the number gets larger.
Summary & Key Takeaways
• The video introduces the repeated subtraction method as a simple option for finding square roots but states that it is not suitable for large numbers.
• It explains that the method involves subtracting odd numbers from the given number until the result becomes 0.
• The video highlights that this method works only for perfect square numbers and provides an example of finding the square root of 25 in five steps.
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Proving $x-1$ is a Factor of $P(x)$ with Polynomials
• MHB
• anemone
In summary, when we say that $x-1$ is a factor of $P(x)$ with polynomials, it means that when we divide $P(x)$ by $x-1$, the remainder is equal to zero. To prove this, we can use the factor theorem and substitute $x=1$ to see if the result is equal to zero. Alternatively, we can use synthetic division with $a=1$ and if the remainder is zero, then $x-1$ is a factor of $P(x)$. Proving this is important because it allows us to factorize $P(x)$, find other factors, and understand the behavior of the graph near $x=1$.
anemone
Gold Member
MHB
POTW Director
If $P(x),\,Q(x),\,R(x),\,S(x)$ are polynomials such that $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$, prove that $x-1$ is a factor of $P(x)$.
Let $P(x)=a_nx^n+\cdots+a_0x^0$ with $a_n\ne 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_n(n-2m)(n-1)=0$, so $n=1$ or $n=2m$.
If $n=1$, one easily verifies that $P(x)=x$ is a solution, while $P(x)=1$ is not. Since the given condition is linear in $P$, this means that the linear solutions are precisely $P(x)=tx$ for $t\in \mathbb{R}$.
Now, assume that $n=2m$. The polynomial $xP(x+1)-(x+1)P(x)=(n-1)a_nx^n+\cdots$ has degree $n$, and therefore it has at least one (possibly complex) root $r$. If $r\ne \{0,\,-1\}$, define $k=\dfrac{P(r)}{r}=\dfrac{P(r+1)}{r+1}$. If $r=0$, let $k=P(1)$. If $r=-1$, let $k=-P(-1)$. We now consider the polynomial $S(x)=P(x)-kx$. It also satisfies (1) because $P(x)$ and $kx$ satisfy it. Additionally, it has the useful property that $r$ and $r+1$ are roots.
Let $A(x)=x^3-mx^2+1$ and $B(x)=x^3+mx^2+1$. Plugging in $x=s$ into (1) implies that
If $s-1$ and $s$ are roots of $S$ and $s$ is not a root of $A$, then $s+1$ is a root of $S$.
If $s$ and $s+1$ are roots of $S$ and $s$ is not a root of $B$, then $s-1$ is a root of $S$.
Let $a\ge 0$ and $b\ge 1$ be such that $r-a,\,r-a+1,\,\cdots, r,\,r+1,\.\cdots,\,r+b-1,\,r+b$ are roots of $S$, while $r-a-1$ and $r+b+1$ are not. The two statements above imply that $r-a$ is a root of $B$ and $r+b$ is a root of $A$.
Since $r-a$ is a root of $B(x)$ and of $A(x+a+b)$, it is also a root of their greatest common divisior $C(x)$ as integer polynomials. If $C(x)$ was a non-trivial divisor of $B(x)$, then $B$ would have a rational root $\alpha$. Since the first and last coefficients of $B$ are 1, $\alpha$ can only be 1 or $-1$, but $B(-1)=m>0$ and $B(1)=m+2>0$ since $n=2m$.
Therefore $B(x)=A(x+a+b)$. Writing $c=a_b\ge 1$, we compute
$0=A(x+c)-B(x)=(3c-2m)x^2+c(3c-2m)x^2+c(3c-2m)x+c^2(c-m)$
Then we must have $3c-2m=c-m=0$, which gives $m=0$, a contradiction. We could conclude that $f(x)=tx$ is the only solution.
1. How do you prove that $x-1$ is a factor of $P(x)$ with polynomials?
To prove that $x-1$ is a factor of $P(x)$, we can use the Remainder Theorem. This theorem states that if we divide $P(x)$ by $x-1$, the remainder will be equal to $P(1)$. Therefore, if $P(1) = 0$, then $x-1$ is a factor of $P(x)$.
2. Can you provide an example of proving $x-1$ is a factor of $P(x)$ with polynomials?
For example, let's say we have the polynomial $P(x) = 3x^3 - 2x^2 + 5x - 2$. To prove that $x-1$ is a factor of $P(x)$, we divide $P(x)$ by $x-1$ using long division or synthetic division. The remainder is $P(1) = 4$, therefore $x-1$ is not a factor of $P(x)$. However, if $P(x) = 2x^3 + 4x^2 - 3x + 1$, the remainder is $P(1) = 4$, which means $x-1$ is a factor of $P(x)$.
3. What is the significance of proving $x-1$ is a factor of $P(x)$ with polynomials?
Proving that $x-1$ is a factor of $P(x)$ allows us to simplify the polynomial and potentially find its roots. This makes it easier to work with and solve equations involving the polynomial.
4. Are there any other methods to prove that $x-1$ is a factor of $P(x)$ with polynomials?
Yes, there are other methods such as using the Factor Theorem or the Rational Root Theorem. These theorems provide conditions for a polynomial to have a certain factor, and can be used to prove that $x-1$ is a factor of $P(x)$.
5. What are some common mistakes to avoid when proving $x-1$ is a factor of $P(x)$ with polynomials?
One common mistake is forgetting to check the remainder after dividing by $x-1$. It is important to remember that if the remainder is not equal to zero, then $x-1$ is not a factor of $P(x)$. Another mistake is assuming that just because $P(1) = 0$, $x-1$ is automatically a factor. It is important to use long division or synthetic division to confirm this.
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# Impulse and Linear momentum
### (1) Introduction
• It becomes difficult to use Newton's law of motion as it is while studying complex problems like collision of two objects,motion of the molecules of the gas,rocket propulsion system etc
• Thus a further study of newton's law is required to find some theorem or principles which are direct consequences of Newton's law
• We have already studied one such principle which is principle of conservation of energy.Here in this chapter we will define momentum and learn about the principle of conservation of momentum .
• Thus we begin this chapter with the concept of impluse and momentum which like work and energy are developed from Newton's law of motion
### (2) Impulse and Linear momentum
• To explain the terms impulse and momentum consider a particle of mass m is moving along x-axis under the action of constant force F as shown below in the figure
• If at time t=0 ,velocity of the particle is v0 then at any time t velocity of particle is given by the equation
v = v0 + at
where a = F/m
can be determined from the newton's second law of motion .Putting value of acceleration in above equation\
we get
mv = mv0 + Ft
or
Ft = mv - mv0 -(1)
• right side of the equation Ft, is the product of force and the time during which the force acts and is known as the impluse
Thus
Impulse= Ft
• If a constant force acts on a body during a time from t1 and t2,then impulse of the force is
I = F(t2-t1) -(2)
Thus impulse recieved during an impact is defined as the product of the force and time interval during which it acts
• Again consider left hand side of the equation (1) which is the difference of the product of mass and velocity of the particle at two different times t=0 and t=t
• This product of mass and velocity is known as linear momentum and is represented by the symbol p. Mathematically
p = mv --(3)
• physically equation (1) states that the impulse of force from time t=0 to t=t is equal to the change in linear momentum during
• If at time t1 velocity of the particle is v1 and at time t2 velocity of the particle is v2,then
F(t2-t1)=mv2-mv1 -(4)
• so far we have considered the case of the particle moving in a straight line i.e along x-axis and quantities involved F,v, and a were all scalars
• If we call these quantities as components of the vectors F,v and a along x-axis and generalize the definations of momentum and impulse so that the motion now is not constrained along one -direction ,Thus we got
Impulse=I=F(t2-t1) -(5)
Linear momentum=p=mv -(6)
where
I=Ixi+Iyj+Izk
F=Fxi+Fyj+Fzk
p=pxi+pyj+pzk
v=vxi+vyj+vzk
are expressed in terms of their components along x,y and z axis and also in terms of unit vectors
• On generalizing equation (4) using respective vectors quantities we get the equation
F(t2-t1) =mv2-mv1 -(7)
• So far while discussing Impulse and momentum we have considered force acting on particle is constant in direction and maagnitude
• In general ,the magnitude of the force may vary with time or both the direction and magnitude may vary with time
• Consider a particle of mass m moving in a three-dimensional space and is acted upon by the varying resultant force F. Now from newtons second law of motion we know that
F=m(dv/dt)
or Fdt=mdv
• If at time t1 velocity of the particle is v1 and at time t2 velocity of the particle is v2,then from above equation we have
• Integral on the left hand side of the equation (8) is the impulse of the force F in the time interval (t2-t1) and is a vector quantity,Thus
Above integral can be calculated easily if the Force F is some known function of time t i.e.,
F=F(t)
• Integral on the right side is when evaluated gives the product of the mass of the particle and change in the velocity of the particle
• using equation (9) and (10) to rewrite the equation (8) we get
• Equivalent equations of equation (11) for particle moving in space are
• Thus we conclude that impulse of force F during the time interval t2-t1 is equal to the change in the linear momentum of the body on which its acts
• SI units of impulse is Ns or Kgms-1
Watch this tutorial for more information on How to solve momentum/Impulse problems
|
# Calculus Squeeze Theorem
In this section, ask-math explains you the calculus squeeze theorem.
The limit of function that is squeezed between two other functions, each function will have same limit at given value of x. If two functions squeeze together at a particular point, then any function trapped between them will get squeezed to that same point.
1) If two functions sandwich together at a particular point, then any function trapped between them will get squeezed to that same point.
2) The squeeze or sandwich theorem deals with the limit values rather instead of function values.
3) Squeeze theorem or sandwich theorem is also called Pinch theorem.
If h(x) $\leq$ f(x) $\leq$ g(x), for all x in open interval containing 'c' except possibly at 'c' itself, and if
$\lim_{x->c} h(x) = L = \lim_{x->c} g(x)$ then
$\lim_{x->c}f(x)$ exists and is equal to 'L'.
## Examples on calculus squeeze theorem
1) Evaluate $\lim_{x->0} x * sin(\frac{1}{x})$
Solution : As we know that the sine function lies between -1 and 1.
$-1 \leq sin(\frac{1}{x})\leq 1$ for all x
So, $-|x| \leq sin(\frac{1}{x})\leq |x|$ for all x
We know that
$\lim_{x->0}|x|$ = 0 and $\lim_{x->0}-|x|$ = 0
So according to squeeze theorem $\lim_{x->0} x * sin(\frac{1}{x})$ = 0
2) Use sandwich theorem to find $\lim_{x->c}f(x)$, c = 0 , $4 - x^{2}\leq f(x)\leq 4 + x^{2}$
Solution : Let h(x) = $4 - x^{2}$ and g(x) = $4 + x^{2}$
Now we will use squeeze theorem to find lim f(x) as x approaches to 0
h(x) $\leq$ f(x) $\leq$ g(x)
$\lim_{x->0}4 - x^{2}$ = 4 and $\lim_{x->0}4 + x^{2}$ = 4
$4 \leq \lim_{x->0}f(x)\leq 4$
So, $\lim_{x->0}f(x)$ = 4
3) Evaluate $\lim_{x->0} x * cos(\frac{1}{x})$
Solution : As we know that the sine function lies between -1 and 1.
$-1 \leq cos(\frac{1}{x})\leq 1$ for all x
So, $-|x| \leq cos(\frac{1}{x})\leq |x|$ for all x
We know that
$\lim_{x->0}|x|$ = 0 and $\lim_{x->0}-|x|$ = 0
So according to squeeze theorem $\lim_{x->0} x * cos(\frac{1}{x})$ = 0
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# Long Answers - Force and Laws of Motion, Science, Class 9 Class 9 Notes | EduRev
## Class 9 : Long Answers - Force and Laws of Motion, Science, Class 9 Class 9 Notes | EduRev
The document Long Answers - Force and Laws of Motion, Science, Class 9 Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Science by VP Classes.
All you need of Class 9 at this link: Class 9
Q1. Explain Newton’s second law of motion and with the help of an example show how it is used in sports.
Ans.
Newton’s second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Let us assume : Object of mass m, is moving along a straight line with an initial velocity ‘u’, It is uniformly accelerated to velocity v in time ‘t’ by the application of force, F throughout the time ‘t’.
Initial momentum of the object = P1 = mu
Final momentum of the object = P2 = mv
The change in momentum ∝ P2 – P1
∝ mv – mu = m (v – u)
The rate of change of momentum
∴ Applied force
∴ F = k ma
k = constant of proportionality F = kg m/s2 = Newton
Use of second law of motion in sports: In cricket field, the fielder gradually pulls his hands backward while catching a ball. The fielder catches the ball and gives swing to his hand to increase the time during which the high velocity of the moving ball decreases to zero.
The acceleration of the ball is decreased and therefore the impact of catching the fast moving ball is reduced.
If not done so, then the fast moving ball will exert large force and may hurt the fielder.
Q2. State all 3 Newton’s law of motion. Explain inertia and momentum.
Ans. Newton’s I law of motion:
An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.
Newton’s II law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Newton’s III law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies.
Inertia: The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia.
Momentum: The momentum of an object is the product of its mass andvelocity and has the same direction as that of the velocity.
Its S.I. unit is kg m/s. p = m × v
Q3. Define force. Give its unit and define it. What are different types of forces?
Ans. Force:
It is a push or pull on an object that produces acceleration in the body on which it acts.
A force can do 3 things on a body
(a) It can change the speed of a body.
(b) It can change the direction of motion of a body.
(c) It can change the shape of the body.
The S.I. unit of force is Newton.
Newton: A force of one Newton produces an acceleration of 1 m/s2 on an object of mass 1kg.
1N = 1 kg m/s2 Force = mass × acceleration
Types of forces: (i) Balanced force: When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
(ii) Unbalanced force: When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced force.
(iii) Frictional force: The force that always opposes the motion of object is called force of friction.
Q4. What is inertia? Explain different types of inertia. Give three examples in daily life which shows inertia.
Ans.
Inertia: The natural tendency of an object to resist change in their state of rest or of motion is called inertia.
The mass of an object is a measure of its inertia. Its S.I. unit is kg.
Types of inertia: Inertia of rest: The object at rest will continue to remain at rest unless acted upon by an external unbalanced force.
Inertia of motion: The object in the state of uniform motion will continue to remain in motion with same speed and direction unless it is acted upon by an external unbalanced force.
Three examples of inertia in daily life are:
(i) When we are travelling in a vehicle and sudden brakes are applied we tend to fall forward.
(ii) When we shake the branch of a tree vigorously, leaves fall down.
(iii) If we want to remove the dust from carpet we beat the carpet so that dust fall down.
Q5. Consider the two forces acting on a person standing on the ground; the downward pull of gravity and the upward push of the ground.
(a) Explain whether the forces are equal and opposite.
(b) Explain whether the forces are an action-reaction pair.
Ans.
(a) They are equal and opposite but the forces are not acting on two different bodies, the forces in question are on the same body.
(b) There are two action-reaction pairs which combine to make the effect – gravity provides the Earth-person attraction and the normal perpendicular force provides the Earth-person repulsion it is equal and opposite to prevent the person sinking into the Earth.
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# What are the 4 representations of a function?
## What are the 4 representations of a function?
Note We will typically represent related data in four ways or from four viewpoints. Numerically (using a chart or table of data) • Graphically (using a scatter plot or continuous graph) • Verbally (using a word description) • Algebraically (using a mathematical model).
What are types of function in math?
There are various types of functions in mathematics which are explained below in detail. Quadratic Function. Rational Function. Algebraic Functions.
### What are the types of the functions?
Types of Functions
Based on Elements One-One Function Many-One Function Onto Function One-One and Onto Function Into Function Constant Function
Based on the Equation Identity Function Linear Function Quadratic Function Cubic Function Polynomial Functions
What are the 4 ways that a math problem can be represented?
Key Takeaways
• A function can be represented verbally. For example, the circumference of a square is four times one of its sides.
• A function can be represented algebraically. For example, 3x+6 3 x + 6 .
• A function can be represented numerically.
• A function can be represented graphically.
#### What is general mathematical function?
In mathematics, a function is a relation between a set of inputs and a set of permissible outputs. Functions have the property that each input is related to exactly one output. For example, in the function f(x)=x2 f ( x ) = x 2 any input for x will give one output only. We write the function as:f(−3)=9 f ( − 3 ) = 9 .
What are the different types of functions in discrete mathematics?
Types of functions:
• One to one function(Injective): A function is called one to one if for all elements a and b in A, if f(a) = f(b),then it must be the case that a = b.
• Onto Function (surjective): If every element b in B has a corresponding element a in A such that f(a) = b.
## How many functions are there?
Explanation: From a set of m elements to a set of 2 elements, the total number of functions is 2m. Out of these functions, 2 functions are not onto (If all elements are mapped to 1st element of Y or all elements are mapped to 2nd element of Y). So, number of onto functions is 2m-2.
How do you identify functions?
Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function.
### What is an example of a function in math?
How do you write a function in math?
You write functions with the function name followed by the dependent variable, such as f(x), g(x) or even h(t) if the function is dependent upon time. You read the function f(x) as “f of x” and h(t) as “h of t”. Functions do not have to be linear. The function g(x) = -x^2 -3x + 5 is a nonlinear function.
#### How many functions are there in the following mathematical relation?
How many functions are there in the following mathematical relation? Explanation: There are only 2 functions used to describe a meaningful relationship between p and x. They are the sin() and log(x).
How do you find the number of functions?
If a set A has m elements and set B has n elements, then the number of functions possible from A to B is nm. For example, if set A = {3, 4, 5}, B = {a, b}. If a set A has m elements and set B has n elements, then the number of onto functions from A to B = nm – nC1(n-1)m + nC2(n-2)m – nC3(n-3)m+…. – nCn-1 (1)m.
## What are the four types of function in mathematics?
One – one function (Injective function) If each element in the domain of a function has a distinct image in the co-domain,the function is said to be one –
• Many – one function.
• Onto – function (Surjective Function) A function is called an onto function if each element in the co-domain has at least one pre – image in the domain.
• Into – function.
• What are the most important functions in mathematics?
Pythagoras. The life of the famous Greek Pythagoras is somewhat mysterious.
• David Hilbert. The German mathematician David Hilbert is one of the most influential figures from the field in the 19th and 20th centuries.
• Sir Isaac Newton.
• Hypatia.
|
# How do you simplify sqrt10*sqrt20?
Jan 24, 2017
$10 \sqrt{2}$
#### Explanation:
Using the property that states that $\sqrt{a} \sqrt{b} = \sqrt{a b}$,
$\sqrt{10} \cdot \sqrt{20} = \sqrt{10 \cdot 20} = \sqrt{200} = \sqrt{25 \cdot 4 \cdot 2}$
$= \sqrt{25} \sqrt{4} \sqrt{2} = 5 \cdot 2 \cdot \sqrt{2} = 10 \sqrt{2}$.
Jan 24, 2017
Multiply the two radicals together, then simplify the result. Details below...
The result is $10 \sqrt{2}$
#### Explanation:
First, since square root is the same as using the exponent $\frac{1}{2}$ the two radicals obey the rule
${a}^{x} \cdot {b}^{x} = {\left(a \cdot b\right)}^{x}$
So ${10}^{\frac{1}{2}} \cdot {20}^{\frac{1}{2}} = {200}^{\frac{1}{2}}$ or $\sqrt{200}$
Next, look for the largest factor of 200 that is a perfect square. This is $100 \times 2$
So, $\sqrt{200} = \sqrt{100} \cdot \sqrt{2} = 10 \sqrt{2}$
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NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3
# NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3
## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3.
### Ex 11.3 Class 7 Maths Question 1.
Find the circumference of the circles with the following radius. (Take π = 22/7)
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution:
(a) Given: Radius (r) = 14 cm
Circumference of a circle = 2Ï€r = 2 × 22/7 × 14
= 88 cm
(b) Given: Radius (r) = 28 mm
Circumference of a circle = 2Ï€r = 2 × 22/7 × 28
= 176 mm
(c) Given: Radius (r) = 21 cm
Circumference of a circle = 2Ï€r = 2 × 22/7 × 21
= 132 cm
### Ex 11.3 Class 7 Maths Question 2.
Find the area of the following circles, given that (Take π = 22/7)
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm
Solution:
(a) Given, r = 14 mm
Area of a circle = πr2
= Ï€ × 14 × 14 = 22/7 × 14 × 14
= 616 mm2
(b) Given, diameter = 49 m
(c) Given, radius = 5 cm
### Ex 11.3 Class 7 Maths Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7)
Solution:
Given, circumference = 154 m
2Ï€r = 154
### Ex 11.3 Class 7 Maths Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds offence. Also find the cost of the rope, if it costs ₹ 4 per metre. (Take Ï€ = 22/7)
Solution:
Given, diameter of the circular garden = 21 m
Radius = 21/2 m
Circumference = 2Ï€r = 2 × 22/7 × 21/2
= 66 m
Length of the rope needed for 2 rounds
= 2 × 66 m = 132 m
Cost of the rope = ₹4 × 132 = ₹ 528
### Ex 11.3 Class 7 Maths Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Solution:
Radius of the circular sheet = 4 cm
Area of the circular sheet = Ï€r2 = Ï€ × 4 × 4 = 16Ï€ cm2
Radius of the circular sheet to be removed = 3 cm
Area of the circular sheet removed = πr2 = 9π cm2
Area of the remaining sheet
= (16Ï€ – 9Ï€) cm2 = 7Ï€ cm2
= 7 × 3.14 cm2 = 21.98 cm2
Hence, the area of the remaining sheet is 21.98 cm2.
### Ex 11.3 Class 7 Maths Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one metre of the lace costs ₹15. (Take Ï€ = 3.14)
Solution:
Given, diameter of the table cover = 1.5 m
Radius = 1.5/2 = 0.75 m
Length of the lace = 2Ï€r = 2 × 3.14 × 0.75
= 4.710 m
Cost of the lace = ₹ 15 × 4.710 = ₹ 70.65
### Ex 11.3 Class 7 Maths Question 7.
Find the perimeter of the given figure, which is a semicircle including its diameter.
Solution:
Given, diameter = 10 cm
Hence, the required perimeter is 25.7 cm (approx.).
### Ex 11.3 Class 7 Maths Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15 per m2. (Take Ï€ = 3.14)
Solution:
Given, diameter = 1.6 m
Radius = 1.6/2 = 0.8 m
Area of the table-top = πr2
= 3.14 × 0.8 × 0.8 m2
= 2.0096 m2
Cost of polishing = ₹ 15 × 2.0096
= ₹ 30.14 (approx.)
### Ex 11.3 Class 7 Maths Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π = 22/7)
Solution:
Length of the wire to be bent into a circle = 44 cm
2Ï€r = 44
Now, the length of the wire is bent into a square.
Here, perimeter of square = Length of the wire
4 × Side = 44
Side = 44/4 = 11 cm
Area of the square = (Side)2 = (11)2 = 121 cm2
Since, 154 cm2 > 121 cm2
Thus, the circle encloses more area than the square.
### Ex 11.3 Class 7 Maths Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure below). Find the area of the remaining sheet. (Take π = 22/7)
Solution:
Radius of the circular sheet = 14 cm
Area of the circular sheet = Ï€r2 = 22/7 × 14 × 14 cm2
= 616 cm2
Area of 2 small circles = 2 × Ï€r2
= 2 × 22/7 × 3.5 × 3.5 cm2
= 77 cm2
Area of the rectangle = l × b
= 3 × 1 cm2 = 3 cm2
Area of the remaining sheet after removing the 2 circles and 1 rectangle
= 616 cm2 – (77 + 3) cm2
= 616 cm2 – 80 cm2 = 536 cm2
### Ex 11.3 Class 7 Maths Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
Solution:
Given, side of the square sheet = 6 m
Area of the sheet = (Side)2 = (6)2 = 36 cm2
Radius of the circle = 2 cm
Area of the circle to be cut out = πr2
= 3.14 × 2 × 2 = 12.56 cm2
Area of the left over sheet = 36 cm2 12.56 cm2
= 23.44 cm2
### Ex 11.3 Class 7 Maths Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take π = 3.14)
Solution:
Circumference of the circle = 31.4 cm
2Ï€r = 31.4
r = 31.4/(2 × 3.14) = 5 cm
Area of the circle = Ï€r2 = 3.14 × 5 × 5 = 78.5 cm2
Hence, the radius of the circle is 5 cm and area is 78.5 cm2.
### Ex 11.3 Class 7 Maths Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (Take π = 3.14)
Solution:
Given, diameter of the flower bed = 66 m
Radius = 66/2 = 33 m
Let r1 = 33 m
Width of the path = 4 m
Radius of the flower bed including path
= 33 m + 4 m = 37 m
Let r2 = 37 m
Area of the circular path = π(r22r12)
= 3.14 (372 – 332)
= 3.14 × (37 + 33) (37 – 33) [Using a2 – b2 = (a + b)(a – b)]
= 3.14 × 70 × 4 = 879.20 m2
Hence, the area of the path is 879.20 m2.
### Ex 11.3 Class 7 Maths Question 14.
A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler can water the entire garden? [Take π = 3.14]
Solution:
Area of the flower garden = 314 m2
Radius of the circular portion covered by the sprinkler = 12 m
Area of the circular portion covered by the sprinkler
= Ï€r2 = 3.14 × 12 × 12
= 3.14 × 144 m2 = 452.16 m2
Since 452.16 m2 > 314 m2
Yes, the sprinkler can water the entire garden.
### Ex 11.3 Class 7 Maths Question 15.
Find the circumference of the inner and the outer circles, shown in the given figure. (Take π = 3.14)
Solution:
Radius of the outer circle = 19 m
Circumference of the outer circle = 2Ï€r
= 2 × 3.14 × 19 = 3.14 × 38 m
= 119.32 m
Radius of the inner circle = 19 m – 10 m = 9 m
Circumference of the inner circle = 2Ï€r = 2 × 3.14 × 9
= 56.52 m
Hence, the required circumferences are 56.52 m and 119.32 m.
### Ex 11.3 Class 7 Maths Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
Solution:
Given, radius of the wheel = 28 cm
Circumference = 2Ï€r = 2 × 22/7 × 28 = 176 cm
Number of rotations made by the wheel in 1 rotation = 176 cm
Number of rotations made by the wheel in going 352 m or 35200 cm
= 35200/176 = 200
Hence, the number of rotations made by the wheel is 200.
### Ex 11.3 Class 7 Maths Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
Solution:
Given, length of the minute hand = 15 cm
Radius = 15 cm
Circumference = 2Ï€r
= 2 × 3.14 × 15 cm = 94.2 cm
Since the minute hand covers the distance in 1 hour equal to the circumference of the circle. Hence, the required distance covered by the minute hand is 94.2 cm.
You can also like these:
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12
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# Grade 8 Introduction to Graphs Worksheets
## Grade 8 Maths Introduction to Graphs Multiple Choice Questions (MCQs)
1. The coordinate of A in the adjacent graph is:
(a) (-7, 3)
(b) (7, -7)
(c) (-6, -1)
(d) (2, -3)
2. The coordinate of B in the adjacent graph is:
(a) (-7, 3)
(b) (7, -7)
(c) (-6, -1)
(d) (2, -3)
3. The coordinate of C in the adjacent graph is:
(a) (-7, 3)
(b) (7, -7)
(c) (-6, -1)
(d) (2, -3)
4. The coordinate of D in the adjacent graph is:
(a) (-7, 3)
(b) (7, -7)
(c) (-6, -1)
(d) (2, -3)
5. The coordinate of E in the above graph is:
(a) (9, -3)
(b) (-4, 2)
(c) (-3, -4)
(d) (-7, 9)
6. If y – coordinate of a point is zero, then this point always lies:
(c) x-axis
(d) y-axis
7. If x – coordinate of a point is zero, then this point always lies:
(c) x-axis
(d) y-axis
8. The point (-4, -3) means:
(a) x = -4, y = -3
(b) x = -3, y = -4
(c) x = 4, y = 3
(d) None of these
9. On joining points (0, 0), (0, 2), (2, 2) and (2, 0) we obtain a:
(a) Square
(b) Rectangle
(c) Rhombus
(d) Parallelogram
10. Point (-2, 3) lies in the:
11. Point (0, -2) lies:
(a) on the x-axis
(c) on the y-axis
12. Abscissa of the all the points on x-axis is:
(a) 0
(b) 1
(c) -1
(d) any number
13. Ordinate of the all the points on x-axis is
(a) 0
(b) 1
(c) -1
(d) any number
14. Ordinate of the all the points on y-axis is
(a) 0
(b) 1
(c) -1
(d) any number.
15. The point whose ordinate is 4 and which lies on y-axis is:
(a) (4, 0)
(b) (0, 4)
(c) (1, 4)
(d) (4, 2)
16. The perpendicular distance of the point P(3, 4) from the y-axis is:
(a) 3
(b) 4
(c) 5
(d) 7
17. The number of coordinate axis is:
(a) 1
(b) 2
(c) 4
(d) 3
18. The y-axis is a:
(a) horizontal line
(b) vertical line
(c) oblique line
(d) vertical seqment
19. The abscissa of the point (-3, -2) is
(a) -3
(b) -2
(c) -5
(d) -1
20. Y-coordinate is also called:
(a) Abscissa
(b) Ordinate
(c) y-axis
(d) None of these
### Class 8 Maths Introduction to Graphs Fill In The Blanks
1. The point where X-axis and Y-axis meet is called ……………….. .
2. Who is considered to be father of cartesian system …………….. .
3. The X-coordinate of the point A (3, 7) is ……………… .
4. Perpendicular distance of the point (2, 3) from X-axis is …………….. .
5. The Y-coordinate of the point B (4, 1) is ………………… .
6. Perpendicular distance of the point (5, 2) from Y-axis is ……………… .
7. Point (-6, 4) lies in the …………… quadrant.
8. Point (0,4) lies in the ……………… quadrant.
9. Point (5,0) lies in the ………………. quadrant.
10. Abscissa of all the points on y-axis is ………………. .
From the given graph, locate the position of points A, B, C, D and E.
For an experiment in science, Vikram and Rohit grew one plant each under similar lab conditions. Their heights were measured at the end of each week for 5 weeks, the result is shown by the follow:
1. Write height of plant A and plant B after
(a) 2 weeks
(b) 5 weeks.
2. Write difference of height of the two plants after 5 weeks.
3. During which week plant A grew most?
4. Druing which week plant B grew least?
The quantity of petrol filled in a car and the cost of petrol are given in the following table:
1. Draw the graph of above data find the cost 12 l of petrol using graph. How much petrol can be purchased for ‚¹ 800?
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# How do you simplify sqrt7-6sqrt7?
May 19, 2017
$- 5 \sqrt{7}$
#### Explanation:
Let $x = \sqrt{7}$
$x - 6 x = - 5 x$
Substituting $\sqrt{7}$ will give us $- 5 \sqrt{7}$
May 19, 2017
$- 5 \sqrt{7}$
#### Explanation:
Remember it is actually $1 \sqrt{7} - 6 \sqrt{7}$
There are two terms with a common factor of $\sqrt{7}$
$1 \sqrt{7} - 6 \sqrt{7}$
$= \sqrt{7} \left(1 - 6\right)$
$= - 5 \sqrt{7}$
Or just treat it as adding like terms:
$1 \sqrt{7} - 6 \sqrt{7} = - 5 \sqrt{7}$
|
### Congruence of Triangles - Solutions 2
CBSE Class –VII Mathematics
NCERT Solutions
Chapter 7 Congruence of Triangles
(Ex. 7.2)
Question 1. Which congruence criterion do you use in the following?
(a) Given: AC = DF, AB = DE, BC = EF
So $\mathrm{\Delta }$ABC $\cong \mathrm{\Delta }$DEF
(b) Given: RP = ZX, RQ = ZY, $\mathrm{\angle }$PRQ = $\mathrm{\angle }$XZY
So $\mathrm{\Delta }$PQR $\cong \mathrm{\Delta }$XYZ
(c) Given: $\mathrm{\angle }$MLN = $\mathrm{\angle }$FGH, $\mathrm{\angle }$NML = $\mathrm{\angle }$HFG, ML = FG
So $\mathrm{\Delta }$LMN $\cong \mathrm{\Delta }$GFH
(d) Given: EB = BD, AE = CB, $\mathrm{\angle }$A = $\mathrm{\angle }$C = ${90}^{\circ }$
So $\mathrm{\Delta }$ABE $\cong \mathrm{\Delta }$CDB
Answer: (a) By SSS congruence criterion, since it is given that AC = DF, AB = DE, BC = EF
The three sides of one triangle are equal to the three corresponding sides of another triangle.
Therefore, $\mathrm{\Delta }$ABC $\cong$$\mathrm{\Delta }$DEF
(b) By SAS congruence criterion, since it is given that RP = ZX, RQ = ZY and $\mathrm{\angle }$PRQ = $\mathrm{\angle }$XZY
The two sides and one angle in one of the triangle are equal to the corresponding sides and the angle of other triangle.
Therefore, $\mathrm{\Delta }$PQR $\cong$$\mathrm{\Delta }$XYZ
(c) By ASA congruence criterion, since it is given that $\mathrm{\angle }$MLN = $\mathrm{\angle }$FGH, $\mathrm{\angle }$NML = $\mathrm{\angle }$HFG, ML = FG.
The two angles and one side in one of the triangle are equal to the corresponding angles and side of other triangle.
Therefore, $\mathrm{\Delta }$LMN $\cong$$\mathrm{\Delta }$GFH
(d) By RHS congruence criterion, since it is given that EB = BD, AE = CB, $\mathrm{\angle }$A = $\mathrm{\angle }$C = ${90}^{\circ }$
Hypotenuse and one side of a right angled triangle are respectively equal to the hypotenuse and one side of another right angled triangle.
Therefore, $\mathrm{\Delta }$ABE $\cong$$\mathrm{\Delta }$CDB
Question 2. You want to show that $\mathrm{\Delta }$ART $\cong \mathrm{\Delta }$PEN:
If you have to use SSS criterion, then you need to show:
(i) AR = (ii) RT = (iii) AT =
If it is given that $\mathrm{\angle }$T = $\mathrm{\angle }$N and you are to use SAS criterion, you need to have:
(i) RT = and(ii) PN =
If it is given that AT = PN and you are to use ASA criterion, you need to have:
(i) ? (ii) ?
Answer: (a) Using SSS criterion, $\mathrm{\Delta }$ART $\cong$$\mathrm{\Delta }$PEN
(i) AR = PE (ii) RT = EN (iii) AT = PN
(b) Given: $\mathrm{\angle }$T = $\mathrm{\angle }$N
Using SAS criterion, $\mathrm{\Delta }$ART $\cong$$\mathrm{\Delta }$PEN
(i) RT = EN (ii) PN = AT
(c) Given: AT = PN
Using ASA criterion, $\mathrm{\Delta }$ART $\cong$$\mathrm{\Delta }$PEN
(i) $\mathrm{\angle }$RAT = $\mathrm{\angle }$EPN (ii) $\mathrm{\angle }$RTA = $\mathrm{\angle }$ENP
Question 3. You have to show that $\mathrm{\Delta }$AMP $\cong \mathrm{\Delta }$AMQ. In the following proof, supply the missing reasons:
Steps Reasons PM = QM ∠$\mathrm{\angle }$PMA = ∠$\mathrm{\angle }$QMA AM = AM Δ$\mathrm{\Delta }$AMP ≅Δ$\cong \mathrm{\Delta }$AMQ __________ __________ __________ __________
Steps Reasons PM = QM ∠$\mathrm{\angle }$PMA = ∠$\mathrm{\angle }$QMA AM = AM Δ$\mathrm{\Delta }$AMP ≅Δ$\cong \mathrm{\Delta }$AMQ Given Given Common SAS congruence rule
Question 4. In $\mathrm{\Delta }$ABC, $\mathrm{\angle }$A = ${30}^{\circ },$$\mathrm{\angle }$B = ${40}^{\circ }$ and $\mathrm{\angle }$C = ${110}^{\circ }.$
In $\mathrm{\Delta }$PQR, $\mathrm{\angle }$P = ${30}^{\circ },$$\mathrm{\angle }$Q = ${40}^{\circ }$ and $\mathrm{\angle }$R = ${110}^{\circ }$.
A student says that $\mathrm{\Delta }$ABC $\cong$$\mathrm{\Delta }$PQR by AAA congruence criterion. Is he justified? Why or why not?
Answer: No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be an enlarged copy of the other.
Question 5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write $\mathrm{\Delta }$RAT $\cong$ ?
Answer: In the figure, given two triangles are congruent. So, the corresponding parts are:
$↔$ O, R $↔$ W, T $↔$ N.
We can write, $\mathrm{\Delta }$RAT $\cong$$\mathrm{\Delta }$WON [By SAS congruence rule]
Question 6. Complete the congruence statement:
$\mathrm{\Delta }$BCA $\cong$ ? $\mathrm{\Delta }$QRS $\cong$ ?
Answer: In $\mathrm{\Delta }$BAT and $\mathrm{\Delta }$BAC, given triangles are congruent so the corresponding parts are:
$↔$ B, A $↔$ A, T $↔$ C
Thus, $\mathrm{\Delta }$BCA $\cong$$\mathrm{\Delta }$BTA [By SSS congruence rule]
In $\mathrm{\Delta }$QRS and $\mathrm{\Delta }$TPQ, given triangles are congruent so the corresponding parts are:
$↔$ R, T $↔$ Q, Q $↔$ S
Thus, $\mathrm{\Delta }$QRS $\cong$$\mathrm{\Delta }$TPQ [By SSS congruence rule]
Question 7. In a squared sheet, draw two triangles of equal area such that:
(i) the triangles are congruent.
(ii) the triangles are not congruent.
What can you say about their perimeters?
Answer: In a squared sheet, draw $\mathrm{\Delta }$ABC and $\mathrm{\Delta }$PQR.
When two triangles have equal areas and
(i) these triangles are congruent, i.e., $\mathrm{\Delta }$ABC $\cong$$\mathrm{\Delta }$PQR
[By SSS congruence rule]
Then, their perimeters are same because length of sides of first triangle are equal to the length of sides of another triangle by SSS congruence rule.
(ii) But, if the triangles are not congruent, then their perimeters are not same because lengths of sides of first triangle are not equal to the length of corresponding sides of another triangle.
Question 8. Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Answer: Let us draw two triangles PQR and ABC.
All angles are equal, two sides are equal except one side. Hence, $\mathrm{\Delta }$PQR are not congruent to $\mathrm{\Delta }$ABC.
Question 9. If $\mathrm{\Delta }$ABC and $\mathrm{\Delta }$PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
Answer: $\mathrm{\Delta }$ABC and $\mathrm{\Delta }$PQR are congruent. Then one additional pair is $\overline{\text{BC}}$ = $\overline{\text{QR}}.$
Given: $\mathrm{\angle }$B = $\mathrm{\angle }$Q = ${90}^{\circ }$
$\mathrm{\angle }$C = $\mathrm{\angle }$R
$\overline{\text{BC}}$ = $\overline{\text{QR}}$
Therefore, $\mathrm{\Delta }$ABC $\cong$$\mathrm{\Delta }$PQR [By ASA congruence rule]
Question 10. Explain, why $\mathrm{\Delta }$ABC $\cong$$\mathrm{\Delta }$FED.
Answer: Given: $\mathrm{\angle }$A = $\mathrm{\angle }$F, BC = ED, $\mathrm{\angle }$B = $\mathrm{\angle }$E
In $\mathrm{\Delta }$ABC and $\mathrm{\Delta }$FED,
$\mathrm{\angle }$B = $\mathrm{\angle }$E = ${90}^{\circ }$
$\mathrm{\angle }$A = $\mathrm{\angle }$F
BC = ED
Therefore, $\mathrm{\Delta }$ABC $\cong$$\mathrm{\Delta }$FED [By RHS congruence rule]
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# RD Sharma Class 10 Ex 5.1 Solutions Chapter 5 Trigonometric Ratios
In this chapter, we provide RD Sharma Class 10 Ex 5.1 Solutions Chapter 5 Trigonometric Ratios for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 5.1 Solutions Chapter 5 Trigonometric Ratios pdf, Now you will get step by step solution to each question.
# Chapter 5: Trigonometric Ratios Exercise – 5.1
### Question: 1
Find the value of Trigonometric ratios in each of the following provided one of the six trigonometric ratios are given.
(i) sin A = 2/3
(ii) cos A = 4/5
(iii) tan θ = 11/1
(iv) sin θ = 11/15
(v) tan α = 5/12
(vi) sin θ = √3/2
(vii) cos θ = 7/25
(viii) tan θ = 8/15
(ix) cot θ = 12/5
(x) sec θ = 13/5
(xi) cosec θ = √10
(xii) cos θ =12/15
### Solution:
(i) sin A = 2/3
Given: sin A = 2/3 … (1)
By definition,
By Comparing (1) and (2)
We get,
Perpendicular side = 2 and Hypotenuse = 3
Therefore, by Pythagoras theorem,
AC= AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
Therefore,
32 = AB2 + 22
AB= 32 – 22
AB2 = 9 – 4
AB= 5
AB = √5
Hence, Base = √5
Therefore,
Therefore,
Therefore,
cot A = √5/2
(ii) cos A = 4/5
Given: cos A = 4/5 … (1)
By Definition,
By comparing (1) and (2)
We get,
Base = 4 and
Hypotenuse = 5
Therefore,
By Pythagoras theorem,
AC= AB+ BC2
Substituting the value of base (AB) and hypotenuse (AC) and get the perpendicular side (BC)
52 = 42+ BC2
BC2 = 52 – 42
BC= 25 – 16
BC2 = 9
BC = 3
Hence, Perpendicular side = 3
Now,
Therefore,
sin A = 3/5
Now, cosec A = 1/(sin A)
Therefore,
cosec A = 1/(sin A)
Therefore,
Therefore,
Therefore,
tan A = 3/4
Now, cot A = 1/(tan A)
Therefore,
By definition,
By Comparing (1) and (2)
We get,
Base = 1 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base side (AB) and perpendicular side (BC) and get hypotenuse (AC)
AC2 = 12 + 112
AC2 = 1 + 121
AC= 122
AC = √122
Therefore,
Sin θ = 11√122
Therefore,
Therefore,
Therefore,
By definition,
By Comparing (1) and (2)
We get,
Perpendicular Side = 11 and
Hypotenuse = 15
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
152 = AB2 +112
AB2 = 15– 112
AB2= 225 – 121
AB= 104
Therefore,
Therefore,
Therefore,
Therefore,
Therefore,
(v) tan α = 5/12
Given: tan α = 5/12 … (1)
By definition,
By comparing (1) and (2)
We get,
Base = 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base side (AB) and the perpendicular side (BC) and gte hypotenuse (AC)
AC2 = 122 + 52
AC2 = 144 + 25
AC= 169
AC = 13
Hence Hypotenuse = 13
Therefore,
sin α = 5/13
Therefore,
Therefore,
By definition,
By comparing (1) and (2)
We get,
Perpendicular side = √3
Hypotenuse = 2
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
AB2 = 4 – 3
AB2 = 1
AB = 1
Hence Base = 1
Therefore,
cos θ = 1/2
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
Therefore,
Therefore,
(vii) cos θ = 7/25
Given: cos θ = 7/25 … (1)
By definition,
By comparing (1) and (2)
We get,
Base = 7 and
Hypotenuse = 25
Therefore
By Pythagoras theorem,
AC = AB2 + BC2
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
252 = 7+ BC2
BC2 = 252 – 72
BC2 = 625 – 49
BC = 576
BC = √576
BC = 24
Hence, Perpendicular side = 24
Therefore,
Sin θ = 24/25
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
Therefore,
Therefore,
By definition,
By comparing (1) and (2)
We get,
Base = 15 and Perpendicular side = 8
Therefore,
By Pythagoras theorem,
AC2= 152 + 82
AC2= 225 + 64
AC2 = 289
AC = √289
AC = 17
Hence, Hypotenuse = 17
Therefore,
sin θ = 8/17
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
cos θ =15/17
Now, sec θ = 1/cos θ
Therefore,
Therefore,
(ix) cot θ = 12/5
Given: cot θ = 12/5 … (1)
By definition,
By comparing (1) and (2)
We get,
Base = 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
AC= AB2 + BC2
Substituting the value of base side (AB) and perpendicular side (BC) and get the hypotenuse (AC)
AC2 = 122 + 52
AC= 144 + 25
AC2 = 169
AC = √169
AC = 13
Hence, Hypotenuse = 13
Therefore,
sin θ = 5/13
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
cos θ = 12/13
Now, sec θ = 1/cos θ
Therefore,
Therefore,
(x) sec θ = 13/5
Given: sec θ = 13/5… (1)
By definition,
By comparing (1) and (2)
We get,
Base = 5
Hypotenuse = 13
Therefore,
By Pythagoras theorem,
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
132 = 52 + BC2
BC2 = 13– 52
BC= 169 – 25
BC= 144
BC = √144
BC = 12
Hence, Perpendicular side = 12
Therefore,
sin θ = 12/13
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
Therefore,
tan θ = 12/5
Now, cot θ = 1/tan θ
Therefore,
(xi) cosec θ = √10
Given: cosec θ = √10/1 … (1)
By definition
By comparing (1) and (2)
We get,
Perpendicular side = 1 and
Hypotenuse = √10
Therefore,
By Pythagoras theorem,
AC= AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
AB= 10 – 1
AB = √9
AB = 3
Hence, Base side = 3
Therefore,
Therefore,
Therefore,
cot θ = 3
(xii) cos θ = 12/15
Given: cos θ = 12/15 … (1)
By definition,
By comparing (1) and (2)
We get,
Base = 12 and Hypotenuse = 15
Therefore,
By Pythagoras theorem,
AC2 = AB+ BC2
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
152 = 122 + BC2
BC2 = 152 – 122
BC2 = 225 – 144
BC 2= 81
BC = √81
BC = 9
Hence, Perpendicular side = 9
Therefore,
Therefore,
Therefore,
Therefore,
Therefore,
### Question: 2
In a ΔABC, right angled at B, AB – 24 cm, BC = 7cm, Determine
(i) sin A, cos A
(ii) sin C, cos C
### Solution:
(i) The given triangle is below:
Given: In ΔABC, AB = 24 cm
BC = 7cm
∠ABC = 90°
To find: sin A, cos A
In this problem, Hypotenuse side is unknown
Hence we first find hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
AC2 = AB2 + BC2
AC2 = 242 + 72
AC2 = 576 + 49
AC= 625
AC = √625
AC = 25
Hypotenuse = 25
By definition,
By definition,
(ii) The given triangle is below:
Given: In Δ ABC, AB = 24 cm
BC = 7cm
∠ABC = 90°
To find: sin C, cos C
In this problem, Hypotenuse side is unknown
Hence we first find hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
AC2 = AB2 + BC2
AC2 = 242 + 72
AC2 = 576 + 49
AC= 625
AC = √625
AC = 25
Hypotenuse = 25
By definition,
By definition,
By definition,
### Question: 3
In the below figure, find tan P and cot R. Is tan P = cot R? To find, tan P, cot R
### Solution:
In the given right angled ΔPQR, length of side OR is unknown
Therefore, by applying Pythagoras theorem in ΔPQR
We get,
PR2 = PQ2 + QR2
Substituting the length of given side PR and PQ in the above equation
13= 122 + QR2
QR2 = 132 – 122
QR2 = 169 – 144
QR2= 25
QR =√25
By definition, we know that,
Also, by definition, we know that
Comparing equation (1) ad (2), we come to know that that R.H.S of both the equation are equal.
Therefore, L.H.S of both equations is also equal
tan P = cot R
### Question: 4
If sin A = 9/41, Compute cos A and tan A.
### Solution:
Given: sin A = 9/41 … (1)
To find: cos A, tan A
By definition,
By comparing (1) and (2)
We get,
Perpendicular side = 9 and Hypotenuse = 41
Now using the perpendicular side and hypotenuse we can construct ΔABC as shown figure.
Length of side AB is unknown is right angled ΔABC,
To find the length of side AB, we use Pythagoras theorem,
Therefore, by applying Pythagoras theorem in ΔABC,
We get,
AC2 = AB2 + BC2
412 = AB2 + 92
AB2 = 412 – 92
AB2 = 168 – 81
AB= 1600
AB = √1600
AB = 40
Hence, length of side AB = 40
Now
By definition,
Now,
By definition,
### Question: 5
Given 15 cot A = 8, find sin A and sec A.
### Solution:
Given: 15 cot A = 8
To find: sin A, sec A
Since 15 cot A =8
By taking 15 on R.H.S
We get,
Since 15 cot A = 8
By taking 15 on R.H.S
We get,
cot A = 8/15
By definition,
cot A = 1/(tan A)
Hence,
Comparing equation (1) and (2)
We get,
Base side adjacent to ∠A = 8
Perpendicular side opposite to ∠A = 15
ΔABC can be drawn below using above information
Hypotenuse side is unknown.
Therefore, we find side AC of ΔABC by Pythagoras theorem.
So, by applying Pythagoras theorem to ΔABC
We get,
AC2 = AB2 +BC2
Substituting values of sides from the above figure
AC2 = 82 + 152
AC2 = 64 + 225
AC2 = 289
AC = √289
AC = 17
Therefore, hypotenuse =17
Now by definition,
Substituting values of sides from the above figure
Sin A = 15/17
By definition,
sec A = 1/cos A
Hence,
Substituting values of sides from the above figure
Sec A = 17/8
sin A =15/17, sec A = 17/8
### Question: 6
In ΔPQR, right angled at Q, PQ = 4cm and RQ = 3 cm .Find the value of sin P, sin R, sec P and sec R.
### Solution:
Given: ΔPQR is right angled at vertex Q.
PQ = 4cm
RQ = 3cm
To find,
sin P, sin R, sec P, sec R
Given ΔPQR is as shown figure.
Hypotenuse side PR is unknown.
Therefore, we find side PR of ΔPQR by Pythagoras theorem
By applying Pythagoras theorem to ΔPQR
We get,
PR2 = PQ2 + RQ2
Substituting values of sides from the above figure
PR2 = 42 +32
PR2 = 16 + 9
PR2 = 25
PR = √25
PR = 5
Hence, Hypotenuse =5
Now by definition,
sin P = RQ/PR
Substituting values of sides from the above figure
sin P = 3/5
Now by definition,
sin R = PQ/PR
Substituting the values of sides from above figure
sin R = 4/5
By definition,
sec P = 1/cos P
Substituting values of sides from the above figure
sec P = PR/PQ
sec P = 5/4
By definition,
sec R = 1/cos R
Substituting values of sides from the above figure
sec R = PR/RQ
sec R = 5/3
sin P = 3/5,
sin R = 4/5,
sec P = 5/4,
sec R = 5/3
### Question: 7
If cot θ = 7/8, evaluate
(ii) cotθ cot2θ
### Solution:
Given: cot θ = 7/8
We know the following formula
(a + b)(a – b) = a2 – b2
By applying the above formula in the numerator of equation (1)
We get,
(1 + sin θ) × (1 – sin θ) = 1 – sin2θ …. (2) (Where, a = 1 and b = sin θ)
Similarly,
By applying formula (a + b) (a – b) = a2 – b2 in the denominator of equation (1).
We get,
(1 + cos θ)(1 – cos θ) = 1 – cosθ (1 + cos θ)(1 – cos θ) = 1 – cos2θ … (Where a = 1 and b = cos θ cos θ
Substituting the value of numerator and denominator of equation (1) from equation (2), equation (3).
Therefore,
Since,
cos2θ + sin2 θ = 1 cos2 θ + sin2 θ = 1
Therefore,
cos2 θ = 1 – sin2Θcos2 θ =1 – sin2 θ
Also, sin2 θ = 1 – cos2 θ sin2 θ = 1 – cos2 θ
Putting the value of 1 – sin2 θ and 1 – cos2 θ in equation (4)
We get,
We know that,
Since, it is given that cot θ = 7/8
Therefore,
(ii) Given: cot θ = 7/8
To evaluate: cot2 θ
cot θ =7/8
Squaring on both sides,
We get,
49/64
### Question: 8
If 3 cot A = 43 cot A = 4, check whether
### Solution:
Given: 3 cot A = 4
To check whether cos2A –sin2A or not.
3 cot A = 4
Dividing by 3 on both sides,
We get,
cot A = 4/3 … (1)
By definition,
cot A = 1/tan A
Therefore,
Comparing (1) and (2)
We get,
Base side adjacent to ∠A = 4
Perpendicular side opposite to ∠A = 3
Hence ΔABC is as shown in figure.
In ΔABC , Hypotenuse is unknown
Hence, it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem in ΔABC
We get
AC= AB2 + BC2
Substituting the values of sides from the above figure
AC2 = 42 + 32
AC2 = 16 + 9
AC2 = 25
AC = √25
AC = 5
Hence, hypotenuse = 5
To check whethercos2A – sin2A or not.
We get thee values of tan A, cos A, sin A
By definition,
tan A =1/(cot A)
Substituting the value of cot A from equation (1)
We get,
tan A = 1/4
tan A = 3/4 …. (3)
Now by definition,
Substituting the values of sides from the above figure
Now we first take L.H.S of equation
Substituting value of tan A from equation (3)
We get,
Taking L.C.M on both numerator and denominator
We get,
Now we take R.H.S of equation whether
R.H.S = cos2A – sin2A
Substituting value of sin A and cos A from equation (4) and (5)
We get,
Comparing (6) and (7)
We get.
### Question: 9
If tan θ = a/b, find the value of
### Solution:
Given: tan θ = a/b … (1)
Now, we know that
Therefore equation (1) become as follows
Now, by applying invertendo
We get,
Now by applying Componendo – dividendo
We get,
Therefore,
### Question: 10
If 3 tan θ = 4, find the value of
### Solution:
Given: If 3 tan θ = 4/3 tan θ = 4
Therefore,
tan θ = 4/3 … (1)
Now, we know that
Therefore equation (1) becomes
Now, by applying Invertendo to equation (2)
We get,
Now, multiplying by 4 on both sides
We get
Therefore
Now, multiplying by 2 on both sides of equation (3)
We get,
Now by applying componendo in above equation
We get,
Therefore,
Therefore, on L.H.S sin θ sin θ cancels and we get,
Therefore,
4 cos θ – sin θ = 44 cos θ – sin θ = 4
### Question: 11
If 3 cot θ = 2, find the value of
### Solution:
Given: 3 cot θ = 2
Therefore,
Now, we know that
Therefore equation (1) becomes
Now, by applying invertendo to equation (2)
Now, multiplying by 4/3 on both sides,
We get,
Therefore, 3 cancels out on R.H.S and
We get,
Now by applying invertendo dividendo in above equation
We get,
Now, multiplying by 2/6 on both sides of equation (3)
We get,
Therefore, 2 cancels out on R.H.S and
We get,
Now by applying componendo in above equation
We get,
Now, by dividing equation (4) by (5)
We get,
Therefore,
Therefore, on L.H.S (3 sin θ) cancels out and we get,
Now, by taking 2 in the numerator of L.H.S on the R.H.S
We get,
Therefore, 2 cancels out on R.H.S and
We get,
### Question: 12
If tan θ = a/b, prove that
### Solution:
Given:
Now, we know that
Therefore equation (1) becomes
Now, by multiplying by a/b on both sides of equation (2)
We get,
Therefore,
Now by applying dividendo in above equation (3)
We get,
Now by applying componendo in equation (4)
We get,
Now, by dividing equation (4) by equation (5)
We get,
Therefore,
Therefore, b cos θ and bcancels on L.H.S and R.H.S respectively
Hence, it is proved that
### Question: 13
If sec θ = 13/5, show that
### Solution:
Given: sec θ = 13/5
To show that
Now, we know that
Therefore,
Therefore,
Now, we know that
Now, by comparing equation (1) and (2)
We get,
Base side adjacent to ∠θ = 5
And
Hypotenuse =13
Therefore from above figure
Base side BC = 5
Hypotenuse AC = 13
Side AB is unknown. It can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2 = AB2 + BC2
Therefore by substituting the values of known sides
We get,
132 = AB2 + 52
Therefore,
AB2= 132 – 52
AB2= 169 – 25
AB2 = 144
AB = √144
Therefore,
AB = 12 …. (3)
Now, we know that
sin θ = AB/AC
sin θ = 12/13 … (4)
Now L.H.S of the equation to be proved is as follows
Substituting the value cos θ of sin θ and from equation (1) and (4) respectively
We get,
Therefore,
L.H.S = 3
Hence proved that,
### Question: 14
If cos θ = 12/13, show that sin θ(1 – tan θ) = 35/156
### Solution:
To show that
Now we know that
Therefore, by comparing equation (1) and (2)
We get,
Base side adjacent to ∠θ = 12
And
Hypotenuse = 13
Therefore from above figure
Base side BC = 12
Hypotenuse AC = 13
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2= AB2 + BC2
Therefore by substituting the values of known sides
We get,
132= AB2 + 122
Therefore,
AB2= 13– 122
AB2= 169 – 144
AB = 25
AB = √25
AB = 5 …. (3)
Now, we know that
Now from figure (a)
We get,
sin θ = AB/AC
Therefore,
sin θ = 5/12… (5)
Now L.H.S of the equation to be proved is as follows
L.H.S of the equation to be proved is as follows
L.H.S = sin θ (1 – tan θ] …. (6)
Substituting the value of sin θ and tan θ from equation (4) and (5)
We get,
Taking L.C.M inside the bracket
We get,
Therefore,
Now by opening the bracket and simplifying
We get,
From equation (6) and (7),it can be shown that
### Solution:
Now, we know that
Therefore,
Therefore,
Comparing Equation (1) and (2)
We get.
Base side adjacent to ∠θ = 1
Perpendicular side opposite to ∠θ = √3
Therefore, triangle representing angle √3 is as shown below
Therefore, by substituting the values of known sides
We get,
Therefore,
AC2 = 3 + 1
AC= 4
AC = √4
Therefore,
AC = 2 … (3)
Now, we know that
Now from figure (a)
Therefore from figure (a) and equation (3),
Now we know that
Now from figure (a)
We get, BC/AC
Therefore from figure (a) and equation (3),
Now, L.H.S of the equation to be proved is as follows
Substituting the value of from equation (4) and (5)
We get,
Now by taking L.C.M in numerator as well as denominator
We get,
Therefore,
Therefore,
### Solution:
To show that
Now, we know that
Therefore,
Comparing equation (1) and (2)
We get.
Perpendicular side opposite to ∠θ = 1
Base side adjacent to ∠θ = √7
Therefore, Triangle representing ∠ θ is shown figure.
Hypotenuse AC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC= AB2 + BC2
Therefore by substituting the values of known sides
We get,
Therefore,
AC 2 = 1 +7
AC2 = 8
AC = √8
Therefore,
Now we know that
Now, we know that
Therefore, from equation (4)
We get,
Now, we know that
Now from figure (a)
We get,
cos θ = BC/AC
Therefore from figure (a) and equation (3)
Now we know that
Therefore, from equation (6)
We get,
Now, L.H.S of the equation to be proved is as follows
Substituting the value of cosec θ and sec θ from equation (6) and (7)
We get,
Therefore,
Therefore,
L.H.S = 48/64
L.H.S = 3/4 = R.H.S
Therefore,
Hence proved that
### Question: 17
If sec θ = 5/4, find the value of
### Solution:
To find the value of
Now we know that
Therefore,
Therefore from equation (1)
Also, we know that cos2θ + sin2θ = 1
Therefore,
Substituting the value of cos θ cos θ from equation (2)
We get,
Therefore,
Also, we know that
sec2θ =1 + tanθ secθ = 1 + tanθ
Therefore,
Therefore,
tan θ = 3/4 … (4)
Also, cot θ = 1/tan θ
Therefore from equation (4)
We get,
cot θ = 4/3 … (5)
Substituting the value of cos θ, cot θ and tan θ from the equation (2), (3), (4) and (5) respectively in the expression below
We get,
= 12/7
Therefore,
### Question: 18
If sin θ = 12/13, find the value of
### Solution:
To find the value of
Now, we know the following trigonometric identity
cosecθ = 1 + tan2θ
Therefore, by substituting the value of tan θ from equation (1)
We get,
By taking L.C.M on the R.H.S
We get,
Therefore
Therefore
Now, we know that
Therefore
Now, we know the following trigonometric identity
cos2θ + sin2θ = 1
Therefore,
cos2θ = 1 – sin2θ
Now by substituting the value of sin θ from equation (3)
We get,
Therefore, by taking L.C.M on R.H.S
We get,
Now, by taking square root on both sides
We get,
Therefore,
Substituting the value of sin θ and cos θ from equation (3) and (4) respectively in the equation below
Therefore,
Therefore,
### Question: 19
If cos θ = 3/5, find the value of
### Solution:
To find the value of
Now we know the following trigonometric identity
cos2θ + sin2θ = 1
Therefore by substituting the value of cos θ from equation (1)
We get,
Therefore,
Therefore by taking square root on both sides
We get,
Now, we know that
Therefore by substituting the value of sin θ and cos θ from equation (2) and (1) respectively
We get,
Now, by substituting the value of sin θ and of tan θ from equation (2) and equation (4) respectively in the expression below
We get,
Therefore,
### Question: 20
If sin θ = 3/5, find the value of
### Solution:
Given:
Now, we know the following trigonometric identity
cos2θ + sin2θ = 1
Therefore by substituting the value of cos θ from equation (1)
We get,
Therefore,
Now by taking L.C.M
We get,
Therefore, by taking square roots on both sides
We get,
cos θ = 4/5
Therefore,
cos θ = 4/5 … (2)
Now we know that
tan θ = sin θ/cos θ
Therefore by substituting the value of sin θ and cos θ from equation (1) and (2) respectively
We get,
Also, we know that
Therefore from equation (3)
We get,
Now by substituting the value of cos θ, tan θ and cot θ from equation (2), (3) and (4) respectively from the expression below
Therefore,
### Question: 21
If tan θ = 24/7, find that sin θ + cos θ
### Solution:
Given: tan θ =24/7 … (1)
To find, Sin θ + cos θ
Now we know that tan θ is defined as follows
Now by comparing equation (1) and (2)
We get,
Perpendicular side opposite to ∠θ = 24
Base side adjacent to ∠θ = 7
Therefore triangle representing ∠θ is as shown figure.
Side AC is unknown and can be found by using Pythagoras theorem
Therefore,
AC= AB2 + BC2
Now by substituting the value of unknown sides from figure
We get,
AC2= 242 +72
AC = 576 + 49
AC = 625
Now by taking square root on both sides,
We get,
AC = 25
Therefore,
Hypotenuse side AC = 25 …. (3)
Now we know sin θ is defined as follows
Therefore from figure (a) and equation (3)
We get,
Now we know that cos θ is defined as follows
Therefore by substituting the value of sin θ and cos θ from equation (4) and (5) respectively, we get
### Question: 21
If sin θ = a/b, find sec θ + tan θ in terms of a and b.
### Solution:
Given:
sin θ = a/b … (1)
To find: sec θ + tan θ
Now we know, sin θ is defined as follows
Now by comparing equation (1) and (2)
We get,
Perpendicular side opposite to ∠θ = a
Hypotenuse = b
Therefore triangle representing ∠θ is as shown figure.
Hence side BC is unknown
Now we find BC by applying Pythagoras theorem to right angled ΔABC
Therefore,
AC2 = AB2 + BC2
Now by substituting the value of sides AB and AC from figure (a)
We get,
b2 = a2 + BC2
Therefore,
BC2 = b2 – a2
Now by taking square root on both sides
We get,
Therefore,
Now we know cos θ is defined as follows
Therefore from figure (a) and equation (3)
We get,
Now we know,
Therefore,
Now we know,
Now by substituting the values from equation (1) and (3)
We get,
Therefore,
Now we need to find sec θ + tan θ
Now by substituting the values of sec θ and tan θ from equation (5) and (6) respectively
We get,
We get,
Now by substituting the value in above expression
We get,
Now,present in the numerator as well as denominator of above denominator of above expression gets cancels we get,
Square root is present in the numerator as well as denominator of above expression. Therefore we can place both numerator and denominator under a common square root sign
Therefore,
### Question: 23
If 8 tan A = 15, find sin A – cos A
### Solution:
Given:
8 tan A = 15
Therefore,
To find: sin A – cos A
Now we know tan A is defined as follows
Now by comparing equation (1) and (2)
We get
Perpendicular side opposite to ∠A = 15
Base side adjacent to ∠A = 8
Therefore triangle representing angle A is as shown figure.
Side AC = is unknown and can be found by using Pythagoras theorem
Therefore,
AC= AB2 + BC2
Now by substituting the value of known sides from figure (a)
We get,
AC= 152 + 82
AC2 = 225 + 64
AC = 289
Now by taking square root on both sides
We get,
AC = √289
AC = 17
Therefore Hypotenuse side AC = 17 … (3)
Now we know, sin A is defined as follows
Therefore from figure (a) and equation (3)
We get,
Now we know, cos A is defined as follows
Therefore from figure (a) and equation (3)
We get,
Now we find the value of expression sin A – cos A
Therefore by substituting the value the value of sin A and cos A from equation (4) and (5) respectively, we get,
### Question: 24
If tan θ = 20/21, show that
### Solution:
Given: tan θ = 20/21
Now we know that
Therefore,
tan θ = 20/21
Side AC be the hypotenuse and can be found by applying Pythagoras theorem
Therefore,
AC= AB2 + BC2
AC2= 212 + 202
AC2 = 441 + 400
AC2 = 841
Now by taking square root on both sides
We get,
AC = √841
AC = 29
Therefore Hypotenuse side AC = 29
Now we know, sin θ is defined as follows,
Therefore from figure and above equation
We get,
Now we know cos θ is defined as follows
Therefore from figure and above equation
We get,
Now we need to find the value of expression
Therefore by substituting the value of sin θ and cos θ from above equations, we get
Therefore after evaluating we get,
Hence,
### Question: 25
If cosec A = 2, find
### Solution:
Given: cosec A = 2
Here BC is the adjacent side,
By applying Pythagoras theorem,
AC= AB2 + BC2
4 = 1 + BC2
BC= 3
BC = √3
Now we know that
Substitute all the values of sin A, cos A and tan A from the equations (1), (2) and (3) respectively
We get.
= 2
Hence,
### Question: 26
If ∠A and ∠B are acute angles such that cos A =cos B, then show that ∠A = ∠B
### Solution:
Given:
∠A and ∠B are acute angles
cos A = cos B such that ∠A = ∠B
Let us consider right angled triangle ACB
Now since cos A = cos B
Therefore
Now observe that denominator of above equality is same that is AB
Therefore AC = BC
We know that when two sides of triangle are equal, then opposite of the sides are also
Equal.
Therefore
We can say that
Angle opposite to side AC = angle opposite to side BC
Therefore,
∠B = ∠A
Hence, ∠A = ∠B
### Question: 27
In a ΔABC, right angled triangle at A, if tan C = √3, find the value of sin B cos C + cos B sin C.
### Solution:
Given: ΔABC
To find: sin B cos C + cos B sin C
The given a ΔABC is as shown in figure
Side BC is unknown and can be found using Pythagoras theorem,
Therefore,
BC= AB2 + AC2
BC= 3 +1
BC2 = 4
Now by taking square root on both sides
We get,
BC = √4
BC = 2
Therefore Hypotenuse side BC = 2 … (1)
Now,
Therefore,
Now by substituting the values from equation (1) and figure
We get,
Now,
Therefore,
Now substituting the value from equation
Similarly
Now by definition,
So by evaluating
Now, by substituting the value of sin B, cos B, sin C and cos C from equation (2), (3), (4) and (5) respectively in sin B cos C + cos B sin C
Sin B cos C + cos B sin C
= 1
Hence,
sin B cos C + cos B sin C = 1
### Question: 28
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of ∠A.
(iii) cos A is the abbreviation used for the cosecant of ∠A.
(iv) sin θ = 4/3 for some angle θ.
### Solution:
(i) tan A < 1
Value of tan A at 45° i.e… tan 45 = 1
As value as A increases to 90°
Tan A becomes infinite
So given statement is false.
(ii) sec A = 12/5 for some value of angle if
M-I
sec A = 2.4
sec A > 1
So given statements is true.
M- II
For sec A = 12/5 we get adjacent side = 13
Subtending 9i at B.
So, given statement is true.
(iii) Cos A is the abbreviation used for cosecant of angle A.
The given statement is false.
As such cos A is the abbreviation used for cos of angle A, not as cosecant of angle A.
(iv) cot A is the product of cot A and A
Given statement is false? cot A is a co-tangent of angle A and co-tangent of angle
(v) sin θ = 4/3 for some angle θ.
Given statement is false
Since value of sin θ is less than (or) equal to one.
Here value of sin θ exceeds one,
So given statement is false.
### Solution:
As shown in figure
Here BC is the adjacent side,
By applying Pythagoras theorem,
AC= AB+ BC2
169 = 144 + BC2
BC= 169 – 144
BC= 25
BC = 5
Now we know that,
We also know that,
tan θ = sin θ/cos θ
Therefore, substituting the value of sin θ and cos θ from above equations
We get,
tan θ = 12/5
Now substitute all the values of sin θ, cos θ and tan θ from above equations in
We get,
Therefore by further simplifying we get,
Therefore,
Hence,
### Question: 30
If cos θ = 5/13, find the value of
### Solution:
Given: If cos θ = 5/13 ….. (1)
To find: The value of expression
Now we know that
Now when we compare equation (1) and (2)
We get,
Base side adjacent to ∠θ = 5
Hypotenuse = 13
Therefore, Triangle representing ∠θ is as shown figure.
Perpendicular side AB is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2= AB2 + BC2
Therefore by substituting the values of known sides,
AB2 = 132 – 52
AB2= 169 -25
AB2 = 144
AB = 12 …. (3)
Now we know from figure and equation,
Now we know that,
Now w substitute all the values from equation (1), (4) and (5) in the expression below,
Therefore
We get,
Therefore by further simplifying we get,
Therefore,
Hence,
### Solution:
Given: sec A = 17/8
Now we know that
Now, by substituting the value of sec A
We get,
Now we also know that,
sin2A + cos2A = 1
Therefore
sin2A = 1 – cos2A
Now by taking square root on both sides,
We get,
sin A = 15/17
We also know that,
Now by substituting the value of all the terms,
We get,
tan A = 15/8
Now from the expression of above equation which we want to prove:
Now by substituting the value of cos A ad sin A from equation (3) and (4)
We get,
From expression
Now by substituting the value of tan A from above equation
We get,
Therefore,
We can see that,
If sin θ = 3/4,
Prove that
### Solution:
To prove:
By definition,
By comparing (1) and (3)
We get,
Perpendicular side = 3 and
Hypotenuse = 4
Side BC is unknown.
So we find BC by applying Pythagoras theorem to right angled ΔABC
Hence,
AC2 = AB2 + BC2
Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)
Therefore,
4= 32 + BC2
BC2 = 16 – 9
BC= 7
BC = √7
Hence, Base side BC = √7 … (3)
Now cos A = BC/AC
√7/4 … (4)
Now, cosec A = 1/(sin A)
Therefore, from fig and equation (1)
cosec A = 4/3 … (5)
Now, similarly
Sec A = 4/√7 … (6)
Further we also know that
Therefore by substituting the values from equation (1) and (4),
We get,
Now by substituting the value of cosec A, sec A and cot A from the equations (5), (6), and (7) in the L.H.S of expression (2)
Hence it is proved that,
### Question: 33
If sec A = 17/8,
Verify that
### Solution:
Given: sec A = 17/8 … (1)
To verify:
Now we know that
sec A = 1/(cos A)
Therefore,
cos A = 1/(sec A)
We get,
cos A = 8/17 … (3)
Similarly we can also get,
sin A = 15/17 …. (4)
An also we know that
tan A = (sin A)/(cos A)
tan A = 15/8 … (5)
Now from the expression of equation (2)
L.H.S: Missing close
Now by substituting the value of cos A and sin A from equation (3) and (4)
We get,
Now by substituting the value of tan A from equation (5)
We get,
Now by comparing equation (6) and (7)
We get,
### Question: 34
If cot θ = 3/4, prove that
### Solution:
Given: cot θ = 3/4
Prove that:
Now we know that
Here AC is the hypotenuse and we can find that by applying Pythagoras theorem
AC= AB2 +BC2
AC2 = 16 +9
AC= 25
AC = 5
Similarly
Now on substituting the values in equations we get,
Therefore,
### Question: 35
If 3 cos θ – 4 sin θ = 2 cos θ + sin θ, find tan θ tan θ
### Solution:
Given: 3 cos θ – 4 sin θ = 2 cos θ + sin θ
To find: tan θ tan θ
We can write this as:
3 cos θ – 4 sin θ = 2 cos θ + sin θ cos θ = 5 sin θ
Dividing both the sides by cos θ,
We get,
1 = 5 tan θ
tan θ = 1
Hence,
tan θ = 1
### Question: 36
If ∠A and ∠P are acute angles such that tan A = tan P, then show ∠A = ∠P
### Solution:
Given: A and P are acute angles tan A = tan P
Prove that: ∠A = ∠P
Let us consider right angled triangle ACP
We know
∴ tan A = tan P
PC = AC [∵ Angle opposite to equal sides are equal]
∠A = ∠P
All Chapter RD Sharma Solutions For Class10 Maths
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# 5.6: Solving System of Inequalities on the Cartesian Plane
Difficulty Level: At Grade Created by: CK-12
## Solving a System of Linear Inequalities
Introduction
In a previous lesson you learned how to solve a system of linear equations graphically. The solution was determined by identifying the intersection point of the two linear equations. When the system of equations was solved by graphing, the system could have one point of intersection, no point of intersection (the lines were parallel), or an infinite number of intersection points (the equations were multiples of each other).
In this lesson you will learn to solve a system of linear inequalities by identifying the regions of intersection. The region of intersection or the common solution of a system of linear equation is where the shading of the inequalities overlaps. This area of overlapping is known as the feasible region and this region contains the common solution of the system of inequalities.
Objectives
The lesson objectives for Solving Systems Inequalities on the Cartesian Plane are:
• Plotting two or more inequalities on the same Cartesian grid.
• Showing the overlap of the shading of the inequalities.
• Identifying the feasible region on the graph.
Introduction
When a linear inequality is graphed on a Cartesian plane, the region that contains the points that will satisfy the inequality is identified by shading. If two inequalities are graphed on the same grid and if they share common points that will satisfy both linear equations, an area will be created that contains shading from both inequalities. The area that contains this shading is the region that contains the common solution for the system of inequalities.
Watch This
Guidance
Graph the following system of linear inequalities on the same Cartesian grid.
{y>12x+2y2x3}\begin{align*}\begin{Bmatrix} y>-\frac{1}{2}x+2 \\ y \le 2x-3 \end{Bmatrix}\end{align*}
Both inequalities are in slope-intercept form. Begin by graphing y>12x+2\begin{align*}\boxed{y > -\frac{1}{2}x+2}\end{align*}.
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
y>12x+2(1)>12(1)+21>12+21>112Is it true?\begin{align*}& y > -\frac{1}{2}x+2\\ & ({\color{red}1}) > -\frac{1}{2}({\color{red}1})+2\\ &{\color{red}1} > {\color{red}-\frac{1}{2}}+2\\ & \boxed{1>1\frac{1}{2}} \quad \text{Is it true?}\end{align*}
No, one is not greater than one and one-half. Therefore, the point (1, 1) does not satisfy the inequality and will not lie in the shaded area. The shaded area is above the dashed line.
Now, the inequality y2x3\begin{align*}y \le 2x-3\end{align*} will be graphed on the same Cartesian grid.
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
y2x3(1)2(1)31231111Is it true?\begin{align*}& y \le 2x-3\\ & ({\color{red}1}) \le 2({\color{red}1})-3\\ & {\color{red}1} \le {\color{red}2}-3\\ & {\color{red}1} \le {\color{red}-1}\\ & \boxed{1 \le -1} \quad \text{Is it true?}\end{align*}
No, one is not less than or equal to negative one. Therefore, the point (1, 1) does not satisfy the inequality and will not lie in the shaded area. The shaded area is below the solid line.
The region that is indicated as the feasible region is the area on the graph where the shading from each line overlaps. This region contains all the points that will satisfy both inequalities. Another way to indicate the feasible region is to shade the entire region a different color. If you were to do this exercise in your notebook using colored pencils, the feasible region would be very obvious.
The feasible region is the area shaded in yellow.
Example A
Solve the following system of linear inequalities by graphing:
{2x6y12x+2y>4}\begin{align*}\begin{Bmatrix} -2x-6y \le 12\\ -x +2y > -4 \end{Bmatrix}\end{align*}
Write each inequality in slope-intercept form.
2x6y122x+2x6y2x+126y2x+126y62x6+1266y626x+1226Simplify the slope to lowest terms. 26=13y 13x2\begin{align*}& -2x-6y \le 12\\ & -2x{\color{red}+2x}-6y \le {\color{red}2x}+12\\ & -6y \le {\color{red}2x}+12\\ & \frac{-6y}{{\color{red}-6}} \le \frac{2x}{{\color{red}-6}} + \frac{12}{{\color{red}-6}}\\ & \frac{\cancel{-6}y}{\cancel{-6}} \le -\frac{2}{6}x+\frac{\overset{{\color{red}-2}}{\cancel{12}}}{\cancel{-6}} \quad \text{Simplify the slope to lowest terms.} \ \boxed{-\frac{2}{6}=-\frac{1}{3}}\\ & \boxed{y \ {\color{red}\ge} \ -\frac{1}{3}x-2}\end{align*}
x+2y>4x+x+2y>x42y>x42y2>x2422y2>12x422y>12x2\begin{align*}& -x+2y > -4\\ & -x{\color{red}+x}+2y > {\color{red}x}-4\\ & 2y > {\color{red}x}-4\\ & \frac{2y}{{\color{red}2}} > \frac{x}{{\color{red}2}} -\frac{4}{{\color{red}2}}\\ & \frac{\cancel{2}y}{\cancel{2}} > \frac{1}{2}x - \frac{\overset{{\color{red}2}}{\cancel{4}}}{\cancel{2}}\\ & \boxed{y > \frac{1}{2}x-2}\end{align*}
Graph: y13x2\begin{align*}y \ge -\frac{1}{3}x-2\end{align*}.
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
2x6y2(1)6(1)26812121212Is it true?\begin{align*}-2x-6y &\le 12\\ -2({\color{red}1})-6({\color{red}1}) &\le 12\\ {\color{red}-2-6} &\le 12\\ {\color{red}-8} &\le 12 \quad \text{Is it true?}\end{align*}
Yes, negative eight is less than or equal to twelve. The point (1, 1) satisfies the inequality and will lie within the shaded region.
Now graph y>12x2\begin{align*}y>\frac{1}{2}x-2\end{align*} on the same Cartesian grid.
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
x+2y(1)+2(1)1+21>4>4>4>4Is it true?\begin{align*}-x+2y &> -4\\ -({\color{red}1})+2({\color{red}1}) &> -4\\ {\color{red}-1} + {\color{red}2} &> -4\\ {\color{red}1} &> -4 \quad \text{Is it true?}\end{align*}
Yes, one is greater than negative four. The point (1, 1) satisfies the inequality and will lie within the shaded region.
The region that is indicated as the feasible region is the area on the graph where the shading from each line overlaps. This region contains all the points that will satisfy both inequalities.
The feasible region is the area shaded in blue.
Example B
Solve the following system of linear inequalities by graphing:
{x1y>2}\begin{align*}\begin{Bmatrix} x \ge 1\\ y > 2 \end{Bmatrix}\end{align*}
These are the special lines that need to be graphed. The first line is a line that has an undefined slope. The graph is a vertical line parallel to the y\begin{align*}y-\end{align*}axis. The second line is a line that has a slope of zero. The graph is a line parallel to the x\begin{align*}x-\end{align*}axis.
Graph: x1\begin{align*}x \ge 1\end{align*}.
On the graph, every point to the right of the vertical line has an x\begin{align*}x-\end{align*}value that is greater than one. Therefore, the graph must be shaded to the right of the vertical line.
Now graph y>2\begin{align*}y>2\end{align*} on the same Cartesian grid.
On the graph, every point above the horizontal line has a y\begin{align*}y-\end{align*}value that is greater than two. Therefore, the graph must be shaded above the horizontal line.
The region that is indicated as the feasible region is the area on the graph where the shading from each line overlaps. This region contains all the points that will satisfy both inequalities.
The feasible region is the area shaded in purple.
Example C
More than two inequalities can be shaded on the same Cartesian plane. The solution set is all of the coordinates that lie within the shaded regions that overlap. When more than two inequalities are being shaded on the same grid, the shading must be done accurately and neatly.
Solve the following system of linear inequalities by graphing:
y<x+1y2x+5y>0\begin{align*}\begin{Bmatrix} y < x+1\\ y \le -2x+5\\ y > 0 \end{Bmatrix}\end{align*}
Graph: y<x+1\begin{align*}y:
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
y(1)11<x+1<(1)+1<1+1<2Is it true?\begin{align*}y &< x+1\\ ({\color{red}1}) &< ({\color{red}1}) +1\\ {\color{red}1} &< {\color{red}1}+1\\ 1 &< {\color{red}2} \quad \text{Is it true?}\end{align*}
Yes, one is less than two. The point (1, 1) satisfies the inequality and will lie within the shaded region.
y2x+4\begin{align*}y \ge -2x+4\end{align*}
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
y(1)112x+42(1)+42+42Is it true?\begin{align*}y &\ge -2x+4\\ ({\color{red}1}) &\ge -2({\color{red}1})+4\\ {\color{red}1} &\ge {\color{red}-2}+4\\ 1 &\ge {\color{red}2} \quad \text{Is it true?}\end{align*}
No, one is not greater than or equal to two. The point (1, 1) does not satisfy the inequality and will not lie within the shaded region.
y>0\begin{align*}y>0\end{align*}
The graph of y>0\begin{align*}y>0\end{align*} is a horizontal line along the x\begin{align*}x-\end{align*}axis. Every point above the horizontal line has a y\begin{align*}y-\end{align*}value greater than zero. Therefore, the shaded area will be above the graphed line.
The region that is indicated as the feasible region is the area on the graph where the shading from each line overlaps. This region contains all the points that will satisfy both inequalities.
The feasible region is the area shaded in pink.
Vocabulary
Feasible region
The feasible region is the part on the graph where the shaded areas of the inequalities overlap. This area contains the all the solution sets for the inequalities.
Guided Practice
1. Solve the following system of linear inequalities by graphing:
{4x+5y203x+y6}\begin{align*}\begin{Bmatrix} 4x+5y \le 20\\ 3x + y \le 6\\ \end{Bmatrix}\end{align*}
2. Solve the system of linear inequalities by graphing:
2x+y82x+3y<12x0y0\begin{align*}\begin{Bmatrix} 2x+y \le 8\\ 2x+3y < 12\\ x \ge 0\\ y \ge 0\\ \end{Bmatrix}\end{align*}
3. Determine and prove three points that satisfy the following system of linear inequalities:
{y<2x+7y3x4}\begin{align*}\begin{Bmatrix} y < 2x+7\\ y \ge -3x-4\\ \end{Bmatrix}\end{align*}
1. {4x+5y203x+y6}\begin{align*}\begin{Bmatrix} 4x+5y \le 20\\ 3x + y \le 6\\ \end{Bmatrix}\end{align*}
Write each inequality in slope-intercept form.
4x+5y204x4x+5y4x+205y4x+205y54x5+2055y545x+2045y45x+43x+y63x3x+y3x+6y3x+6y3x+6\begin{align*}& 4x+5y \le 20 && 3x+y \le 6\\ &4x{\color{red}-4x}+5y \le {\color{red}-4x}+20 && 3x{\color{red}-3x}+y \le {\color{red}-3x}+6\\ &5y \le {\color{red}-4x}+20 && y \le {\color{red}-3x}+6\\ & \frac{5y}{{\color{red}5}} \le \frac{-4x}{{\color{red}5}}+\frac{20}{\color{red}5} && \boxed{y \le-3x+6}\\ & \frac{\cancel{5}y}{\cancel{5}} \le -\frac{4}{5}x+\frac{\overset{{\color{red}4}}{\cancel{20}}}{\cancel{5}}\\ & \boxed{y \le -\frac{4}{5}x+4}\end{align*}
Graph: y45x+4\begin{align*}y \le -\frac{4}{5}x+4\end{align*}
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
4x+5y4(1)+5(1)4+5920202020Is it true?\begin{align*}4x+5y &\le 20\\ 4({\color{red}1})+5({\color{red}1}) &\le 20\\ {\color{red}4}+{\color{red}5} &\le 20\\ {\color{red}9} &\le 20 \quad \text{Is it true?}\end{align*}
Yes, nine is less than or equal to twenty. The point (1, 1) satisfies the inequality and will lie within the shaded region.
Graph y3x+6\begin{align*}y \le -3x+6\end{align*}
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
3x+y3(1)+(1)3+146666Is it true?\begin{align*}3x+y &\le 6\\ 3({\color{red}1})+({\color{red}1}) &\le 6\\ {\color{red}3}+{\color{red}1} &\le 6\\ {\color{red}4} &\le 6 \quad \text{Is it true?}\end{align*}
Yes, four is less than or equal to six. The point (1, 1) satisfies the inequality and will lie within the shaded region.
The feasible region is the area shaded in pink.
2. 2x+y82x+3y<12x0y0\begin{align*}\begin{Bmatrix} 2x+y \le 8\\ 2x + 3y < 12 \\ x \ge 0\\ y \ge 0 \end{Bmatrix}\end{align*}
Write the first two inequalities in slope-intercept form.
2x+3y<122x2x+3y<2x+123y<2x+123y3<2x3+1233y3<23x+1243y<23x+42x+y82x2x+y2x+8y2x+8y2x+8\begin{align*}& 2x+3y <12 && 2x+y \le 8\\ & 2x{\color{red}-2x}+3y < {\color{red}-2x}+12 && 2x{\color{red}-2x}+y \le {\color{red}-2x}+8\\ & 3y < {\color{red}-2x}+12 && y \le {\color{red}-2x}+8\\ & \frac{3y}{{\color{red}3}} < \frac{-2x}{{\color{red}3}} + \frac{12}{{\color{red}3}} && \boxed{y \le -2x+8}\\ & \frac{\cancel{3}y}{\cancel{3}} < -\frac{2}{3}x + \frac{\overset{{\color{red}4}}{\cancel{12}}}{\cancel{3}}\\ & \boxed{y < -\frac{2}{3}x+4}\end{align*}
Graph: y<23x+4\begin{align*}y<-\frac{2}{3}x+4\end{align*}
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
2x+3y2(1)+3(1)2+35<12<12<12<12Is it true?\begin{align*}2x+3y &< 12\\ 2({\color{red}1})+3({\color{red}1}) &< 12\\ {\color{red}2}+{\color{red}3} &< 12\\ {\color{red}5} &< 12 \quad \text{Is it true?}\end{align*}
Yes, five is less than twelve. The point (1, 1) satisfies the inequality and will lie within the shaded region.
Graph: y2x+8\begin{align*}y \le -2x+8\end{align*}
The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.
2x+y2(1)+(1)2+138888Is it true?\begin{align*}2x+y &\le 8\\ 2({\color{red}1})+({\color{red}1}) &\le 8\\ {\color{red}2}+{\color{red}1} &\le 8\\ {\color{red}3} &\le 8 \quad \text{Is it true?}\end{align*}
Yes, three is less than or equal to eight. The point (1, 1) satisfies the inequality and will lie within the shaded region.
Graph: x0\begin{align*}x \ge 0\end{align*}
The graph will be a vertical line that will coincide with the y\begin{align*}y-\end{align*}axis. All x\begin{align*}x-\end{align*}values to the right of the line are greater than or equal to zero. The shaded area will be to the right of the vertical line.
Graph: y0\begin{align*}y \ge 0\end{align*}
The graph will be a horizontal line that will coincide with the x\begin{align*}x-\end{align*}axis. All y\begin{align*}y-\end{align*}values above the line are greater than or equal to zero. The shaded area will be above the horizontal line.
The feasible region is the area shaded in green.
3. {y<2x+7y3x4}\begin{align*}\begin{Bmatrix} y < 2x+7\\ y \ge -3x-4 \\ \end{Bmatrix}\end{align*}
Graph the system of inequalities to determine the feasible region.
Three points in the feasible region are (-1, 3); (4, -2); and (6, 5). These points will be tested in each of the linear inequalities. All of these points should satisfy both inequalities.
Test (-1, 3)
\begin{align*}& y < 2x+7 \qquad \qquad \qquad and && y \ge -3x-4\\ & y < 2x+7 && y \ge -3x-4\\ & ({\color{red}3})<2({\color{red}-1})+7 && ({\color{red}3}) \ge -3({\color{red}-1})-4\\ & {\color{red}3} < {\color{red}-2}+7 && {\color{red}3} \ge {\color{red}3}-4\\ & 3 < {\color{red}5} && 3 \ge {\color{red}-1}\end{align*}
The point (-1, 3) satisfies both inequalities. In the first inequality, three is less than five. In the second inequality three is greater than or equal to negative one. Therefore, the point lies within the feasible region and is a solution for the system of linear inequalities.
Test (4, -2)
\begin{align*}& y < 2x+7 && y \ge -3x-4\\ & ({\color{red}-2}) < 2({\color{red}4})+7 && ({\color{red}-2}) \ge -3({\color{red}4})-4\\ & {\color{red}-2}<{\color{red}8}+7 && {\color{red}-2} \ge {\color{red}-12}-4\\ & -2 < {\color{red}15} && -2 \ge {\color{red}-16}\end{align*}
The point (4, -2) satisfies both inequalities. In the first inequality, negative two is less than fifteen. In the second inequality negative two is greater than or equal to negative sixteen. Therefore, the point lies within the feasible region and is a solution for the system of linear inequalities.
Test (6, 5)
\begin{align*}& y < 2x+7 && y \ge -3x-4\\ & ({\color{red}5})< 2({\color{red}6})+7 && ({\color{red}5}) \ge -3({\color{red}6}) -4\\ & {\color{red}5} < {\color{red}12}+7 && {\color{red}5} \ge {\color{red}-18}-4\\ & 5 < {\color{red}19} && 5 \ge {\color{red}-22}\end{align*}
The point (6, 5) satisfies both inequalities. In the first inequality, five is less than nineteen. In the second inequality five is greater than or equal to negative twenty-two. Therefore, the point lies within the feasible region and is a solution for the system of linear inequalities.
Summary
In this lesson you have learned to solve a system of linear inequalities on a Cartesian plane. The solution set was all points in the area that was in the overlapping, shaded region of the graphs. To determine where the shaded area should be with respect to each line, a point, which was not on the line, was tested in the original inequality. If the point made the inequality true, then the area containing the point was shaded. If the point did not make the inequality true, then the tested point did not lie within the shaded region.
Problem Set
Solve the following systems of linear inequalities by graphing.
\begin{align*}\begin{Bmatrix} 3x+5y>15\\ 2x-7y \le 14\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 3x+2y \ge 10\\ x-y < -1\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} x-y > 4\\ x+y > 6\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} y>3x-2\\ y < -2x+5\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 3x-6y > -6\\ 5x+9y \ge -18\\ \end{Bmatrix}\end{align*}
Solve the following systems of linear inequalities by graphing.
\begin{align*}\begin{Bmatrix} 2x-y<4\\ x \ge -1\\ y \ge -2 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2x+y>6\\ x +2y \ge 6\\ x \ge 0\\ y \ge 0 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} x \le 3\\ x \ge -2\\ y \le 4\\ y \ge -1 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} y < x+1\\ y \ge -2x+3\\ y > 0 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} x+y > -1\\ 3x -2y \ge 2\\ x < 3\\ y \ge 0 \end{Bmatrix}\end{align*}
Solve the following systems...
\begin{align*}& \begin{Bmatrix} 3x+5y>15\\ 2x-7y \le 14\\ \end{Bmatrix}\\ & 3x+5y>15 && 2x-7y \le 14\\ & 3x-3x+5y > -3x+15 && 2x-2x-7y \le -2x+14\\ & 5y > -3x+15 && -7y \le -2x+14\\ & \frac{5y}{5}>\frac{-3x}{5}+\frac{15}{5} && \frac{-7y}{-7} \le \frac{-2x}{-7}+\frac{14}{-7}\\ & \frac{\cancel{5}y}{\cancel{5}} > -\frac{3}{5}x + \frac{\overset{{\color{red}3}}{\cancel{15}}}{\cancel{5}} && \frac{\cancel{-7}y}{\cancel{-7}} \le \frac{2}{7}x + \frac{\overset{{\color{red}-2}}{\cancel{14}}}{\cancel{-7}}\\ & \boxed{y>-\frac{3}{5}x+3} && \boxed{y \ge \frac{2}{7}x-2}\end{align*}
Test (1, 1)
\begin{align*}3x+5y &>15\\ 3(1)+5(1) &> 15\\ 3+5 &> 15\\ 8 &> 15\end{align*}
It is not true. The point (1, 1) will not lie within the shaded region.
Test (1, 1)
\begin{align*}2x-7y &\le 14\\ 2(1)+7(1) &\le 14\\ 2-7 &\le 14\\ -5 &\le 14\end{align*}
It is true. The point (1, 1) will lie within the shaded region.
The feasible region is the area shaded in purple.
\begin{align*}& \begin{Bmatrix} x-y > 4\\ x+y > 6\\ \end{Bmatrix}\\ & x-y > 4 && x+y>6\\ & x-x-y > -x+4 && x-x+y>-x+6\\ & -y > -x+4 && \boxed{y>-x+6}\\ & \frac{-y}{-1}> \frac{-x}{-1}+\frac{4}{-1}\\ & \frac{\cancel{-1}y}{\cancel{-1}}> \frac{\cancel{-1}x}{\cancel{-1}}+\frac{\overset{{\color{red}-4}}{\cancel{4}}}{\cancel{-1}}\\ & \boxed{y < x-4}\end{align*}
Test (1, 1)
\begin{align*}x+y &> 4\\ (1)+(1) &> 4\\ 1+1 &> 4\\ 2 &> 4\end{align*}
It is not true. The point (1, 1) will not lie within the shaded region.
Test (1, 1)
\begin{align*}x+y &> 6\\ (1)+(1) &> 6\\ 1+1 &> 6\\ 2 &> 6\end{align*}
It is not true. The point (1, 1) will not lie within the shaded region.
The feasible region is the area shaded in gold.
\begin{align*}& \begin{Bmatrix} 3x-6y > -6\\ 5x+9y \ge -18\\ \end{Bmatrix}\\ & 3x-6y > -6 && 5x+9y \ge -18\\ & 3x-3x-6y > -3x-6 && 5x-5x+9y \ge -5x-18\\ & -6y > -3x-6 && 9y \ge -5x-18\\ & \frac{-6y}{-6} > \frac{-3x}{-6} -\frac{6}{-6} && \frac{9y}{9} \ge \frac{-5x}{9}-\frac{18}{9}\\ & \frac{\cancel{-6}y}{\cancel{-6}} > \frac{3}{6}x - \frac{\overset{{\color{red}-1}}{\cancel{6}}}{\cancel{-6}} && \frac{\cancel{9}y}{\cancel{9}} \ge -\frac{5}{9}x - \frac{\overset{{\color{red}2}}{\cancel{18}}}{\cancel{9}}\\ & \boxed{y < \frac{1}{2}x+1} && \boxed{y \ge -\frac{5}{9}x-2}\end{align*}
Test (1, 1)
\begin{align*}3x-6y &> -6\\ 3(1)-6(1) &> -6\\ 3-6 &> -6\\ -3 &> -6\end{align*}
It is true. The point (1, 1) will lie within the shaded region.
Test (1, 1)
\begin{align*}5x+9y &\ge -18\\ 5(1)+9(1) &\ge -18\\ 5+9 &\ge -18\\ 14 &\ge -18\end{align*}
It is true. The point (1, 1) will lie within the shaded region.
The feasible region is the area shaded in blue.
Solve the following systems...
\begin{align*}& \begin{Bmatrix} 2x-y<4\\ x \ge -1\\ y \ge -2 \end{Bmatrix}\\ & 2x-y < 4\\ & 2x-2x-y < -2x+4\\ & -y < -2x+4\\ & \frac{-1y}{-1} < \frac{-2x}{-1}+\frac{4}{-1}\\ & \frac{\cancel{-1}y}{\cancel{-1}} < \frac{\overset{{\color{red}2}}{\cancel{-2}x}}{\cancel{-1}} + < \frac{\overset{{\color{red}-4}}{\cancel{4}}}{\cancel{-1}}\\ & \boxed{y > 2x-4}\end{align*}
The feasible region is the area shaded in red.
\begin{align*}\begin{Bmatrix} x \le 3\\ x \ge -2\\ y \le 4\\ y \ge -1 \end{Bmatrix}\end{align*}
The feasible region is the area shaded in yellow.
\begin{align*}& \begin{Bmatrix} x+y > -1\\ 3x -2y \ge 2\\ x < 3\\ y \ge 0 \end{Bmatrix}\\ & x+y > -1 && 3x-2y \ge 2\\ & x-x+y > -x-1 && 3x-3x-2y \ge -3x+2\\ & \boxed{y>-x-1} && -2y \ge -3x+2\\ & && \frac{-2y}{-2} \ge \frac{-3x}{-2}+\frac{2}{-2}\\ & && \frac{\cancel{-2}y}{\cancel{-2}} \ge \frac{3}{2}x+\frac{\overset{{\color{red}-1}}{\cancel{2}}}{\cancel{-2}}\\ & && \boxed{y \le \frac{3}{2}x-1}\end{align*}
The feasible region is the area shaded in gold.
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The Parabola
The Northern Cross Radiotelescope
The parabola is defined in analytic geometry as the set of all points P in a plane that are the same distance from a given line and a fixed point not on a line. The fixed point is called the focus and the fixed line is called the directrix.
Activity
A fun classroom activity involving parabolas is paper folding with wax paper. The idea was obtained from an article by Scott Smith titled Paper Folding and Conic Sections.
Instructions
On the sheet of wax paper draw a fixed line and a fixed point. Fold the fixed point onto any point on the fixed line and crease the paper. Repeat this many times. The creases form tangents to and envelope a parabola. The following is a GSP simulation of what takes place with the paper folding.
This exercise leads to a nice presentation of the derivation of the standard equation of a parabola whose vertex is at the origin.
The figure below shows point P called the Focus as the point that is folded onto point P’ (located on The Directrix). At P’ a perpendicular line to the Directrix was constructed. This perpendicular line hits the crease line at point A. The crease line is the perpendicular bisector of PP’, thus point A is equidistant from fixed point A and fixed the fixed line. Point A is a point on the parabola whose focus is P and whose directrix is L. (this can be proved by showing that the crease line through A is tangent to the parabola).
By definition of a parabola the distance from point A to P is equal to the distance from A to P’. Recalling the distance formula
we can write the following
the length of = the length of
=
or by squaring both sides and multiplying out
or
the standard form of an up/down parabola whose vertex is at the origin
The following activity was obtained from Virginia Laird, Rockwell High School, Richardson, Texas.
Students should be provided with the below information through direct teaching and then complete the following problems.
• The standard form of a parabola with vertex is and.
• A parabola is the set of all points P in a plane that are the same distance from a given line and a fixed point not on the line.
• The fixed point is called the focus and the fixed line is called the directrix.
• The focus and the directrix are each located c units from the vertex, but lie on opposite sides of the parabola.
• The parabola cannot intersect the directrix.
• The width of the parabola at its focus is 4c units and is called the latus rectum.
• A parabola’s direction is dependent upon which term is squared in the standard form.
After the above information has been given to the students they should be grouped in pairs. They should then explore the effects of (h,k) and c upon the parabola in
each of the forms.
Graph the following parabolas on the computer graphics program and draw conclusions about the direction of the curve, the effect of (h,k) and c. Find the length of the latus rectum for each curve and verify from the graph. Sketch the graph of each on your graph paper. Also multiply each of the equations out and solve for x or y accordingly.
1. 2. 3. 4.
Based on the work that you have done above complete the following paragraph to enable you to make some conjectures about parabolas.
A parabola in the form of (x-h)^2=4c(y-k) turns __________ or __________. When 4c is a negative number, the parabola turns __________, but if 4c is positive, the parabola turns __________. (h,k) represents the __________ of the parabola. A parabola in the form of (y-k)^2=4c(x-h) turns __________ or __________. If 4c is positive the parabola turns __________, but if 4c is negative the parabola turns __________. The focus is located__________ units from the vertex and the latus rectum is __________ times this distance.
Describe in paragraph form the following parabolas. Check your results by graphing each parabola on the computer.
1. 2.
Put the following equations in standard form. Use above conjectures to construct each parabola on graph paper and use Graphing Calculator to verify graph.
1. 2.
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# RD Sharma Solutions for Class 7 Maths Chapter 12 Profit And Loss
The PDF of RD Sharma Solutions for Class 7 Maths Chapter 12 Profit and Loss are downloaded from the given links. Class 7 is a stage where several new topics are introduced. Our experts formulate RD Sharma Solutions to help students in their exam preparation to achieve excellent marks in Maths. The solutions are stepwise and detailed to make learning easy for students. RD Sharma Solutions for Class 7 help students to get a good score in the examinations. It also provides extensive knowledge about the subject, as Class 7 builds a foundation in their academic career. Let us have a look at some important topics being discussed in this Chapter.
• Definition and meaning of profit and loss
• Definition and meaning of cost price
• Definition and meaning of selling price
• Profit percent and loss percent
• On finding profit or loss when C.P. and S.P. are given
• On computing C.P. or S.P. when profit or loss is given
• On finding profit or loss percent
• On finding the S.P. when C.P. and profit or loss percent are given
• On finding the cost price when S.P. and profit or loss are given
## Download the PDF of RD Sharma Solutions For Class 7 Maths Chapter 12 Profit And Loss
### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 12 – Profit And Loss
Exercise 12.1 Page No: 12.8
1. Given the following values, find the unknown values:
(i) C.P. = Rs 1200, S.P. = Rs 1350 Profit/Loss?
(ii) C.P. = Rs 980, S.P. = Rs 940 Profit/Loss =?
(iii) C.P. = Rs 720, S.P. =?, Profit = Rs 55.50
(iv) C.P. =? S.P. = Rs 1254, Loss = Rs 32
Solution:
(i) Given CP = Rs. 1200, SP = Rs. 1350
Clearly CP < SP. So, profit.
Profit = SP – CP
= Rs. (1350 – 1200)
= Rs. 150
(ii) Given CP = Rs. 980, SP = Rs. 940
Clearly CP > SP. So, loss.
Loss = CP – SP
= Rs. (980 – 940)
= Rs. 40
(iii) CP = Rs. 720, SP =?, profit = Rs. 55.50
Profit = SP – CP
55.50 = SP – 720
SP = (55.50 + 720)
= Rs. 775.50
(iv) CP =?, SP = Rs. 1254, loss = Rs. 32
Loss = CP – SP
32 = CP – 1254
CP = (1254 + 32)
= Rs. 1286
2. Fill in the blanks in each of the following:
(i) C.P. = Rs 1265, S.P. = Rs 1253, Loss = Rs …….
(ii) C.P. = Rs……., S.P. = Rs 450, Profit = Rs 150
(iii) C.P. = Rs 3355, S.P. = Rs 7355,……. = Rs……
(iv) C.P. = Rs……., S.P. = Rs 2390, Loss = Rs 5.50
Solution:
(i) Loss = Rs 12
Explanation:
Given CP = Rs. 1265, SP = Rs. 1253
Loss = CP – SP
= Rs. (1265 – 1253)
= Rs. 12
(ii) C.P. = Rs 300
Explanation:
Given CP = ?, SP = Rs. 450, profit = Rs. 150
Profit = SP – CP
150 = 450 – CP
CP = Rs. (450 – 150)
= Rs. 300
(iii) Profit = Rs 4000
Explanation:
Given CP = Rs. 3355, SP = Rs. 7355,
Here SP > CP, so profit.
Profit = SP – CP
Profit = Rs. (7355 – 3355)
= Rs. 4000
(iv) C. P. = Rs 2395.50
Explanation:
Given CP = ?, SP = Rs. 2390, loss = Rs. 5.50
Loss = CP – SP
5.50 = CP – 2390
= Rs. (5.50 + 2390)
= Rs. 2395.50
3. Calculate the profit or loss and profit or loss percent in each of the following cases:
(i) C.P. = Rs 4560, S.P. = Rs 5000
(ii) C.P. = Rs 2600, S.P. = Rs 2470
(iii) C.P. = Rs 332, S.P. = Rs 350
(iv) C.P. = Rs 1500, S.P. = Rs 1500
Solution:
(i) Given CP = Rs. 4560, SP = Rs. 5000
Here, clearly SP > CP. So, profit.
Profit = SP – CP
= Rs. (5000 – 4560)
= Rs. 440
Profit % = {(Profit/CP) x 100} %
= {(440/4560) x 100} %
= {0.0965 x 100} %
Profit % = 9.65%
(ii) Given CP = Rs. 2600, SP = Rs. 2470.
Here, clearly CP > SP. So, loss.
Loss = CP – SP
= Rs. (2600 – 2470)
= Rs. 130
Loss % = {(Loss/CP) x 100} %
= {(130/2600) x 100} %
= {0.05 x 100} %
Loss % = 5%
(iii) Given CP = Rs. 332, SP= Rs. 350.
Here, clearly SP > CP. So, profit.
Profit = SP – CP
= Rs. (350 – 332)
= Rs. 18
Profit% = {(Profit/CP) x 100} %
= {(18/332) x 100} %
= {0.054 x 100} %
Profit % = 5.4%
(iv) Given CP = Rs. 1500, SP = Rs. 1500
Here clearly SP = CP.
So, neither profit nor loss.
4. Find the gain or loss percent, when:
(i) C.P. = Rs 4000 and gain = Rs 40.
(ii) S.P. = Rs 1272 and loss = Rs 328
(iii) S.P. = Rs 1820 and gain = Rs 420.
Solution:
(i) Given CP = Rs. 4000, gain = Rs. 40
Gain % = {(Gain/CP) x 100) %
= {(40/4000) x 100} %
= (0.01 x 100) %
Gain % = 1%
(ii) Given SP = Rs. 1272, loss = Rs. 328
Loss = CP – SP
Hence, CP = Loss+ SP
= Rs. 328 + Rs. 1272
= Rs. 1600
Loss % = {(Loss/CP) x 100} %
= {(328/1600) x 100%
Loss % = 20.5%
(iii) Given SP = Rs. 1820, gain = Rs. 420
Gain = SP – CP
CP = 1820 – 420
= Rs. 1400
Gain % = {(Gain/CP) x 100} %
= {(420/1400) x 100 %
Gain % = 30%
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5. Find the gain or loss percent, when:
(i) C.P. = Rs 2300, Overhead expenses = Rs 300 and gain = Rs 260.
(ii) C.P. = Rs 3500, Overhead expenses = Rs 150 and loss = Rs 146
Solution:
(i) Given CP = Rs. 2300, overhead expenses = Rs. 300 and gain = Rs. 260
We know that Gain % = {(Gain/ (CP + overhead expenses)} x 100
= {260/ (2300 + 300} x 100
= {260/2600} x 100
Gain = 10%
(ii) Given CP = Rs. 3500, overhead expenses = Rs. 150 and loss = Rs. 146
We know that Loss % = {(Loss/ (CP + overhead expenses)} x 100
= {146/ (3500+ 150)} x 100
= {146/3650} x 100
= 14600/3650
Loss = 4%
6. A grain merchant sold 600 quintals of rice at a profit of 7%. If a quintal of rice cost him Rs 250 and his total overhead charges for transportation, etc. were Rs 1000 find his total profit and the selling price of 600 quintals of rice.
Solution:
Given Cost of 1 quintal of rice = Rs. 250
Cost of 600 quintals of rice = 600 x 250 = Rs. 150000
CP = Rs. (150000 + 1000) = Rs. 151000
Profit % = (Profit/CP) x 100
7 = (Profit /151000) x 100
Profit = 1510 x 7
Profit = Rs. 10570
Now SP = CP + profit
= Rs. (151000 + 10570)
SP = Rs. 161570
7. Naresh bought 4 dozen pencils at Rs 10.80 a dozen and sold them for 80 paise each. Find his gain or loss percent.
Solution:
Given Cost of 1 dozen pencils = Rs. 10.80
Therefore cost of 4 dozen pencils = 4 x 10.80
= Rs. 43.2
Also given that selling price of each pencil = 80 paise
Total number of pencils = 12 x 4 = 48
SP of 48 pencils = 48 x 80 paise
= 3840 paise
= Rs. 38.40
Here, clearly SP < CP.
Loss = CP – SP
= Rs. (43.2 – 38.4)
= Rs. 4.8
Loss % = (Loss/CP) x 100
= (4.8/43.2) x 100
= 480/43.2
Loss = 11.11%
8. A vendor buys oranges at Rs 26 per dozen and sells them at 5 for Rs 13. Find his gain percent.
Solution:
Given CP of 1 dozen oranges = Rs. 26
CP of 1 orange = 26/12
= Rs. 2.16
CP of 5 oranges = 2.16 x 5
= Rs. 10.8
Now, SP of 5 oranges = Rs. 13
Gain = SP – CP
= Rs. (13- 10.8)
= Rs. 2.2
Gain %= (Gain/CP) x 100
= (2.2/10.8) x 100
Gain = 20.3%
9. Mr Virmani purchased a house for Rs 365000 and spent Rs 135000 on its repairs. If he sold it for Rs 550000, find his gain percent.
Solution:
Given Mr. Virmani spent to purchase the house = Rs. 365000
Amount he spent on repair = Rs. 135000
Total amount he spent on the house (CP) = Rs. (365000 + 135000)
= Rs. 500000
Given SP of the house = Rs. 550000
Gain = SP – CP
= Rs. (550000 – 500000)
= Rs. 50000
Gain % = (Gain/CP) x 100
= (50000/500000) x 100
= 5000000/500000
Gain = 10%
10. Shikha purchased a wrist watch for Rs 840 and sold it to her friend Vidhi for Rs 910. Find her gain percent.
Solution:
Given CP of the wristwatch that Shikha purchased, CP = Rs. 840
The price at which she sold it, SP = Rs. 910
Gain = SP – CP
= (910 – 840)
= Rs. 70
Now Gain % = (Gain/CP) x 100
= (70/840) x 100
= 7000/840
Gain = 8.3%
11. A business man makes a 10% profit by selling a toy costing him Rs 120. What is the selling price?
Solution:
CP = Rs. 120
Profit % = 10
We now that
SP = {(100 + profit %) /100} x CP = {(100+ 10)/100} x 120
= {(110/100)} x 120 = 1.1 x 120
= Rs. 132
12. Harish purchased 50 dozen bananas for Rs 135. Five dozen bananas could not be sold because they were rotten. At what price per dozen should Harish sell the remaining bananas so that he makes a profit of 20%?
Solution:
Given cost price of 50 dozens bananas that Harish purchased, CP = Rs. 135
Bananas left after removing 5 dozen rotten bananas = 45 dozens
Effective CP of one dozen bananas = Rs. 135/45 = Rs. 3
Calculating the price at which Harish should sell each dozen bananas to make a profit of 20% (or 1/5), we get
Profit % = (Gain/CP) x100
To get a gain of 20% we give profit % = 20
And substitute 20 = (gain/135) x100
Gain = 270/10 = 27
We know; SP = CP + Gain
SP = 27 + 135
SP = 162
Now that SP is for 45 Dozens of bananas
Calculating for one dozen
= 162/45
= Rs. 3.6
Harish should sell the bananas at Rs. 3.60 a dozen in order to make a profit of 20%.
13. A woman bought 50 dozen eggs at Rs 6.40 a dozen. Out of these 20 eggs were found to be broken. She sold the remaining eggs at 55 paise per egg. Find her gain or loss percent.
Solution:
Given cost of one dozen eggs = Rs. 6.40
Cost of 50 dozen eggs = 50 x 6.40 = Rs. 320
Total number of eggs = 50 x 12 = 600
Number of eggs left after removing the broken ones = 600 – 20 = 580
SP of 1 egg = 55 paise
So, SP of 580 eggs = 580 x 55 = 31900 paise
= Rs. 31900/100
= Rs. 319
Loss = CP – SP
= Rs. (320 – 319) = Rs. 1
Loss % = (Loss/CP) x 100
= (1/320) x 100
Loss = 0.31%
14. Jyotsana bought 400 eggs at Rs 8.40 a dozen. At what price per hundred must she sell them so as to earn a profit of 15%?
Solution:
Given cost of eggs per dozen = Rs. 8.40
Cost of 1 egg = 8.40/12
= Rs. 0.7
Cost of 400 eggs = 400 x 0.7 = Rs. 280
Calculating the price at which Jyotsana should sell the eggs to earn a profit of 15%,
We get 15% of 280 + 280
= {(15/100) x 280} + 280
= {4200/100} + 280
= 42 + 280
= Rs. 322
So, Jyotsana must sell the 400 eggs for Rs. 322 in order to earn a profit of 15%. Therefore, the SP per one hundred eggs = Rs. 322/4 = Rs. 80.50.
15. A shopkeeper makes a profit of 15% by selling a book for Rs 230. What is the C.P. and the actual profit?
Solution:
Given that the SP of a book = Rs. 230
Profit % = 15
Since
CP = (SP x 100)/ (100 + profit %)
CP = (230x 100)/ (100 + 15)
CP = 23000/ 115 = Rs. 200
Also, Profit = SP – CP = Rs. (230 – 200) = Rs. 30
Actual profit = Rs. 30
16. A bookseller sells all his books at a profit of 10%. If he buys a book from the distributor at Rs 200, how much does he sell it for?
Solution:
Given profit % = 10% CP = Rs. 200
Since SP = {(100 + profit %) /100} x CP
= {(100 + 10)/100} x 200
= {110/100} x 200 = Rs. 220
The bookseller sells the book for Rs. 220.
17. A floweriest buys 100 dozen roses at Rs 2 a dozen. By the time the flowers are delivered, 20 dozen roses are mutilated and are thrown away. At what price should he sell the rest if he needs to make a 20% profit on his purchase?
Solution:
Given cost of 1 dozen roses = Rs. 2
Number of roses bought by the floweriest = 100 dozens
Thus, cost price of 100 dozen roses = 2 x 100 = Rs. 200
Roses left after discarding the mutilated ones = 80 dozens
Calculating the price at which the floweriest should sell the 80 dozen roses in order to make a profit of 20%, we have
Profit % = ((SP-CP)/CP) x100 = ((SP-200)/200) x 100
40 = SP – 200
SP = Rs. 240
Therefore, the SP of the roses should be Rs. 240/80 = Rs. 3 per dozen.
18. By selling an article for Rs 240, a man makes a profit of 20%.What is his C.P.? What would his profit percent be if he sold the article for Rs 275?
Solution:
Let CP = Rs. x SP = Rs. 240
Let profit be Rs. P.
Now, profit % = 20%
Since Profit % = (Profit/CP) x 100
20 = (P/x) x 100
P = 20x/100 = x/5
Profit = SP – CP = 240 – x
P = 240 – x
x /5 = 240 – x
240 = x + x/5
240 = 6×/5
x = 1200/6
x = 200
So, CP = Rs. 200
New SP = Rs. 275 and CP = Rs. 200
Profit % = {(SP – CP)/CP} x 100
{(275 – 200)/200} x 100 = (75/200) x 100
= 7500/200
= 37.5%
1. Jai Prakash
Very nice
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# Solving composite functions
When Solving composite functions, there are often multiple ways to approach it. Math can be a challenging subject for many students.
## Solve composite functions
In this blog post, we will explore one method of Solving composite functions. Primary School of Linshu sub district paid special attention to the training of students in speaking mathematics Baohua middle school organized teachers to publish the lecture notes, courseware, homework and answers on the class teaching platform in time periods and in a planned way after the completion of online teaching. Teachers corrected homework online. Teachers in each class used the class QQ group to supplement learning materials in time and solve problems and worries for students at the first time. Huayang middle school requires the same as the homework at the beginning of the class. The teacher in class prepares the homework.
The onepiece determinant layout arranges the parallel buildings in a staggered manner: The small knowledge points in the college entrance examination, such as matrix, determinant, program block diagram, linear programming, etc., are not difficult, but if students are not familiar with them and find that they have not seen them in the examination room at ordinary times, they will be nervous. Lao Xiao will comb it for you. The difference between a determinant and a matrix is that a determinant is a definite number and algebraic expression, while a matrix is only a table of numbers. The number of rows and columns of a determinant must be the same, and the number of rows and columns of a matrix can be different.
According to the formula and the options, item a is selected for this question. Fill in the blanks with sequence to test your understanding of the essence of the summation formula of the isometric sequence. You can use the general term and summation formula of the sequence, the properties of the isometric sequence, and the image of the quadratic function to solve it. The vector blank filling questions, the question design and the expression of the conclusion are very familiar to the examinee, but the choice of the solution path is diverse. Through the explanation of the above questions, I believe that we have a better grasp of solving such problems.
At the same time, we should report the situation to the leaders in time, so as to study ways to solve the problems. We should do a good job in persuading and educating those who put forward unreasonable demands, and report the situation to the leaders in charge in a timely manner. In terms of people, in combination with the change of the two committees of the community and the open recruitment, 285 community workers were allocated, a one community one volunteer service team was established, and 69 retired cadres who were enthusiastic about public welfare undertakings and volunteer community services were selected as community consultants. Focusing on solving people's livelihood issues such as water, electricity, gas and heating, the district has selected 33 section level leaders from housing construction, municipal and other departments as service specialists to solve people's problems. Since last year, various urban communities in Hanting have continuously carried out more than 280 resident service days of talk to you if you have something to do.
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# NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.1 – Algebraic Expressions & Identities
NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.1 – Algebraic Expressions & Identities, has been designed by the NCERT to test the knowledge of the student on the following topics:-
• What are Expressions?
• Terms, Factors and Coefficients
• Monomials, Binomials and Polynomials
• Like and Unlike Terms
• Addition and Subtraction of Algebraic Expressions
### NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.1 – Algebraic Expressions & Identities
NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.1 – Algebraic Expressions & Identities
Q.1 Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 − 3zy
(ii) 1 +
x + x2
(iii) 4
x2y2 − 4x2y2z2 + z2
(iv) 3 − pq + qrrp
(v) x/2 +y/2 – xy
(vi) 0.3
a − 0.6ab + 0.5b
Solution:
The terms and their coefficients of the given expressions:
Q.2 Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y − 3y2, 2y − 3y2 + 4y3, 5x − 4y + 3xy, 4z − 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q
Solution:
The given polynomials are classified as
Monomials: 1000, pqr
Binomials: x + y, 2y − 3y2, 4z − 15z2, p2q + pq2, 2p + 2q
Trinomials: 7 + y + 5x, 2y − 3y2 + 4y3, 5x − 4y + 3xy
Polynomials that do not fit in any of these categories: x + x2 + x3 + x4 , ab + bc + cd + da
(i) ab − bc, bc − ca, ca − ab
(ii) a − b + ab, b − c + bc, c − a + ac
(iii) 2p2q2 − 3pq + 4, 5 + 7pq – 3p2q2
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Solution:
Writing the expressions in separate rows, with like terms one below the other, we have
(i)
Therefore, sum of the given expressions is 0
(ii)
Therefore, sum of the given expressions is ab + bc + ac
(iii)
Therefore, sum of the given expressions is −p2q2 + 4pq + 9
(iv)
Therefore, sum of the given expressions is 2(l2 + m2 + n2 + lm + mn + nl).
Q.4
(a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3
(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
(c) Subtract 4p2q − 3pq + 5pq2 − 8p + 7q − 10 from 18 − 3p − 11q + 5pq − 2pq2 + 5p2q
Solution:
Writing the expressions in separate rows, with like terms one below the other, we have
(a)
(b)
(c)
The next Exercise for NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.2 – Algebraic Expressions and Identities can be accessed by clicking here
NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.1 – Algebraic Expressions & Identities
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# Add 3 Single Digit Numbers
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
## Overview
By now you should be comfortable with adding two numbers.
In this lesson, we are going to learn how to add three numbers together.
Look at the example below:
In this case, we can add 6 + 3.
6 + 3 = 9. Now, we will take our sum of 9 and join it with our last addend.
9 + 2 = 11
Therefore,
6 + 3 + 2 = 11
Let’s try a few more problems together! Use the pictures to help you solve.
In this case, we can first add 3 + 1.
3 + 1 = 4. Now, we will take our sum of 4 and join it with our last addend.
4 + 5 = 9
Therefore,
3 + 1 + 5 = 9
We can first add 4 + 5.
4 + 5 = 9. Now, we will take our sum of 9 and join it with our last addend.
9 + 7 = 16
Therefore,
4 + 5 + 7 = 16
We can first add 2 + 5.
2 + 5 = 7. Now, we will take our sum of 7 and join it with our last added.
7 + 1 = 8
Therefore,
2 + 5 + 1 = 8
We can first add 6 + 3.
6 + 3 = 9. Now, we will take our sum of 9 and join it with our last addend.
9 + 4 = 13
Therefore,
6 + 3 + 4 = 13
We can also add the numbers in any order and we will still get the same sum!
For example: We could add 6 + 4 first and then join the last addend of 3 with it.
6 + 4 = 10
10 + 3 = 13
6 + 3 + 4 = 13
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Relevant Topics:
Share good content with friends and get 15% discount for 12-month subscription
### Practice Question 1
Let’s practice! Add the three numbers together to find the correct sum.
### Practice Question 4
Write an addition sentence and then add the numbers on the petals to find the sum.
### Practice Question 5
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# Lesson 7: One Hundred Percent
Let’s solve more problems about percent increase and percent decrease.
## 7.1: Notice and Wonder: Double Number Line
What do you notice? What do you wonder?
## 7.2: Double Number Lines
For each problem, complete the double number line diagram to show the percentages that correspond to the original amount and to the new amount.
1. The gas tank in dad’s car holds 12 gallons. The gas tank in mom’s truck holds 50% more than that. How much gas does the truck’s tank hold?
2. At a movie theater, the size of popcorn bags decreased 20%. If the old bags held 15 cups of popcorn, how much do the new bags hold?
3. A school had 1,200 students last year and only 1,080 students this year. What was the percentage decrease in the number of students?
4. One week gas was \$1.25 per gallon. The next week gas was \$1.50 per gallon. By what percentage did the price increase?
5. After a 25% discount, the price of a T-shirt was \\$12. What was the price before the discount?
6. Compared to last year, the population of Boom Town has increased 25%.The population is now 6,600. What was the population last year?
## 7.3: Representing More Juice
Two students are working on the same problem:
A juice box has 20% more juice in its new packaging. The original packaging held 12 fluid ounces. How much juice does the new packaging hold?
• Here is how Priya set up her double number line.
• Here is how Clare set up her double number line.
Do you agree with either of them? Explain or show your reasoning.
## 7.4: Protecting the Green Sea Turtle
Green sea turtles live most of their lives in the ocean, but come ashore to lay their eggs. Some beaches where turtles often come ashore have been made into protected sanctuaries so the eggs will not be disturbed.
1. One sanctuary had 180 green sea turtles come ashore to lay eggs last year. This year, the number of turtles increased by 10%. How many turtles came ashore to lay eggs in the sanctuary this year?
2. At another sanctuary, the number of nesting turtles decreased by 10%. This year there were 234 nesting turtles. How many nesting turtles were at this sanctuary last year?
## Summary
We can use a double number line diagram to show information about percent increase and percent decrease:
The initial amount of cereal is 500 grams, which is lined up with 100% in the diagram. We can find a 20% increase to 500 by adding 20% of 500:
\begin{align}500+(0.2)\boldcdot 500 &= (1.20)\boldcdot 500\\&=600\end{align}
In the diagram, we can see that 600 corresponds to 120%.
If the initial amount of 500 grams is decreased by 40%, we can find how much cereal there is by subtracting 40% of the 500 grams:
\begin{align}500−(0.4)\boldcdot 500 &= (0.6)\boldcdot 500\\&=300\end{align}
So a 40% decrease is the same as 60% of the initial amount. In the diagram, we can see that 300 is lined up with 60%.
To solve percentage problems, we need to be clear about what corresponds to 100%. For example, suppose there are 20 students in a class, and we know this is an increase of 25% from last year. In this case, the number of students in the class last year corresponds to 100%. So the initial amount (100%) is unknown and the final amount (125%) is 20 students.
Looking at the double number line, if 20 students is a 25% increase from the previous year, then there were 16 students in the class last year.
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Number of ways to distribute the coins?
There are 3 pots and 4 coins. All these coins are to distributed into these pots where any pot contain any number of coins.
In how many ways can all these coins be distributed if out of 4 coins 2 coins are identical and all pots are different?
-
You really need to start putting (and showing) a bit more effort into your questions. You are merely copying statements, providing absolutely no evidence of any work on your part. The whole point is for you to learn how to solve these problems, and if you don't specify why you are unable to solve them (where you are stuck, what you have tried and why it doesn't seem to you to work, etc), you won't learn anything and you will soon exhaust any good will regular members of this forum extend to newcomers. – Arturo Magidin Apr 24 '11 at 11:58
Edit: I mistakenly computed with 4 pots rather than 3. Correct computation below.
• With 4 pots:
If both identical coins are together, then you select which pot the two coins go in (4 ways), and select where the other two, distinguishable coins go (4 ways for each), for a total of $4^3=64$ ways.
If the two identical coins are in different pots, then you select which two pots contain the coins ($\binom{4}{2}$ ways), then select where to put the other two, distinguishable coins (4 ways each), for a total of $\binom{4}{2}\times 16 = 96$ ways.
This gives a total of $64+96 = 160$ ways.
• With 3 pots:
As above: for both identical coins in the same pot, you have $3^3 = 27$ ways of distributing the coins.
For the identical coins in different pots, you have $\binom{3}{2}\times 3^2 = 27$ ways as well.
So the total is $27+27 = 54$ ways when there are three pots.
Added 2. Another way of thinking about it, which may be better for generalization purposes, is to think of the two identical coins as 3-sided dice that you are rolling, with the outcome telling you in which pots to put the coins. You want to count the total number of distinct rolls with two 3-sided dice to find out the number of ways of distributing those two coins. The other, distinguishable, coins each have 3 ways to be placed.
For two dice with three sides each, you are counting combinations with repetitions, so the formula is $$\binom{n+r-1}{n}$$ where $n$ is the number of dice, $r$ the number of faces in each die. So with $n=2$ and $r=3$, you get $\binom{3+2-1}{2} = \binom{4}{2} = 6$ ways of doing it. Then we have three ways for each of the other coins, giving you $6\times3\times 3 = 54$ total ways.
This method is easier to generalize to, say, four identical coins and three other different coins with 5 pots, without having to consider different cases. Such a problem would give you $\binom{4+5-1}{4} = \binom{8}{4}$ ways of placing the four identical coins, and $5^3$ for the remaining three different coins, for a total of $$\binom{8}{4}\times 5^3 = 8750$$ ways. Or if you have four coins and three pots, with 2 sets of two equal coin (say, two pennies and two dimes), then you would have $\binom{2+3-1}{2} = \binom{4}{2}=6$ ways for each set of two equal coins, for a total of $36$ ways. Simpler than considering the case where each set is together, each set is separated, etc.
-
(4,0,0)->3 ways (3,1,0)-> 18 ways (2,2,0)-> 12 ways (2,1,1)-> 21 ways , so totel is 3+18+12+21=54.....What do you think....is this right logic...if coins are(A,A,B,C) – prem shekhar Apr 24 '11 at 9:58
you've computed with 4 pots, rather than three. – Thomas Andrews Apr 24 '11 at 11:09
@Thomas: Sigh. Thanks. – Arturo Magidin Apr 24 '11 at 11:47
@prem: I first did the wrong number of pots. Your method does work if you keep in mind the two separate coins, but you end up having to consider way more cases than I did. – Arturo Magidin Apr 24 '11 at 11:51
I am very glad.....to get such a simple solution.....Thanks.......My way of calculation was bit messy – prem shekhar Apr 24 '11 at 12:11
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Contact The Learning Centre
# Probability
## Example 4: using the probability rules
It is known that the probability that a child will buy a phantom comic is $$0.3$$. If $$3$$ children purchase comics at the same time, what is the probability that:
a) all will purchase a phantom comic?
b) none will purchase a phantom comic?
c) only one child will purchase a phantom comic?
Solution:
Let $$A$$ represent a child will purchase a Phantom comic.
a) \begin{eqnarray*}
P(\mbox{All purchase}) &=& P(A) \times P(A) \times P(A) \\
&=& 0.3 \times 0.3 \times 0.3 \\
&=& 0.027
\end{eqnarray*}
b) \begin{eqnarray*}
P(\mbox{none purchase}) &=& P(\bar{A})\times P(\bar{A})\times P(\bar{A}) \\
&=& 0.7 \times 0.7 \times 0.7 \\
&=& 0.343
\end{eqnarray*}
c) \begin{eqnarray*}
&& P(\mbox{1 purchase}) \\
&=& P(A\,,\ \bar{A}\,,\ \bar{A}) + P(\bar{A}\,,\ {A}\,,\ \bar{A}) + P(\bar{A}\,,\ \bar{A}\,,\ {A}) \\
&=& 0.3\times 0.7 \times 0.7 + 0.7 \times 0.3\times 0.7 + 0.7\times0.7\times0.3 \\
&=& 3 \times (0.3\times0.7\times0.7) \\
&=& 0.441
\end{eqnarray*}
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Area bounded by given curves
• Jan 30th 2010, 09:27 AM
DBA
Area bounded by given curves
Find the area of the region by the given curves.
y=5ln(x) and y=xln(x)
This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
$\displaystyle \int\! u\,dv = uv - \int\! v\,du$
I have no idea how to start here. I am not even sure how the regon should look like.
Thanks for any help.
• Jan 30th 2010, 09:42 AM
e^(i*pi)
Quote:
Originally Posted by DBA
Find the area of the region by the given curves.
y=5ln(x) and y=xln(x)
This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
$\displaystyle \int\! u\,dv = uv - \int\! v\,du$
I have no idea how to start here. I am not even sure how the regon should look like.
Thanks for any help.
FYI that is the integration by parts formula: Integration by parts - Wikipedia, the free encyclopedia and you're expected to use it on both equations
$\displaystyle 5\int ln(x)$
To find the integral of ln(x) note that $\displaystyle \ln(x) = 1 \times \ln(x)$
Now to integrate by parts. Since we don't know the integral of ln(x) we must say that $\displaystyle u = \ln(x)$ and so $\displaystyle du = \frac{1}{x} dx$
Therefore $\displaystyle dv =\, dx$ giving $\displaystyle v = x$. Using the formula you posted
$\displaystyle \int \ln(x)\,dx = x\ln(x) - \int \left(x \times \frac{1}{x}\right)\, dx = x\ln(x)-x+k = x(ln(x)-1)+k$ where k is a constant
Reintroducing the 5:
$\displaystyle 5x[ln(x)-1]+C$ where $\displaystyle C=5k$
Can you do the second integral? It's much the same but with x instead of 1
• Jan 30th 2010, 10:23 AM
DBA
I still do not understand what I am doing. I am looking for one area and if I use the formula on both equations I get 2 answers.
And what are the limits of the integrals?
I tried to do the second integral
Sorry I am not that good with this LaTex stuff yet, so I write an S for integral
I set u=lnx, du = 1/x dx and dv = x, v = 1/2x^2
S xln(x) = 1/2 x^2 * ln(x) - S 1/2 x^2 * 1/x dx
But I still have a product in my integral? So what do I do now?
• Jan 30th 2010, 11:03 AM
General
Quote:
Originally Posted by DBA
Find the area of the region by the given curves.
y=5ln(x) and y=xln(x)
This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
$\displaystyle \int\! u\,dv = uv - \int\! v\,du$
I have no idea how to start here. I am not even sure how the regon should look like.
Thanks for any help.
First, find the intersection points by setting the two curves equal to each other
$\displaystyle 5ln(x)=xln(x)$
Solving, you get $\displaystyle x=1,5$
The area of the region = the integral from 1 to 5 for the bigger curve - the smaller curve.
To know which the is bigger, take any value for x in $\displaystyle (1,5)$ and substitute in both of them.
Let take $\displaystyle x=e$
substitute this in $\displaystyle y=5ln(x)$ you got 5
substitute this in $\displaystyle y=xln(x)$ you got e
since $\displaystyle 5>e$
Hence, $\displaystyle 5ln(x)>xln(x)$ for all x in $\displaystyle (1,5)$
So the bigger curve is $\displaystyle y=5ln(x)$ and the smaller curve is $\displaystyle y=xln(x)$
The area = $\displaystyle \int_1^5 ( 5ln(x) - xln(x) ) dx = 5 \int_1^5 ln(x) dx - \int_1^5 xln(x) dx$
Now, Solve this by using the integration by parts.
The final answer = the desired area.
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# 1993 AIME Problems/Problem 4
## Problem
How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$?
## Solution
### Solution 1
Let $k = a + d = b + c$ so $d = k-a, b=k-c$. It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$. Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$.
Solve them in terms of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$. The last two solutions don't follow $a < b < c < d$, so we only need to consider the first two solutions.
The first solution gives us $c - 93\geq 1$ and $c + 1\leq 499$ $\implies 94\leq c\leq 498$, and the second one gives us $32\leq c\leq 496$.
So the total number of such quadruples is $405 + 465 = \boxed{870}$.
### Solution 2
Let $b = a + m$ and $c = a + m + n$. From $a + d = b + c$, $d = b + c - a = a + 2m + n$.
Substituting $b = a + m$, $c = a + m + n$, and $d = b + c - a = a + 2m + n$ into $bc - ad = 93$, $$bc - ad = (a + m)(a + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)$$ Hence, $(m,n) = (1,92)$ or $(3,28)$.
For $(m,n) = (1,92)$, we know that $0 < a < a + 1 < a + 93 < a + 94 < 500$, so there are $405$ four-tuples. For $(m,n) = (3,28)$, $0 < a < a + 3 < a + 31 < a + 34 < 500$, and there are $465$ four-tuples. In total, we have $405 + 465 = \boxed{870}$ four-tuples.
### Solution 3
Square both sides of the first equation in order to get $bc$ and $ad$ terms, which we can plug $93$ in for. \begin{align*} (a+d)^2 = (b+c)^2 &\implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \\ &\implies 2bc-2ad = a^2-b^2 + d^2-c^2 \\ &\implies 2(bc-ad) = (a-b)(a+b)+(d-c)(d+c) \end{align*} We can plug $93$ in for $bc - ad$ to get $186$ on the left side, and also observe that $a-b = c-d$ after rearranging the first equation. Plug in $c-d$ for $a-b$.
$186 = (c-d)(a+b) + (d-c)(d+c) \implies 186 = -(d-c)(a+b) + (d-c)(d+c) \implies 186 = (d-c)(d+c-a-b)$
Now observe the possible factors of $186$, which are $1 \cdot 186, 2\cdot 93, 3 \cdot 62, 6\cdot 31$. $(d-c)$ and $(d+c-a-b)$ must be factors of $186$, and $(d+c-a-b)$ must be greater than $(d-c)$.
$1 \cdot 186$ work, and yields $405$ possible solutions. $2 \cdot 93$ does not work, because if $c-d = 2$, then $a+b$ must differ by 2 as well, but an odd number $93$ can only result from two numbers of different parity. $c-d$ will be even, and $a+b$ will be even, so $c+d - (a+b)$ must be even. $3 \cdot 62$ works, and yields $465$ possible solutions, while $6 \cdot 31$ fails for the same reasoning above.
Thus, the answer is $405 + 465 = \boxed{870}$
### Solution 4
Add the two conditions together to get $a+d+ad+93=b+c+bc$. Rearranging and factorising with SFFT, $(a+1)(d+1)+93=(b+1)(c+1)$. This implies that for every quadruple $(a,b,c,d)$, we can replace $a\longrightarrow a+1$, $b\longrightarrow b+1$, etc. and this will still produce a valid quadruple. This means, that we can fix $a=1$, and then just repeatedly add $1$ to get the other quadruples.
Now, our conditions are $b+c=d+1$ and $bc=d+93$. Replacing $d$ in the first equation, we get $bc-b-c=92$. Factorising again with SFFT gives $(b-1)(c-1)=93$. Since $b, we have two possible cases to consider.
Case 1: $b=2$, $c=94$. This produces the quadruple $(1,2,94,95)$, which indeed works.
Case 2: $b=4$, $c=32$. This produces the quadruple $(1,4,32,35)$, which indeed works.
Now, for case 1, we can add $1$ to each term exactly $404$ times (until we get the quadruple $(405,406,498,499)$), until we violate $d<500$. This gives $405$ quadruples for case 1.
For case 2, we can add $1$ to each term exactly $464$ times (until we get the quadruple $(465,468,496,499)$). this gives $465$ quadruples for case 2.
In conclusion, having exhausted all cases, we can finish. There are hence $405+465=\boxed{870}$ possible quadruples.
### Solution 5
Let $r = d-c$. From the equation $a+d = b+c$, we have $$r = d-c = b-a ,$$ so $b = a+r$ and $c = d-r$. We then have $$93 = (a+r)(d-r) - ad = rd - ra - r^2 = r(d-a-r) .$$ Since $c > b$, $d-r > a+r$, or $d-a-r > r$. Since the prime factorization of 93 is $3 \cdot 31$, we must either have $r=1$ and $d-a-r = 93$, or $r=3$ and $d-a-r = 31$. We consider these cases separately.
If $r=1$, then $d-a = 94$, $b= a+1$, and $c = d-1$. Thus $d$ can be any integer between 95 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$. We therefore have $499-95+1 = 405$ possibilities in this case.
If $r=3$, then $d-a = 34$, $b = a+3$, and $c=d-3$. Thus $d$ can be any integer between 35 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$, as before. We therefore have $499-35+1 = 465$ possibilities in this case.
Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are $405 + 465 = \boxed{870}$ four-tuples.
## Solution 6
Assume $d = x+m, a = x-m, c = x+n$, and $b = x-n$. This clearly satisfies the condition that $a+d = b+c$ since ($2x = 2x$) . Now plug this into $bc-ad = 93$. You get $(x+n)(x-n) - (x+m)(x-m) = 93 \Rightarrow m^2 - n^2 = 93 \Rightarrow (m-n)(m+n) = 93$
Since $m>n$ (as given by the condition that $a), $m+n>m-n$ and $m$ and $n$ are integers, there are two cases we have to consider since $93 = 3\cdot 31$. We first have to consider $m-n = 1, m+n = 93$, and then consider $m-n=3, m+n = 31$.
In the first case, we get $m=47, n=46$ and in the second case we get $m=17, n=14$. Now plug these values (in separate cases) back into $a,b,c,d$. Since the only restriction is that all numbers have to be greater than $0$ or less than $500$, we can write two inequalities. Either $x+47 < 500, x-47 > 0$, or $x+17 < 500, x-17 > 0$ (using the inequalities given by $d$ and $a$, and since $b$ and $c$ are squeezed in between $d$ and $a$, we only have to consider these two inequalities).
This gives us either $47 < x < 453$ or $17 < x < 483$, and using simple counting, there are $405$ values for $x$ in the first case and $465$ values for $x$ in the second case, and hence our answer is $405+465 = \boxed{870}$
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# 6.1 Algebraic Expressions III
6.1 Algebraic Expressions III
6.1.1 Expansion
1. The product of an algebraic term and an algebraic expression:
• a(b + c) = ab + ac
• a(bc) = ab ac
2. The product of an algebraic expression and another algebraic expression:
• (a + b) (c + d) = ac + ad + bc + bd
• (a + b)2= a2 + 2ab + b2
• (ab)2= a2 – 2ab + b2
• (a + b) (ab) = a2b2
6.1.2 Factorization
1. Factorize algebraic expressions:
• ab + ac = a(b + c)
• a2b2 = (a + b) (ab)
• a2+ 2ab + b2 = (a + b)2
• ac + ad + bc + bd = (a + b) (c + d)
2. Algebraic fractions are fractions where both the numerator and the denominator or either the numerator or the denominator are algebraic terms or algebraic expressions.
Example:
$\frac{3}{b},\frac{a}{7},\frac{a+b}{a},\frac{b}{a-b},\frac{a-b}{c+d}$
3(a) Simplification of algebraic fractions by using common factors:
$\begin{array}{l}•\text{}\frac{{}^{1}\overline{)4}\overline{)b}c}{{}^{{}_{{}^{3}}}\overline{)12}\overline{)b}d}=\frac{c}{3d}\\ •\text{}\frac{bm+bn}{em+en}=\frac{b\overline{)\left(m+n\right)}}{e\overline{)\left(m+n\right)}}\\ \text{}=\frac{b}{e}\end{array}$
3(b) Simplification of algebraic fractions by using difference of two squares:
$\begin{array}{l}\frac{{a}^{2}-{b}^{2}}{an+bn}=\frac{\overline{)\left(a+b\right)}\left(a-b\right)}{n\overline{)\left(a+b\right)}}\\ \text{}=\frac{a-b}{n}\end{array}$
6.1.3 Addition and Subtraction of Algebraic Fractions
1. If they have a common denominator:
$\frac{a}{m}+\frac{b}{m}=\frac{a+b}{m}$
2.
If they do not have a common denominator:
$\frac{a}{m}+\frac{b}{n}=\frac{an+bm}{nm}$
6.1.4 Multiplication and Division of Algebraic Fractions
1. Without simplification:
$\begin{array}{l}•\text{}\frac{a}{m}×\frac{b}{n}=\frac{ab}{mn}\\ •\text{}\frac{a}{m}÷\frac{b}{n}=\frac{a}{m}×\frac{n}{b}\\ \text{}=\frac{an}{bm}\end{array}$
2.
With simplification:
$\begin{array}{l}•\text{}\frac{a}{c\overline{)m}}×\frac{b\overline{)m}}{d}=\frac{ab}{cd}\\ •\text{}\frac{a}{cm}÷\frac{b}{dm}=\frac{a}{c\overline{)m}}×\frac{d\overline{)m}}{b}\\ \text{}=\frac{ad}{bc}\end{array}$
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# linear least squares fit
One of the most common uses of least squares fitting is fitting a straight line to data. Whilst, in general, it is difficult to determine the curve which best fits the data, in this case there is a relatively simple formula which can be used.
###### Theorem 1.
Suppose we have a data set $(x_{1},y_{1}),\ldots,(x_{n},y_{n})$. Then the straight line which best fits this set is given as
$y={ns-pq\over nr-p^{2}}x+{qr-ps\over nr-p^{2}}$
where
$\displaystyle p$ $\displaystyle=\sum_{k=1}^{n}x_{k}$ (1) $\displaystyle q$ $\displaystyle=\sum_{k=1}^{n}y_{k}$ (2) $\displaystyle r$ $\displaystyle=\sum_{k=1}^{n}x_{k}^{2}$ (3) $\displaystyle s$ $\displaystyle=\sum_{k=1}^{n}x_{k}y_{k}$ (4)
###### Proof.
Being the best fitting line means minimizing the merit function $M$, given as
$M(a,b)=\sum_{k=0}^{n}(ax_{k}+b-y_{k})^{2}$
with respect to the parameters $a$ and $b$. Expanding the square, this can be written as
$M(a,b)=ra^{2}+2pab+nb^{2}-2sa-2qb+t$
where $p,q,r,s$ are as above and
$t=\sum_{k=1}^{n}y_{k}^{2}.$
This function $M$ is a quadratic polynomial; moreover, from its definition as a sum of squares, it is clear that the highest order terms are positive definite, hence it has a minimum and all that remains is to find that minimum. To do this, we set the derivatives equal to zero to obtain the following equations:
$\displaystyle 0={\partial M(a,b)\over\partial a}$ $\displaystyle=2ar+2pb-2s$ (5) $\displaystyle 0={\partial M(a,b)\over\partial b}$ $\displaystyle=2pa+2nb-2q$ (6)
These equations are easily solved to give
$\displaystyle a$ $\displaystyle={ns-pq\over nr-p^{2}}$ (7) $\displaystyle b$ $\displaystyle={qr-ps\over nr-p^{2}};$ (8)
substituting in the equation $y=ax+b$ for a straight line, we obtain the answer given above. ∎
Because of the ease with which one can make a least squares fit of a line, this technique is often adapted to fitting other sorts of curves by making a change of variables. Two common cases of this practice are power laws and exponentials.
Suppose that one wants to fit some data to a curve of the form $y=ce^{kx}$. Making a change of variable $y=e^{u}$ and defining $b=\log c$, the equation of the curve becomes $u=kx+b$. One can therefore fit the data set $(x_{1},\log y_{1}),\ldots(x_{n},\log y_{n})$ to a straight line.
Suppose that one wants to fit some data to a curve of the form $y=cx^{p}$. Making a change of variable $x=e^{v}$, $y=e^{u}$ and defining $b=\log c$, the equation of the curve becomes $u=pv+b$. One can therefore fit the data set $(\log x_{1},\log y_{1}),\ldots(\log x_{n},\log y_{n})$ to a straight line.
Although convenient and common, this procedure can be a cheat because changing variables and making a least squares fit of a line is not the same as making a least squares fit to a curve. The reason for this is that the merit functions are different and will not, in general have a minimum in the same place. However, if the data happen to approximately lie on a power curve or an exponential, then the answer obtained by changing variables and fitting will be an approximation to the correct answer. Depending on what one is doing, this approximation may be good enough or one may use it as a starting point for some algorithm to compute the correct minimum.
Title linear least squares fit LinearLeastSquaresFit 2013-03-22 17:24:16 2013-03-22 17:24:16 rspuzio (6075) rspuzio (6075) 13 rspuzio (6075) Definition msc 15-00 RegressionModel GaussMarkovTheorem
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## Math 7, Lesson 7, Fall 2021, 11/07/2021
zhenli Posted in Teaching Plan
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Section 1: Like Terms and Unlike Terms
1. Terms, coefficient, and constant terms
• The expression 2x – 3y + 8 consists of three terms. They are 2x, -3y and 8. The numerical part, including the sign, of a term is called the coefficient of the variable.
• Term 2x, the coefficient of x is 2
• Term -3y, the coefficient of y is -3
• Term 8, is called constant term
2. Classify the like terms and unlike terms: explained in class
3. Simplify the algebraic expression by combining (or collecting) like terms
• 2x + 3x = 5x
• 8y – 3y = 5y
• 3a + 4b – 2a + 5b = (3a -2a) + (4b +5b) = a + 9b
4. Home Work:
• Handout: two pages
• Workbook:
• page 21: 1, 2
• Page 22: 6, 9, 10
Section 2: Distributive Law, Addition and Subtraction of Linear Algebraic Expressions
1. Use of parentheses and distributive law
• a(x + y) = ax + ay
• (x + y)a = a(x + y) = ax + ay = xa + xy
• a(x – y) = a{x + (-y)] = ax + a(-y) = ax – ay
• a(x + y + z) = ax + ay + az
• x – (a – b) = a -a + b
2. Addition and subtraction of linear algebraic expressions: explained in class
• (2a +3b) + (5a -4b) =
• Find the sum of -2p + 3q – 4 and p + 5q – 3
• (4x – 5) – (7x – 3)
3. Home Work:
• Handout:
• one pages
• Workbook:
• page 21: 3, 4, 5
• Page 23: 11, 12, 13, 14, 15
• Page 24: 21
• Page 25: 22, 23
## Math 7, Lesson 6, Fall 2021, 10/24/2021
zhenli Posted in Teaching Plan
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1. Exam on chapter “Introduction to Algebra”
2. Home Work:
• Redo all the problems you got wrong in last three homework assignment.
## Math 7, Lesson 5, Fall 2021, 10/17/2021
zhenli Posted in Teaching Plan
Comments Off on Math 7, Lesson 5, Fall 2021, 10/17/2021
Section 1: Evaluation of algebraic expressions and formulas
1. Evaluation of algebraic expressions
• The process of replacing each variable with its value to find the actual value of an algebraic expression is called substitution.
2. Formulas
• The area of a rectangle is given by
• Area = Length x Width
• A = lw
• This equality of connecting two or more variables is called a formula. When the values of l and w are known, we can find the value of A in the formula by substitution.
3. Home Work:
• Handout: two pages
• Workbook:
• page 15: 3, 4, 5
• Page 16: 11, 12, 13
• Page 17: 14
• Page 19: 26, 27
• Page 20: 28
Section 2: Writing Algebraic Expressions to Represent Real-world Situation
1. We may use algebraic expressions and formulas to express the relationship between two or more quantities in our daily life
• Lots of examples teaching in the classroom, and lots of exercise on whiteboard
• Visualizing, drawing
• Variables are representing quantities with similar units.
2. Home Work:
• Handout: three pages
• Workbook:
• page 17: 15, 16, 17, 18, 19
• Page 18: 20, 21, 22, 23, 24
• Page 19: 25
## Math 7, Lesson 4, Fall 2021, 10/3/2021
zhenli Posted in Teaching Plan
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The use of letters in algebra
Quiz-2 today.
1. The use of letters
• In general, an algebraic expression involves numbers and letters that are connected with operation symbols such as “+”, “-”, “x” and “/”
• In 10 + 8n, we call n a variable and 10 + 8n an algebraic expression
2. Basic notation in algebra
• In algebra, there are rules for writing algebraic expressions. The operation symbols “+”, “-”, “x”, “/” and “=” have the same meanings in both algebra and arithmetic.
• Add a to b: sum = a + b = b + a
• Subtract c from d: difference = d – c != c -d
• Multiply g by h: product = g x h = h x g = gh
• Divide x by y where y != 0: quotient = x/y
3. Exponential notation
• Teach in the class on white board
4. Simplify algebraic expressions
• Teach in the class on white board
5. Home Work:
• Handout: two pages
• Workbook:
• page 15: 1, 2
• page 16: 6, 7, 8, 9, 10
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# Right n-sided pyramid solid angle
Have:
N-faces right pyramid
value of apex solid angle
value of N
Need to find:
value of flat angle (C on image)
• Anything else is known? The angle will depend on the height of the pyramid. Oct 10, 2017 at 13:32
• @Theorem I thought it meant solid angle. Oct 10, 2017 at 13:32
• @Vasya We do know something else: "value of apex solid angle" Oct 10, 2017 at 13:33
• @Vasya solid angle, number of faces and the fact that pyramid is right is enough to find angle C Oct 10, 2017 at 13:54
• Do you need an exact result, or an approximate solution? Oct 10, 2017 at 14:08
Hint:
Assuming given apex solid angle is defined in the same way as it is on this Wikipedia entry. Assuming the polygon at the base is regular, so that $\angle AOB = 2 \pi / N$. Assume the radius of the circumcircle of base polygon is 1 (w.l.o.g).
Possible steps to a solution are:
• Find $AB$ in the base polygon.
• Find $OF$ in the base polygon.
• Find the height $OS$ of the pyramid using the formula for the solid angle of a right $N$-gonal pyramid with a regular base
• Find $SF$ using $OF$ and $OS$
• Find $c / 2$ using $SF$ and $BF = AB / 2$
The formula referred to on the Wikipedia page for the solid angle of a $n$-gonal right-pyramid with height $h$ and with $r$ as the radius of the circle circumscribing the base is: $$\Omega = 2\pi - 2n \arctan\left(\frac {\tan \left({\pi\over n}\right)}{\sqrt{1 + {r^2 \over h^2}}} \right)$$
Below is an example of the use of this formula for a particular pyramid. Let the base of the pyramid be $A=(1,0,0)$, $B=(0,1,0)$ and $C=(0,0,1)$ and let the apex be $O=(0,0,0)$. This pyramid occupies one eighth of a full sphere. The solid angle of a full sphere is $4 \pi$ so we expect the solid angle for the pyramid to be $\frac{\pi}{2}$.
The point at the centre of the base is $P=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$. The height $h$ of the pyramid is given by $h = |OP| = \frac{1}{\sqrt{3}}$. The radius $r$ of the circumcircle of the base can be obtained from $r = |AP| =\frac{\sqrt{2}}{\sqrt{3}}$. We can then calculate $\sqrt{1 + \frac{r^2}{h^2}} = \sqrt{3}$.
For this pyramid, $n=3$ so $\tan\left(\frac{\pi}{n}\right)= \sqrt{3}$. Putting this together with the above, we can calculate $\Omega$ $$\Omega = 2\pi - 2n \arctan\left( \frac{\sqrt{3}}{\sqrt{3}} \right) = 2\pi - 2 \times 3 \times \frac{\pi}{4} = \frac{\pi}{2}$$ confirming the value we expect for the solid angle.
• ∠OSF is not solid angle. en.wikipedia.org/wiki/Solid_angle Oct 10, 2017 at 13:48
• Thanks Petr, I've adjusted the suggested approach to use that. Oct 10, 2017 at 14:11
• yes, it looks like a solution. It remains only to compile this into one expression. Oct 10, 2017 at 14:20
Complete formula:
$$t = \frac{tan(\pi/n)^2}{tan(\frac{\Omega-2\pi}{2n})^2}-1$$ $$C = ∠ASB = 2arcsin(\frac{\sqrt t * sin(\pi/n)}{\sqrt {t+1}})$$
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# SSC > Straight Lines
Explore popular questions from Straight Lines for SSC. This collection covers Straight Lines previous year SSC questions hand picked by experienced teachers.
General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.
Correct2
Incorrect-0.5
The equation of the straight line joining the origin to the point of intersection of {tex} y-x+7=0\ and\ y+2x-2=0 {/tex}is:
A
{tex} 3x+4y=0 {/tex}
B
{tex} 3x-4y=0 {/tex}
C
{tex} 4x-3y=0 {/tex}
{tex} 4x+3y=0 {/tex}
##### Explanation
The intersection point of {tex} y-x+7=0 {/tex} and {tex} y+2x-2=0 {/tex} is {tex} \left(3,\ -4\right) {/tex}
Therefore, Equation of straight line joining from origin to the point {tex} \left(3,\ -4\right) {/tex} is
{tex} y-0 = \frac{-4}{3} \left(x-0\right) {/tex}
=> {tex} 3y = -4x => 4x+3y=0 {/tex}
Q 2.
Correct2
Incorrect-0.5
The angle between the straight lines {tex} x-y\sqrt{3}=5\ and\ \sqrt{3}x+y=7 {/tex} is:
A
{tex} 90 ^{\circ} {/tex}
{tex} 60 ^{\circ} {/tex}
C
{tex} 75 ^{\circ} {/tex}
D
{tex} 30 ^{\circ} {/tex}
##### Explanation
Given equation is compared with {tex} a_{1}x+b_{1}y=0\ and\ a_{2}x+b_{2}y=0 {/tex}
Now, {tex} a_{1}a_{2}+b_{1}b_{2}= \left(1\right) \left(\sqrt{3}\right)+ \left(-\sqrt{3}\right) \left(1\right)=0 {/tex}
Line are perpendicular
{tex} \theta-90 ^{\circ} {/tex}
Q 3.
Correct2
Incorrect-0.5
The three straight lines {tex} ax+by=c,\ bx+cy=a\ and\ cx+ay=b {/tex} are collinear if:
{tex} a+b+c = 0 {/tex}
B
{tex} c+a=b {/tex}
C
{tex} b+c=a {/tex}
D
{tex} a+b=c {/tex}
##### Explanation
We have {tex} ax+by=c {/tex} ...(i)
{tex} bx+cy=a {/tex} ...(ii)
and {tex} cx+ay=b {/tex} ...(iii)
On adding equation (i), (ii) and (iii), we get
{tex} ax+by+bx+cy+cx+ay = a+b+c {/tex}
=> {tex} \left(a+b+c\right)x+ \left(a+b+c\right)y = \left(a+b+c\right) {/tex}
On comparing with {tex} ox+oy=0 {/tex} (for collinearity)
We get{tex} a+b+c = 0 {/tex}
Q 4.
Correct2
Incorrect-0.5
If a tangent to the curve {tex} y=6x-x^{2} {/tex} is parallel to the line {tex} 4x-2y-1=0 {/tex} then the point of tangency on the curve is: ' _
A
(6, 1)
B
(8, 2)
(2, 8)
D
(4, 2)
##### Explanation
Since, tangent is parallel to {tex} y = 2x-\frac{1}{2} {/tex}
Therefore, Equation of tangent is {tex} y = 2x+h {/tex}
The point of tangency will be the point of intersection of tangent and curve but in the given equation of options only option (a) satisfied the equation of curve then {tex} \left(2,\ 8\right) {/tex} will be the point oftangency.
Q 5.
Correct2
Incorrect-0.5
Distance between the lines {tex} 5x+3y-7=0\ and\ 15x+9y+14=0 {/tex} is:
A
{tex} \frac{35}{\sqrt{34}} {/tex}
B
{tex} \frac{1}{\sqrt{34}} {/tex}
C
{tex} \frac{35}{2\sqrt{34}} {/tex}
{tex} \frac{35}{3\sqrt{34}} {/tex}
##### Explanation
Given equation of lines are
{tex} 5x+3y-7=0 {/tex} ...(i)
and {tex} 15x+9y+14=0\ or\ 5x+3y+\frac{14}{3}=0 {/tex} ...(ii)
Therefore, Lines (i) and (ii) are parallel and {tex} C_{1} {/tex} and {tex} C_{2} {/tex} and of opposite signs, therefore these lines are on opposite sides of the origin, so the distance between them is
{tex} |\frac{C_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}| + |\frac{C_{2}}{\sqrt{a_{1}^{2}|+b_{2}^{2}}}| = |\frac{7}{\sqrt{5^{2}+3^{2}}}| + |\frac{14}{3\sqrt{5^{2}+3^{2}}}|{/tex} {tex} = |-\frac{7}{\sqrt{34}}| + |\frac{14}{3\sqrt{34}}| = \frac{35}{3\sqrt{34}} {/tex}
Q 6.
Correct2
Incorrect-0.5
The equation of the sides of a triangle are {tex} x-3y=0.\ 4x+3y=5\ and\ 3x+y=0 {/tex}. The lines {tex} 3x-4y=0 {/tex} passage through
A
the incentre
B
the centroid
C
the orthocentre
the circumcentre
##### Explanation
Two sides {tex} x-3y=0\ and\ 3x+y=0 {/tex} are perpendicular to each other. Therefore its orthocentre is the point of intersection of {tex} x-3y=0 {/tex} i.e. {tex} \left(0,\ 0\right) {/tex} so, the line {tex} 3x-4y=0 {/tex} passes through the orthocentre of triangle.
Q 7.
Correct2
Incorrect-0.5
If (-4, -5) is one vertex and {tex} 7x-y+8=0 {/tex} is one diagonal of a square then the equation of second diagonal is:
A
{tex} x+3y=21 {/tex}
{tex} 2x-3y=7 {/tex}
C
{tex} x+7y=31 {/tex}
D
{tex} 2x+3y=21 {/tex}
##### Explanation
Equation of perpendicular line to {tex} 7x-y+8=0 {/tex} is {tex} x+7y=h {/tex} which is passes through {tex} \left(-4,\ 5\right) {/tex}
h = 31
So, equation of another diagonal is {tex} x+7y=31 {/tex}
Q 8.
Correct2
Incorrect-0.5
The number of the straight which is equally inclined to both the axes is
4
B
3
C
5
D
1
##### Explanation
There are four possible straight line which are equally inclined in both the axes.
i.e., Ist, IInd, IIIrd and IVth quadrant.
Q 9.
Correct2
Incorrect-0.5
The line passing. through {tex} \left(1-\frac{ \pi }{2}\right) {/tex} and perpendicular to {tex} \sqrt{3} \sin \theta + 2\cos \theta =\frac{4}{r} {/tex}, is:
{tex} 2=\sqrt{3}r\cos \theta-2r\sin \theta {/tex}
B
{tex} 5=2\sqrt{3}r\sin \theta +4r\cos \theta {/tex}
C
{tex} 2=\sqrt{3}r\cos \theta + 2r\sin \theta {/tex}
D
{tex} 5=2\sqrt{3}r\sin \theta + 4r\cos \theta {/tex}
##### Explanation
Any line which is perpendicular to
{tex} \sqrt{3}\sin \theta + 2\cos \theta = \frac{4}{r} {/tex} is
{tex} \sqrt{3}\sin \left(\frac{ \pi }{2}+\theta \right) + 2\cos \left(\frac{ \pi }{2}+\theta \right)
= \frac{k}{r} {/tex} ...(i)
Since, it is passing through {tex} \left(-1,\ \frac{ \pi }{2}\right) {/tex}
{tex} \sqrt{3}\sin \pi + 2\cos \pi = \frac{k}{-1}=k=2 {/tex}
Putting k = 2 in eq. (i) we get
{tex} \sqrt{3}\cos \theta - 2\sin \theta = \frac{2}{r} {/tex}
=> {tex} 2 = \sqrt{3}r \cos \theta - 2r \sin \theta {/tex}
Q 10.
Correct2
Incorrect-0.5
If the straight line {tex} ax+by+c=0 {/tex} always passes through {tex} \left(1-2\right) {/tex} then abc are
in AP
B
in HP
C
in GP
D
none of these
##### Explanation
Since {tex} ax+by+c=0 {/tex} is always passes through {tex} \left(1-2\right) {/tex}
{tex} a-2b+c=0 {/tex}
=> {tex} 2b = a+c {/tex}
Therefore a, b and c are in AP.
Q 11.
Correct2
Incorrect-0.5
The equation of pair of lines joining origin to the points of intersection of {tex} x^{2}+y^{2}=9 {/tex} and {tex} x+y=3 {/tex} is:
A
{tex} x^{2}+ \left(3-x^{2}\right)=9 {/tex}
B
{tex} \left(3+y\right)^{2}+y^{2}=9 {/tex}
{tex} xy=0 {/tex}
D
{tex} \left(x-y\right)^{2}=9 {/tex}
##### Explanation
We have {tex} x^{2}+y^{2}=9 {/tex} ...(i)
and {tex} x+y=3 {/tex} ...(ii)
From equation (i) we get
{tex} x^{2}+y^{2}=3^{2} {/tex}
=> {tex} x^{2}+y^{2}= \left(x+y\right)^{2} {/tex} [using equation. (ii)]
=> {tex} x^{2}+y^{2}=x^{2}+y^{2}+2xy {/tex}
=> {tex} 2xy = 0 {/tex}
=> {tex} xy = 0 {/tex}
Q 12.
Correct2
Incorrect-0.5
If the {tex} \angle \theta {/tex} is acute, then the acute angle between {tex} x^{2} \left(\cos \theta -\sin \theta \right)+2xy \cos \theta +y^{2} \left(\cos \theta +\sin \theta \right)=0 {/tex} is
A
{tex} 2 \theta {/tex}
B
{tex} \frac{ \theta }{3} {/tex}
{tex} \theta {/tex}
D
{tex} \frac{ \theta }{2} {/tex}
##### Explanation
Comparing the given equation, we get
{tex} a=\cos \theta - \sin \theta ,\ b=\cos \theta + \sin \theta ,\ h=\cos \theta {/tex}
{tex} tan\phi =\frac{2\sqrt{h^{2}-ab}}{a+b} {/tex}
=>{tex} \tan \phi = \frac{2\sqrt{\cos^{2} \theta - \left(\cos^{2} \theta - \sin^{2} \theta \right) }}{\cos \theta - \sin \theta +\cos \theta +\sin \theta } = \frac{2\sin \theta }{2\cos \theta } {/tex}
=> {tex} \tan \phi = \tan \theta => \phi = 0 {/tex}
Q 13.
Correct2
Incorrect-0.5
Set of lines {tex} \left(x-2y+1\right)+h \left(x+y\right)=0 {/tex} (where h is a parameter) passing through a fixed point:
A
{tex} \left(\frac{1}{3},\ -\frac{1}{3}\right) {/tex}
{tex} \left(-\frac{1}{3},\ \frac{1}{3}\right) {/tex}
C
(1, 1)
D
none of these
##### Explanation
Set of lines passes through intersection point of
{tex} x-2y+1=0\ and\ x+y=0 {/tex} which is {tex} \left(-\frac{1}{3},\ \frac{1}{3}\right) {/tex}
Q 14.
Correct2
Incorrect-0.5
The equation of the line equidistant from the lines {tex} 2x+3y-5=0\ and\ 4x+6y=11 {/tex} is
A
{tex} 2x+3y-1=0 {/tex}
B
{tex} 4x+6y-1=0 {/tex}
{tex} 8x+12y-1=0 {/tex}
D
none of these
##### Explanation
Since, the lines {tex} 2x+3y+5=0\ and\ 2x+3y-\frac{11}{2}=0 {/tex} are parallel
Let required lines is {tex} 2x+3y+n=0,\ n=\frac{C_{1}+C_{2}}{2} {/tex}
{tex} n=\frac{5-\frac{11}{2}}{2} = \frac{-1}{4} {/tex}
So {tex} 8x+12y-1=0 {/tex} is required line.
Q 15.
Correct2
Incorrect-0.5
The equation of line parallel to lines {tex} L_{1}=x+2y-5=0\ and\ x+2y+9=0 {/tex} and dividing the distance between {tex} L_{1} {/tex} and {tex} L_{2} {/tex} in the ratio 1 : 6 (internally) is:
{tex} x+2y-3=0 {/tex}
B
{tex} x+2y+2=0 {/tex}
C
{tex} x+2y+7=0 {/tex}
D
none of these
##### Explanation
Let line be {tex} x+2y+n=0 {/tex}
{tex} n=\frac{-5 \times 6+1 \times 9}{7}=-3 {/tex} {tex} \ Here \ n = \frac{mc_{2}+nc_{1}}{m+n} {/tex}
So, required line is {tex} x+2y-3=0 {/tex}.
Q 16.
Correct2
Incorrect-0.5
The family of lines making an angle {tex} 30 ^{\circ} {/tex} with the line {tex} \sqrt{3}y=x+1 {/tex} is:
A
{tex} x=h {/tex} (h is parameter)
B
{tex} y=-\sqrt{3}x+h {/tex} (h is parameter)
{tex} y=\sqrt{3}+h {/tex}
D
none of these
##### Explanation
Slope of given line is {tex} \frac{1}{\sqrt{3}} {/tex} it's angle from positive x-axis is {tex} 30 ^{\circ} {/tex}. Now lines making an angle {tex} 30 ^{\circ} {/tex} from it either x-axis (i.e. y = 0) or makes and angle {tex} 60 ^{\circ} {/tex} with positive x-axis (i.e. {tex} y=\sqrt{3}x+n {/tex}
Q 17.
Correct2
Incorrect-0.5
A square of area 25 sq unit is formed by talking two sides as {tex} 3x+4y=k_{1}\ and\ 3x+4y=k_{2} {/tex} then {tex} |k_{1}-k_{2}| {/tex} is:
A
5
B
1
25
D
none of these
##### Explanation
Each side of square is 5 unit, distance between given lines is 5 unit.
i.e., {tex} |\frac{k_{1}-k_{2}}{5}|=5 => |k_{1}-k_{2}|=25 {/tex}
Q 18.
Correct2
Incorrect-0.5
The lines {tex} ax+by+c=0,\ bx+cy+a=0\ and\ cx+ay+b=0 {/tex} {tex} a \ne b \ne c {/tex} are concurrent if:
A
{tex} a^{3}+b^{3}+c^{3}+3abc=0 {/tex}
B
{tex} a^{2}+b^{2}+c^{2}-3abc=0 {/tex}
{tex} a+b+c=0 {/tex}
D
none of these
##### Explanation
Since, the given lines are concurrent
\begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix} = 0 =>{tex} a^{3}+b^{3}+c^{3}-3abc=0 {/tex}
{tex} => \left(a + b + c\right) \left(a^{2} + b^{2} + c^{2} - ab - bc - ca\right) = 0 {/tex}
{tex} => \frac{ \left(a + b + c\right) }{2}\{ \left(a-b\right)^{2} + \left(b - c\right)^{2} + \left(c - a\right)^{2} = 0 \} {/tex}
{tex} => a + b + c = 0 {/tex}
Q 19.
Correct2
Incorrect-0.5
A straight line through the point (2, 2) intersects the lines {tex} \sqrt{3}x+y=0\ and\ \sqrt{3}x-y=0 {/tex} at the points A and B. The equation to the line AB so that the triangle OAB is equilateral is:
A
{tex} x-2=0 {/tex}
{tex} y-2=0 {/tex}
C
{tex} x+y-4=0 {/tex}
D
none of these
##### Explanation
From the given equations, we get
{tex} m^{2}+am+2=0 {/tex}
Since, m is real {tex} a^{2}\ge 8,\ |a| \ge 2\sqrt{2} {/tex}
So least value of {tex} |a|\ is\ 2\sqrt{2} {/tex}
{tex} \sqrt{3}x+y=0 {/tex} makes an angle of {tex} 120 ^{\circ} {/tex} with OX and {tex} \sqrt{3}x-y=0 {/tex} makes an angle {tex} 60 ^{\circ} {/tex} with OX.
So the required line is {tex} y-2=0 {/tex}
Q 20.
Correct2
Incorrect-0.5
Equation to the straight line cutting of an intercept from the negative direction of the axis of y and inclined at {tex} 30 ^{\circ} {/tex} to the positive direction of axis of x is:
A
{tex} y+x-\sqrt{3}=0 {/tex}
B
{tex} y-x+2=0 {/tex}
C
{tex} y-\sqrt{3}x-2=0 {/tex}
{tex} \sqrt{3}y-x+2\sqrt{3}=0 {/tex}
##### Explanation
Let the equation of line is y = mx + C {tex} m = tan 30 ^{\circ} = \frac{1}{\sqrt{3}} {/tex} and c = -2 [Because, It is intercepted in negative axes of y with an angle of {tex} 30 ^{\circ} {/tex}] The required line {tex} y = \frac{x}{\sqrt{3}} - 2 => \sqrt{3}y - x + 2\sqrt{3} = 0 {/tex}
Q 21.
Correct2
Incorrect-0.5
Two consecutive side of parallelogram are {tex} 4x+5y=0\ and\ 7x+2y=0 {/tex}. One diagonal of the parallelogram is {tex} 11x+7y=9 {/tex} the other diagonal is{tex} ax+by+c=0 {/tex}, then
A
{tex} a=-1,\ b=-1,\ c=2 {/tex}
{tex} a=1,\ b=-1,\ c=0 {/tex}
C
{tex} a=-1,\ b=-1,\ c=0 {/tex}
D
{tex} a=1,\ b=1,\ c=1 {/tex}
##### Explanation
Since, the co-ordinates of three vertices A, B and C are {tex} \left(\frac{5}{3},\ -\frac{4}{3}\right),\ \left(0,\ 0\right)\ and\ \left(-\frac{2}{3},\ \frac{7}{3}\right) {/tex} respectively. Also the mid point of AC is {tex} \left(\frac{1}{2},\ \frac{1}{2}\right) {/tex}.
Therefore, the equation of line passing through {tex} \left(\frac{1}{2},\ \frac{1}{2}\right) {/tex} and {tex} \left(0,\ 0\right) {/tex} is given by {tex} x-y=0 {/tex}, which is the required equation of another diagonal. {tex} a=1,\ b=-1\ and\ c=0 {/tex}
Q 22.
Correct2
Incorrect-0.5
The line parallel to the x-axis and passing through the intersection of the lines {tex} ax+2by+3b=0\ and\ bx-2ay-3a=0 {/tex}, where {tex} (a, b) \ne (0, 0) {/tex}, is
A
above the x-axis at a distance of {tex} \frac{2}{3} {/tex} from it.
B
above the x-axis at a distance of {tex} \frac{3}{2} {/tex} from it.
C
below the x-axis at a distance of {tex} \frac{2}{3} {/tex} from it.
below the x-axis at a distance of {tex} \frac{3}{2} {/tex} from it.
##### Explanation
Equation of a line passing through the intersection of lines
{tex} ax+2by+3b=0\ and\ bx-2ay-3a=0 {/tex} is
{tex} \left(ax+2by+3b\right)+n \left(bx-2ay-3a\right)=0 {/tex} ...(i)
Now, this line is parallel to x-axis so cofficient of x = 0
=> {tex} a+nb=0 => n = -\frac{a}{b} {/tex}
On putting this value in equation (i) we get
{tex} b \left(ax+2by+3b\right)-a \left(bx-2ay-3a\right)=0 {/tex}
=> {tex} 2b^{2}y+3b^{2}+2a^{2}y+3a^{2}=0 {/tex}
=> {tex} 2 \left(b^{2}+a^{2}\right)y+3 \left(a^{2}+b^{2}\right) = 0 {/tex}
=> {tex} y = -\frac{3}{2} {/tex}
Q 23.
Correct2
Incorrect-0.5
The number of integral values of m, for which the x-coordinate of the point of intersection of the line {tex} 3x+4y=9\ and\ y=mx+1 {/tex} is also an integer is:
2
B
0
C
4
D
1
##### Explanation
Solving {tex} 3x+4y=9\ and\ y=mx+1 {/tex}, we get
{tex} x=\frac{5}{3+4m} {/tex}
x is an integer
{tex} 3+4m=1,\ -1,\ 5, -5 {/tex}
=> {tex} m=\frac{-2}{4},\ \frac{-4}{4},\ \frac{2}{4},\ \frac{-8}{4} {/tex} so, m has two intergral values.
Q 24.
Correct2
Incorrect-0.5
The arc (in sq unit) of the quadrilateral formed by two pairs of lines {tex} l^{2}x^{2}-m^{2}y^{2}-n \left(lx+my\right)=0\ and\ l^{2}m^{2}-m^{2}y^{2}+n \left(lx-my\right)=0 {/tex} is
{tex} \frac{n^{2}}{2|lm|} {/tex}
B
{tex} \frac{n^{2}}{|lm|} {/tex}
C
{tex} \frac{n}{2|lm|} {/tex}
D
{tex} \frac{n^{2}}{4|lm|} {/tex}
##### Explanation
Given equation can be reduced in product 0 linear equation {tex} \left(lx+my\right) \left(lx-my-n\right)=0\ and\ \left(lx-my\right) \left(lx+my+n\right)=0 {/tex}
and {tex} \left(lx-my\right) \left(lx+my+n\right)=0 {/tex}
=> {tex} lx+my=0,\ lx-my-n=0 {/tex}
and {tex} lx-my=0,\ lx+my+n=0 {/tex}
Area= {tex} |\frac{ \left(c_{1}-d_{1}\right) \left(c_{2}-d_{2}\right) }{ \left(a_{1}b_{2}-a_{2}b_{1}\right) }|=|\frac{ \left(0+n\right) \left(0-n\right) }{ \left(-lm-lm\right) }|=|\frac{n^{2}}{2|lm|}{/tex}
Q 25.
Correct2
Incorrect-0.5
If the pair of lines {tex} ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 {/tex} intersect on the y-axis then:
{tex} 2fgh=bg^{2}+ch^{2} {/tex}
B
{tex} bg^{2} \ne ch^{2}{/tex}
C
{tex} abc = 2fgh {/tex}
D
none of these
##### Explanation
Let {tex} f \left(x,\ y\right)=ax^{2}+by^{2}+2hxy+2gx+2fy+c=0 {/tex} ...(i)
Point of intersection of lines {tex} \frac{af \left(x,\ y\right) }{ax} = 0 {/tex}
=> {tex} 2ax+2hy+2g=0 {/tex}
Since, it intersects on y-axis i.e. x = 0
{tex} y = -\frac{g}{h} {/tex}
Thus, putting this value in equation (i), we get
{tex} \frac{bg^{2}}{h^{2}}+2f \left(-\frac{g}{h}\right)+c=0 {/tex}
=> {tex} bg^{2}+ch^{2}=2fgh {/tex}
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## Geometric Series Worksheet Solution3
In the page geometric series worksheet solution3 you are going to see solution of each questions from the geometric series worksheet.
(6) The second term of the geometric series is 3 and the common ratio is 4/5. Find the sum of first 23 consecutive terms in the given geometric series.
Solution:
second term = 3
t₂ = 3
a r = 3 here r = 4/5 < 1 and n = 23
a (4/5) = 3
a = (3 x 5)/4
a = 15/4
Sum of first 23 terms = a (1-r^n)/(1-r)
= 15/4 (1-(4/5)^23)/(1-(4/5))
= 15/4 (1-(4/5)^23)/((5-4)/5)
= 15/4 (1-(4/5)^23)/(1/5)
= (15/4) x (5/1) (1-(4/5)^23)
= (75/4) (1-(4/5)^23)
= (75/4) (1-(4/5)^23)
(7) A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is 9 and the sum of the last two terms is 36. Find the series.
Solution:
Let a,ar ,ar² and ar³ are the first four terms of the given geometric series
sum of the first two terms = 9
sum of the last two terms = 36
a + a r = 9
a (1+ r) = 9
ar² + ar³ = 36
a r² (1 + r) = 36 --- (1)
Substitute a (1 + r) = 9 in the first equation
r² (9) = 36
r² = 36/9
r² = 4
r = √4
r = ± 2
r = -2 is not admissible r = 2
a (1 + 2) = 9
a (3) = 9
a = 9/3
a = 3
so 3 + 3(2) + 3(2)² + 3(2)³+ .........
Therefore the series is 3 + 6 + 12 + 24 + ......
(8) Find the sum of the first n terms of the geometric series
(i) 7 + 77 + 777 + ..............
Solution:
= [7 + 77 + 777 + ..............to n terms]
= 7 [1 + 11 + 111 + ..............to n terms]
= 7/9[9 + 99 + 999 + .............. to n terms]
= 7/9[(10-1) + (100-1) + (1000-1) + .............. to n terms]
= 7/9[(10+100+1000+.............. to n terms)-(1+1+1+......to n terms)]
a = 10 r = 100/10 r = 10 > 1 a = 1 r = 1/1 r = 1
= 7/9 [10(10^n - 1)/(10-1) - n(1)]
= 7/9 [10(10^n - 1)/9 - n]
= 70/81(10^n - 1) - 7n/9
= 70/81(10^n - 1) - n/9
These are the contents in the page geometric series worksheet solution3.
geometric series worksheet solution3
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# 54 Times Table
As a math teacher, your goal is to instill a love for numbers and foster a deep understanding of mathematical concepts in your students. While the multiplication tables up to 10 or 12 are commonly taught, exploring higher multiplication tables can be a rewarding and enriching experience.
In this blog, we delve into the captivating realm of the 54 times table, uncovering its patterns, making connections, and presenting engaging teaching strategies to help your students master this mathematical marvel.
## Understanding the Structure of the 54 Times Table
The 54 times table comprises the multiples of 54 from 1 to infinity. Let's begin by examining the initial multiples to gain insights into the patterns that emerge:
• 1 x 54 = 54
• 2 x 54 = 108
• 3 x 54 = 162
• 4 x 54 = 216
• 5 x 54 = 270
## Observations and Patterns
### Last Digits
One notable pattern in the 54 times table is the alternating sequence of the last two digits in each multiple.
For instance, the last two digits of the first multiple, 54, are 54. However, the last two digits of the second multiple, 108, are 08. This alternating pattern continues consistently throughout the table.
### Digit Sum
Another intriguing aspect lies in the digit sum, which is the sum of the individual digits of a number. In the 54 times table, the digit sum of each multiple increases by 9 with each subsequent multiple:
• Digit sum of 54: 5 + 4 = 9
• Digit sum of 108: 1 + 0 + 8 = 9
• Digit sum of 162: 1 + 6 + 2 = 9
This pattern of incrementing digit sums by 9 persists indefinitely within the table.
### Relationship with Other Tables
The 54 times table also presents fascinating connections with other multiplication tables.
For instance, any multiple of 54 divided by 2 results in a multiple of 27. This connection between the 54 times table and the 2 times table adds depth and interplay to the exploration of multiplication.
## Teaching Strategies
### Engage with Patterns
Encourage your students to actively engage with the patterns within the 54 times table. Have them observe and record the last two digits and digit sums for several multiples. Explore the consistency of these patterns and discuss their significance.
### Mental Math Practice
Incorporate mental math exercises using the 54 times table. For example, challenge your students to quickly find the product of 54 and a single-digit number or to mentally estimate the value of larger multiples of 54.
This strengthens their mental calculation skills and promotes fluency with the table.
### Connection with Other Tables
Explore the relationship between the 54 times table and other multiplication tables, such as the 2 times table (27 x 2 = 54) or the 9 times table (6 x 9 = 54).
Engage students in discussions about how these connections can simplify calculations and deepen their understanding of number relationships.
### Real-World Applications
Help students see the relevance of the 54 times table by presenting real-world applications.
For instance, discuss scenarios where the concept of scaling by 54 could be useful, such as determining the cost of purchasing multiple items at \$54 each or calculating the total distance covered when traveling at a speed of 54 miles per hour.
## Fifty-four Multiplication Table
Read, Repeat and Learn Fifty-four times table and Check yourself by giving a test below
## Table of 54
54 Times table Test
## Conclusion
Exploring the 54 times table opens up a world of mathematical patterns and connections. As a math teacher, embracing the higher multiplication tables like the 54 times table can enhance your students' numeracy skills, deepen their understanding of number patterns, and foster a sense of curiosity and exploration in mathematics. By incorporating engaging teaching strategies and real-world applications, you can inspire your students to embrace the beauty and relevance of mathematics beyond conventional boundaries. So, embark on this mathematical journey with your students, and together, unlock the mysteries of the captivating 54 times table.
@2024 PrintableMultiplicationTable.net
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Areas and Distances
The fundamental problem that integral calculus seeks to solve involves area. For many applications, we need to be able to find the area of complex shapes, that do not fall under our typical geometric shapes. Typically, this involves finding the area under or between curves that we can represent as functions. In this section, we will take an example function, and build an intuition on how we might be able to get such an area.
Take for example the function f(x) = 2x. Suppose we wanted to find the area of this function between x = 0 and x = 2.
On this graph, I’ve marked the two points that we want to get the area of, to make it clear what we want. The hard part of finding this area is the portion of the shape that is curved. One method we could apply to try to estimate this area is to fill it with a shape that we know how to take the area of. For instance, if we fill the shape with rectangles, we can add the areas of the rectangles to get an approximation of the area.
What I’ve done here is drawn four rectangles, where the left top point of the rectangle touches the curve. You can see that doing this misses a bit of area between each rectangle, however, doing this would still provide us with a fairly good estimate of the area under the curve. What if we drew more rectangles in the area?
From this you can see that are we draw more rectangles, the amount of area missed gets smaller and smaller. From this, one could naturally ask the question, what if we let the number of rectangles drawn under the curve approach infinity? If we do this, we get a very good estimate of the area, and since we know limits, we are able to do exactly that.
First, let’s formalize what we are discussing so that we have a formula to calculate the area given a finite number of rectangles. To do this, let’s call the first endpoint a and the second endpoint b. In our example above, it would mean that a = 0 and b = 2. Next, we need a way to determine the width of our rectangles. If we want to draw 4 rectangles for example, we would have each an equal width inside the curve. Since we start at 0 and end at 2, we can take and make each rectangle half a unit wide. Doing this evenly splits the area we have.
From this, we can conclude that if we want to draw n rectangles, we need to use a width . Using this width ensures that each rectangle is an even size. From here, all we need to do is decide where the rectangle touches the curve. If we start at a, then the next point is going to be exactly units away. So, and . From here, we can keep adding to the previous point to get the next point in the list.
This means in general, if we want to calculate the area under a curve using rectangles, we would do something like this:
When n is the number of rectangles we wish to use, , and , , and so on. You can try this formula on our example above and see how it works.
Example: Estimate the area between a = 0 and b = 2 of , using 4 rectangles.
First, let’s determine what is. .
Next, let’s figure out our x values:
With these values figured out, we simply plug them into the equation and solve for our area.
So, using this formula, we can find estimates for the area easily, especially if the number of rectangles is small. However, as we discussed, the accuracy of the solution increases as the number of rectangles increases. So, the next question we want to answer is, how can we take the limit as this expression approaches infinity.
To answer this, we need to first introduce some notation. We are going to represent the sum formula that we generated using sigma notation. Sigma notation denotes a lower and upper bound and adds up the value that follows between the two bounds. For example:
This gives us a more compact way to write out our summation. We can just add the i index as a placeholder, and the summation tells us to replace it with every number from the lower bound to the upper bound. From this, we can say that the limit to infinity of our expression is:
The focus of integral calculus will be to determine a way to be able to effectively solve this equation and understand how it relates to previous concepts we studied in differential calculus. The next few articles in this series will build up an understanding of how this expression is solved, and then we will look at further techniques and applications to solve these equations
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# Eureka Math Geometry Module 1 Lesson 21 Answer Key
## Engage NY Eureka Math Geometry Module 1 Lesson 21 Answer Key
### Eureka Math Geometry Module 1 Lesson 21 Example Answer Key
Example
ABCD is a square, and $$\overline{A C}$$ is one diagonal of the square. △ABC is a reflection of △ADC across segment AC. Complete the table below, identifying the missing corresponding angles and sides.
a. Are the corresponding sides and angles congruent? Justify your response.
Since a reflection is a rigid transformation, all angles and sides maintain their size.
Yes, since △ADC is a reflection of △ABC, they must be congruent.
### Eureka Math Geometry Module 1 Lesson 21 Exercise Answer Key
Opening Exercise
The figure to the right represents a rotation of △ABC 80° around vertex C. Name the triangle formed by the image of △ABC. Write the rotation in function notation, and name all corresponding angles and sides.
△EFC
RD,80°(△EFC)
Discussion
In the Opening Exercise, we explicitly showed a single rigid motion, which mapped every side and every angle of △ABC onto △EFC. Each corresponding pair of sides and each corresponding pair of angles was congruent. When each side and each angle on the pre-image maps onto its corresponding side or angle on the image, the two triangles are congruent. Conversely, if two triangles are congruent, then each side and angle on the pre-image is congruent to its corresponding side or angle on the image.
Exercises
Each exercise below shows a sequence of rigid motions that map a pre-image onto a final image. Identify each rigid motion in the sequence, writing the composition using function notation. Trace the congruence of each set of corresponding sides and angles through all steps in the sequence, proving that the pre-image is congruent to the final image by showing that every side and every angle in the pre-image maps onto its corresponding side and angle in the image. Finally, make a statement about the congruence of the pre-image and final image.
Exercise 1.
Exercise 2.
Exercise 3.
### Eureka Math Geometry Module 1 Lesson 21 Problem Set Answer Key
Question 1.
Exercise 3 mapped △ABC onto △YXZ in three steps. Construct a fourth step that would map △YXZ back onto △ABC.
Construct an angle bisector for the ∠ABY, and reflect △YXZ over that line.
Question 2.
Explain triangle congruence in terms of rigid motions. Use the terms corresponding sides and corresponding angles in your explanation.
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How to explain to a ~$12-16$ year-old student who is weak at maths, that $\ 38 \times 27 \times 14 = 27 \times 14 \times 38$?
How to explain to a ~$$12-16$$ year-old student who is weak at maths, that $$\ 38 \times 27 \times 14 = 27 \times 14 \times 38,\$$ and that $$\ 3.4 \times 10^{-6} \times 2.1 \times 10^{-5} = 3.4\times 2.1 \times 10^{-6} \times 10^{-5},\$$ and that $$\ 11^4 \times 7^7 \times 3^5 \times 5^2 = 3^5 \times 5^2 \times 7^7 \times 11^4,\$$ and that $$\ \frac{ 8 }{ 14 } \times \frac{ 3 }{ 4 } \times \frac{ 7 }{ 11 } \times \frac{ 22 }{ 9 } = \frac{ 8 \times 3 \times 7 \times 22 }{ 14 \times 4 \times 11 \times 9 } = \frac{ 7 \times 8 \times 22 \times 3 }{ 14 \times 4 \times 11 \times 9 } = \frac{ 7 }{ 14 } \times \frac{ 8 }{ 4 } \times \frac{ 22 }{ 11 } \times \frac{ 3 }{ 9 },\$$ and so on.
I have been in this situation a few times, where it was necessary - or at least extremely helpful - for them to use the fact that if the value of the product of finitely many terms remains unchanged if terms being multiplied are rearranged within the product. [Maybe there's a better way of wording this, but anyway...]
I proceeded to explain that, "you can swap any two numbers without affecting the result due to $$ab = ba,$$ and therefore we can always re-arrange the numbers in the original product $$\ 38 \times 27 \times 14\$$ in any order we want, without affecting the value of the product." However, upon reflection, I think this line of reasoning is too complicated for students who are weak at maths to understand. This is for two reasons. First, they have to understand that swapping any two numbers gives the same result, which is not necessarily obvious, and secondly, they have to understand that there is a way to keep swapping terms to get from the first product $$\ (\ 38 \times 27 \times 14 \ )\$$ to the second product $$\ (\ 27 \times 14 \times 38\ )\$$, which also is not necessarily obvious.
Although, when I was that age, I found both of those steps "obvious", even if I did not know the mathematical language to prove it formally.
And introducing rigor is not advisable for weak students either.
So is there a completely different (perhaps visual?) way to explain this phenomenon/fact that is easy for a weak student to understand, that can also be extended to the product of four, or more numbers? Or can it be explained along the lines I am trying to, but in a way that is easier to understand?
Or is this just one of those "facts" that you tell students of that age that they "just have to accept"?
• "So is there a completely different (perhaps visual?) way to explain this phenomenon/fact that is easy for a weak student to understand". It would be helpful to know what this student's visual concept of multiplication is to start with. Based on your comment to an answer below, some of these students don't know how to find the area of a rectangle. So, what should we assume they can do with multiplication? Commented Dec 12, 2022 at 16:55
• It's worth pointing out that care should be exercised with the words used. Saying that we can always rearrange the numbers in the original product could lead some to posit that $38\times 27 = 32 \times 87$. I would want to say "we can rearrange the factors in the original product" to be careful. Commented Dec 12, 2022 at 17:01
• Every time I see this question I get confused about what is really happening here. We have a young person who is already being perceived as weak at mathematics, being taught by someone who is categorizing an axiom as obvious or not obvious. Then there is the part where repeated uses of two different axioms is also categorized as obvious or not obvious. I think the students' confusion is completely reasonable and not thinking it is anything else is entirely the first step. If motivating axioms and their applications is the question, I wish this question were re-worded. Commented Dec 15, 2022 at 20:11
• However, what I do agree with you on is that my question should clarify some specific things, which I will try to do tonight. Commented Dec 15, 2022 at 22:10
• @Dominique Most students I have tutored are not awful at arithmetic in general, although all students make mistakes with arithmetic: I associate "weaker students" (at arithmetic) as the ones who make mistakes more often than "strong students". But I don't think I have encountered a student who has absolutely no conception or understanding of addition or multiplication when it comes to, for example, (real life) word problems, for example a money calculation problem. Even the weakest student I have tutored had an inkling for when to add and when to multiply. Does this address your question? Commented Feb 9, 2023 at 8:42
For up to three dimensions you could use the concept of "area of a rectangle" or "volume of a rectangular cuboid". Use only numbers (no variables) and let the student pick a convention like "the first factor is the width and the second factor is the height (and the third one the depth)". Then use lots of examples and let the student draw many rectangles (or, if material is available and the task doesn't seem silly, arrange dice or something else in rectangular patterns) and have him or her realise ("accept?") that ab = ba always, because there are the same number of unit squares in both rectangles. Extend to three dimensions and then generalise for arbitrary numbers of factors.
I think this is about "looking at enough examples to accept the commutative law".
• This may work for some weak students. But many weak students don't know how to work out the area of a rectangle, so considerable time would have to be spent working on this in order for your method to be plausible. Also, you haven't said how you get from the student knowing that "the commutative law works" to "you can rearrange the numbers in any order and the product is the same". Commented Dec 12, 2022 at 15:36
• If these students cannot work out the area of a rectangle, what is their mental model of multiplication? Commented Dec 12, 2022 at 18:05
• @AdamRubinson I don't think that your question is answerable until you can figure out what mental model your students are working with. If they are thinking of piles of coin, five piles of three each feels very different from three piles of five each. It is not obvious that piles can be rearranged in this way. So if that is their model, I would suggest that step 1 is to show them how to arrange those piles into rectangular arrays of coins, where commutativity is easier to spot. Commented Dec 12, 2022 at 22:14
• @Adam Rubinson: For weak students, which I interpret as those not planning to attend college except perhaps a community college (which might not be until several years after high school graduation), I wouldn't worry about it. And if you do, forget about induction -- focus on pointing out that this is actually something one might need absolute proof of, and focus on what is at stake. For example, without adding any numbers, how can we be absolutely sure that $(325+228)+(132+947)=[(325+228)+132]+947$? Probably have to describe each in words, describing explicitly what one does and in what order. Commented Dec 13, 2022 at 15:43
• Cuisenaire Rods: teaching.com.au/catalogue/mta/mta-cuisenaire-rods Commented Jan 22, 2023 at 10:47
I think having a student play with examples is a great approach, and that seems the easiest route to having them see that this is true.
If seeing multiple permutations of factors all giving the same result isn't convincing, or if they want a different way to conceptualize the situation, maybe draw a model of a product like $$2\times5\times3$$ and show that pieces can be rearranged. You'd have to choose in advance how to interpret an expression like $$a\times b\times c$$, say where there are $$c$$ objects, each containing $$a\times b$$ smaller units.
Here, each of three given rectangles has $$2\times 5$$ little squares in it. Put them together, shade in different parts (in this case columns), reform the columns into rectangles, and then interpret this new picture as a new product (using the original factors, just reordered).
I'm not totally sure how a conversation like this would go with a student, whether they would be satisfied with it. I guess it would all come down to how student and instructor decide to interpret multiplication. [E.g. If $$a$$ means "number of rows in each rectangle", $$b$$ means "number of columns in each rectangle" and $$c$$ means "number of rectangles", then you could just rotate each rectangle to start with to get other expressions like $$5\times 2 \times 3$$.]
[Edit: I can see that my color choice probably wasn't great, since it looks like red changed size.]
• 2×5×3 are arranged vertically, while 2×3×5 are arranged horizontally. They look different. In fact, they look more like 15×2 :) Commented Dec 24, 2022 at 8:46
• @RustyCore I guess you'll just have to transpose that last row of shapes into a column. Commented Dec 25, 2022 at 14:30
• For a student who cannot represent multiplication of two numbers as an area, that transposition my be the bridge they would not know how to cross. Just sayin'. And you are right with color change. Maybe there should be a "purgatory" stage where all colors change to gray. Commented Jan 17, 2023 at 23:17
• @RustyCore "For a student who cannot represent multiplication of two numbers as an area..." This example presupposes that they can, and that both teacher and student agree where the factors show up in the picture. Commented Jan 17, 2023 at 23:25
Some great answers were provided here, but I wanted to add something which I think is essential: to emphasize to the student that it is a very good thing to question this! Why am I saying that? Because indeed, for many operations in algebra the associative property does not hold! For example, when talking about division: $$8/(4/2) \neq (8/4)/2$$. So in fact I would go so far as to say, that a student who questions this may have a latent / undeveloped talent for Math! Because clearly here is an example of a single operation, namely division, where the associative property does not hold. So I would definitely compliment the student for not taking this for granted, and go on to explain that multiplication is one of the special operations where it holds that $$a\times(b \times c) = (a\times b)\times c$$ and therefore we can drop the parentheses without ambiguity.
A similar example can also explain the commutativity of multiplication: $$2 \times 4 = 4 \times 2$$ but $$2 / 4 \neq 4 / 2$$. Actually I realize that this is actually more relevant to the examples given in the question, and in fact it is important to notice the implication of having these two properties together for a single operation such as multiplication. It can actually be a fun exercise for a student to prove that given:
1. $$(a \times b) \times c = a \times (b \times c)$$
2. $$a \times b = b \times a$$
then also: $$a \times b \times c = c \times b \times a$$ .
Note that given only 1. or 2., this doesn't hold! This is also a source of (justified) confusion: a lot of students implicitly think that commutativity (swapping rule) implies associativity, but those are totally independent properties.
Another thing which occasionally helps here is to ask the student the simplest possible question: "What is bothering you about this rule?". I think I've already established, there is something to be bothered about here. But the student, not always having the faculty with words to explain what is bothering him, just feels "silly" and "inadequate" to deal with the problem. In other words, he feels that it is already wrong to be bothered about something so "simple", which is absolutely bananas! The opposite is true.
• This answer is no less great than the others. Commented Feb 7, 2023 at 2:42
1. List the principle. But NOT in a symbolic (ab) manner. Just "you can change the order when multiplying." [Realize you will need to repeat it, several times. They are not programmable machines. But just keep doing so.]
-and-
1. Each time they miss it, have them do a few simpler examples. 2 x 3 x 4. 2 x 5 x 10. (in a couple orders) [After a while, they will get sick of it. And decide to remember the principle.]
So, net net, I think your instincts are correct in recognizing that weaker students won't learn axiomatically). They are different than you. But in degree...we all share traits of ability and disability in intellect. Unless you're Von Neumann, you need drill, to engrain math tricks. He of course was a Martian. But human brains are more like Harry Potter's hand. Need Umbridge's pen to furrow some lines into it/them.
P.s. I think the "like a rectangular prism volume" is low likely to succeed. Wrong pedagogic insight. It's MUCH more likely that repeated statements and practice will make things click than some magic explanation (in this case a hard one). That doesn't mean, you can't have them, can't try them. After all the "and" principle applies. And you should try lots of things. But. I would not think "key explanation" (like a lock and key) is the way to overcome these hurdles. It's more likely that familiarization is what is needed. And yes, more than you would need. More than what a high IQ student would need.
P.s.s. I don't think of rectangles with multiplication as Xander mentions. I'm not a Ph.D. mathematician. But a decent ex-STEM student. And I think of multiplication as extension of the times table (via long multiplcation). I.e. algebraically. If I really scratch my head and think of the times table itself, I probably think of repeated addition. This is NOT to say that it's the preferred way. Just that rectangles are not something everyone thinks of. I'm always pleasantly surprised by geometric drawings for Pythagoras or the like...since I don't think as geometrically. Again, not even saying this is good. Just saying the impulse to think that everyone thinks in one manner (the way one does) is a mistake. I mean...heck with exponentiation do you think first of repeated multiplication or think first of "more dimensions in Flatland"!?
Most of the answers here have been geometric. Here is an answer for three or more factors that is algebraic. It requires the student to know that multiplication is associative (more understandable to them, "we can put parentheses anywhere because of order of operations", maybe needs additional justification), the commutative law for two numbers (justified well by the rectangles in the other answers) and that two equal expressions can be substituted for one another.
Sample explanation:
Start with $$38 \times 27 \times 14$$. We are not going to calculate this, but if we were, we would need to multiply two of the numbers together first, and then multiply that result times the third number. What that means in symbols is that we might actually calculate $$(38 \times 27) \times 14$$. Because multiplication is associative, we could also do $$38 \times (27 \times 14)$$.
To summarize, $$38 \times 27 \times 14 = (38 \times 27) \times 14 = 38 \times (27 \times 14)$$. Let's say we went with $$(38 \times 27) \times 14$$, and just take a look at $$(38 \times 27)$$ on its own. Because of our commutative law, we know that $$(38 \times 27) = (27 \times 38)$$, and since we can substitute equal expressions into a third expression, our original is now $$38 \times 27 \times 14 = (38 \times 27) \times 14 = (27 \times 38) \times 14$$. We can then move the brackets again (associativity) and switch 38 and 14 to get the final answer.
The bottom line is that by doing this trick, we can see that when we have a bunch of numbers multiplied together, we can switch them around in any order.
• As an aside commenting on whether this answer truly meets the other, overall need of helping the student achieve proficiency in addition to understanding, I am not sure the difference between a student who can successfully recognize the applicability of the commutative law and apply it and one who cannot is that the former has a very clear justification of why this can be done for an arbitrary number of factors. Commented Dec 13, 2022 at 13:13
Many college students have a 2nd issue with a problem like this. If they see 2(3 · 4), they sometimes want to distribute, and write 2·3 · 2·4. Yikes. This one requires seeing the difference between distributive property and a · b · c. We work on that.
First, we go back to areas and volumes. My answer is therefore much like Jasper's. Yes, considerable time must be spent on these basics for further mathematical steps to have any meaning for them.
I ask: What is the area of a rectangle? Someone tells me. Most of them nod. I ask: Why? They look at me like I'm nuts. They think of it as a "formula", and they think of those as givens.
So I have them look at the ceiling tiles (because the floor tiles are 1 foot by 1 foot and don't teach area). We estimate those to be 2 feet by 4 feet. So that's 2 rows of 4 tiles each. We could count, or we could ... multiply, yes.
Once we've done one more problem like that on the board, and they are seeing that the multiplication is how many rows times how much in a row, then we go to volume. I use my hands to "create" (mime) a cardboard box on top of my desk. It's 4 feet across, 3 feet front to back, and 2 feet high. We imagine small 1 cubic foot boxes being packed inside. There are 3 rows of 4 on the bottom layer, and then there's a top layer. How many? 12 and 12 is 24 of the cubic foot boxes. Could we count vertical layers from front to back? Sure. The front 'layer' is 4 by 2, and then they go 3 back. 8 times 3 is also 24. And we could go left to right. 3 by 2 is 6, 4 times that is still 24 cubic feet. Of course.
I also reteach distributive property, using visuals. My goal is always to help them to see why.
Exercise: Following are several products of the same 3 numbers, but in different orders. Calculate these products using a calculator.
14 x 27 x 38
14 x 38 x 27
27 x 14 x 38
27 x 38 x 14
38 x 14 x 27
38 x 27 x 14
What do you notice? Pick three other numbers to test your conjecture.
Does it work with subtraction?
Introduce them some of euclid's axioms. For example:
Things that are equal to the same thing are also equal to one another .
If equals are added to equals, then the wholes are equal (Addition property of equality).
If equals are divided from equals, then the results are equal (division property of equality).
Then, show them with examples how these axioms can be used to prove the desire things ( You can give them a chance prior to this if you wish so). It would not result into any unfairness if they accept it as common sense because these are accepted axioms and not informal shortcuts.
• Why do you prefer an axiomatic approach for a student who is weak at mathematics? Commented Feb 3, 2023 at 17:19
• @DavidSteinberg Because axioms which I mentioned are not uncommon outside math world and hence would be good for newbies to accept as common sense. Commented Feb 3, 2023 at 17:36
When Xander wrote in his comment to his answer, "If these students cannot work out the area of a rectangle, what is their mental model of multiplication?", I think a more relevant comment is, "If these students cannot work out the area of a rectangle, then they shouldn't yet be trying to learn that $$ab = ba,$$ let alone $$abc = cab.$$" I should build multiplication up with geometry at the same level of multiplication that we are working on, with geometry as a helpful tool assisting in the understanding of multiplication, in a similar manner one would use money problems to assist in the teaching of addition. And I shouldn't try to cover all of "numbers" (standard form, HCF/LCM, etc) before moving onto geometry later on; with strong students this might be possible, as we can cover ground more quickly, but with weak students - which is what we are talking about here - it isn't an effective teaching strategy, I guess.
• The intention of my question was not to suggest that areas of rectangles are necessarily the "correct" primitive notion. Rather, the actual question as the important part: What mental model do the students have for multiplication? The way that you approach the various properties of multiplication depends on what they understand multiplication to mean in the first place. Geometry is only one possible approach. Commented Dec 20, 2022 at 22:58
• @XanderHenderson Please can you give me an example of what you're talking about and how it relates to the question. Commented Dec 21, 2022 at 8:39
• Let me clarify: your question seems to be "How do I convince weak students that "abc = cab = bca = acb = cba = bac? That is, how do I convince students that multiplication can be done in any order?" From an axiomatic point of view, this requires both commutativity ($ab = ba$) and associativity ($a(bc) = (ab)c$). But students who are struggling to understand the basic mechanics of multiplication aren't helped by these abstract notions. It is sufficient to say "you can do the multiplication in any order you like", and find some way to justify this, e.g. by permuting the sides of a solid. Commented Dec 21, 2022 at 15:25
• I have five piles of coins, each of which contains seven coins. I am going to make new piles as follows: take one coin from each of my original piles, and put them in a new pile. Then take a second coin from each of the original piles, and put them into a second new pile. Keep doing this. I will end up with piles of five coins (since my new piles contain one coin from each of the original plies), and there will be a total of seven piles (since each of my original piles had seven coins). This gives you commutativity. Commented Dec 21, 2022 at 15:38
• To expand this to more factors, try thinking about putting coins into bags: put $a$ coins into each of $b$ little bags, then put these $b$ little bags into a big bag. Do this $c$ times. This gives you $abc$. Now rearrange. Commented Dec 21, 2022 at 15:39
Start with simpler examples. So, start with explaining why $$\ 1 \times 2 = 2 \times 1$$, then $$\ 2 \times 3 = 3 \times 2$$, etc.
Next, proceed to $$\ 1 \times 2 \times 3 = 2 \times 3 \times 1$$, $$\ 1 \times 2 \times 4 = 2 \times 4 \times 1$$, etc.
Etc.
The examples you give are ridiculously complex and intimidating for a student who doesn't yet understand (or "believe") that multiplication is commutative. But once she gets the above simple examples and understands the concept, she should be able to easily extend it to your examples.
P.S. You repeatedly use the word weak to describe your students and even mention that "when I was that age, I found both of those steps "obvious"". Perhaps your student aren't as smart as you.
But if your goal is to be an effective teacher, perhaps it's better to abandon such beliefs and negative thoughts (satisfying as they may be for your ego):
• Concede that you are not a good teacher and seek to improve (rather than blame your students for their failure to learn).
• Believe it or not, your students probably know that you think they're stupid. This is very demoralizing and intimidating for them and makes it that much harder for them to learn (especially from you).
• You say to explain this and that, but you don't say how to explain it. The original poster was asking for ideas about how to explain things. Also, in my understanding, the complicated examples were examples of calculations where it might come in handy to apply commutativity in solving the kinds of problems students in the 12-16 age group are typically asked to do. I don't think they were intended as examples of how the poster would go about teaching commutativity (which is what the poster wanted to learn how to do). Commented Jan 16, 2023 at 5:36
• @WillOrrick: My answer is meant to complement the other answers already given here. I merely wanted to make one very important point, which is that the teacher should start with simplest possible examples, then work her way up. I did not want this important point to be obscured by other points.
– user18187
Commented Jan 16, 2023 at 6:05
• First of all, this isn't about multiplication being commutative. It's about multiplication being commutative and associative. Am I wrong about this? Either many people have made this mistake in answering my question, or I am wrong about this. For example, you cannot show that $(ab)c = a(bc)$ without the associativity axiom for multiplication, since this is the associativity axiom for multiplication which wouldn't exist if it was redundant, right? That all said, axiom stuff is not applicable to the original question as I have stated many times before. Commented Jan 16, 2023 at 10:18
• Secondly, you say, "The examples you give are ridiculously complex and intimidating for a student who doesn't yet understand (or "believe") that multiplication is commutative. But once she gets the above simple examples and understands the concept, she should be able to easily extend it to your examples". I disagree with literally all of your opinions here. They are just about the simplest examples I can think of. Can you think of simpler ones that aren't trivial? Secondly, just because they believe two or three numbers in a product can be rearranged without changing the result, does not... Commented Jan 16, 2023 at 10:21
• "Concede that you are not a good teacher". How did you determine that I'm not a good teacher? Your "answer" is riddled with unjustified assumptions. Commented Jan 16, 2023 at 14:21
You can teach/remind him to simplify $$(38×27×14)/(27×14×38)$$. If he knows the same number can be eliminated from both sides regardless of the position to simplify a fraction, he'll get 1, meaning they're equal.
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# Which pair of expressions are equivalent
Consider the expressions 3 2 + 1 and 5 × 2. Both are equal to 10. That is, they are equivalent expressions. Now let us consider some expressions that include variables, say 5 x + 2. The expression can be rewritten as 5 x + 2 = x + x + x + x + x + 1 + 1.
Equivalent expressions are expressions that work the same even though they look different. If two algebraic expressions are equivalent, then the two expressions have the same value when we plug in the same value(s) for the variable(s).
## How do you identify equivalent expressions?
Equivalent Expressions Equivalent Expressions are expressions that have the same value. They may look different but will have the same result if calculated. For example, and are equivalent expressions. See why below: The two expressions have the same answer, 27. Therefore, we can say that they are equivalent expressions.
## How do you find the equivalent expression?
equivalent expressionshave the same value but are presented in a differentformat using the properties of numbers eg, ax + bx = (a + b)x are equivalent expressions. Strictly, they are not “equal”, hence we should use 3 parallel lines in the”equal” rather than 2 as shown here.
## How to solve equivalent expressions?
Equivalent equations are algebraic equations that have identical solutions or roots. Adding or subtracting the same number or expression to both sides of an equation produces an equivalent equation. Multiplying or dividing both sides of an equation by the same non-zero number produces an equivalent equation.
## How to write equivalent expression?
equivalentleft (x+x,3xright) equivalent(x+x,3x) 2. Apply the formula: e q u i v a l e n t ( a, b) mathrm {equivalent}left (a,bright) equivalent(a,b) =false, where. a = x + x. a=x+x a= x+x and. b = 3 x. b=3x b =3x.
## What is an example of an equivalent expression?
Examples of Equivalent Expressions 3(x + 2) and 3x + 6 are equivalent expressions because the value of both the expressions remains the same for any value of x. 3x + 6 = 3 × 4 + 6 = 18. and can also be written as 6(x2 + 2y + 1) = 6×2 + 12y + 6. In this lesson, we learn to identify equivalent expressions.
## Which expression is equivalent to FG )( 4 )?
Thus, the expression (f + g)(4) is equal to f(4) + g(4).
## Which expressions are equivalent to the expression square root of 40?
√40 = √(2 × 2 × 2 × 5) = 2√10.
## What expression is equal to 81?
Some expressions that are equivalent to 81 are 9^2, 3\times3^3, and 8^2+17.
## What is this expression equivalent to A -> B?
Hence, a. (a+b) is equivalent to a2+ab.
## Which expression is equivalent to st6?
1 Answer. s(6) × t(6) is equivalent to (st)(6).
## What does it mean for expressions to be equivalent?
What are equivalent expressions? Equivalent expressions are expressions that work the same even though they look different. If two algebraic expressions are equivalent, then the two expressions have the same value when we plug in the same value(s) for the variable(s).
## Which expression is equivalent to the square root of 80?
√80 = 4√5. Therefore, the square root of 80 in radical form is 4√5.
## Which expression is equivalent to square root of 252?
The square root of 252 can be written as √252 and (252)1/2. The square root of 252 can be simplified to 6√7.
## Which power of 8 is equal to 2 to the power 6?
THUS 8 TO POWER 2 IS EQUAL TO 2 TO POWER 6.
## What to the power of 3 equals 81?
0:100:433^x = 81 solve the exponential equation – YouTubeYouTubeStart of suggested clipEnd of suggested clipAnd then times itself a fourth time is equal to 81 so raised to the fourth. 3 raised to the x equalsMoreAnd then times itself a fourth time is equal to 81 so raised to the fourth. 3 raised to the x equals that since they have the same base we can just drop the bases. We get X is equal to 4.
## What is the principal root √ 81?
9√81 = 3×3 = 9. Hence, the value of the square root of 81 is 9.
## Which equation is equivalent to log2n 4?
The correct answer is B) 16.
## Which expression is equivalent to gf3?
If f(x) = 4 – x2 and g(x) = 6x, which expression is equivalent to (g – f)(3)? We know that f(x) and g(x) are both the functions of x and depend on x. Therefore, the expression which is equivalent to (g – f) (3) is 23.
## Which is equivalent to f/g x?
By definition, (F? G)(x)is equal to f(g(x)). This means that every x in f(x) must be replaced with g(x), which is equal to (2/x).
## What is the value of f/g )( 8?
Summary: If f(x) = 3 – 2x and g(x) = 1/x + 5, then the value of (f/g)(8) value is -169.
## How does Sal find equivalent expressions?
Sal finds equivalent expressions by combining like terms and using the distributive property.
## Why do brackets and parentheses have to include the other one?
Brackets [] and Parentheses () both have to include the other one because you could see both. It also reminds us that division and multiplication are on the same level, so generally done left to right. Also, addition and subtraction are on the same level left to right.
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6 September, 06:44
# A plumber has a piece of pipe that is 3-meter long. He needs to cut it into sections that are 10 centimeters long. How many sections will he be able to cut?
+3
1. 6 September, 07:59
0
He will be able to cut the piece of pipe into 30 sections, each 10 cm long.
Step-by-step explanation:
A plumber has a piece of pipe that is 3-meter long.
He needs to cut it into sections that are 10 centimetres long.
This means that he wants to cut the pipe into an equal number of 10-centimetre pipes.
First let us convert 3 metres to centimetres.
1 metre = 100 centimetre
=> 3 metres = 3 * 100 = 300 centimetres.
Since he wants to divide this 300 cm pipe into equal parts of 10 cm, we divide 300 cm by 10 cm. This will yield:
300 / 10 = 30 sections
Therefore, he will be able to cut the piece of pipe into 30 sections, each 10 cm long.
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# Average and Instantaneous Rate of Change
A variable which can assign any value independently is called the independent variable, and the variable which depends on some independent variable is called the dependent variable.
For Example:
If $x = 0,1,2,3, \ldots$ etc, then
We see that as $x$ behaves independently, we call it the independent variable. But the behavior of $y$ or $f(x)$ depends on the variable $x$, so we call it the dependent variable.
Increment:
Literally the word increment means increase, but in mathematics this word covers both increase as well as decrease because the increment may be positive or negative. Simply put, the word increment in mathematics means “the difference between two values of variables.”
The final value minus the initial value is called an increment in the variable. The increment in $x$ is denoted by the symbols $\delta x$ or $\Delta x$ (read as “delta $x$”).
If $y = f(x)$, and $x$ changes from an initial value ${x_0}$ to the final value ${x_1}$, then $y$ changes from an initial value ${y_0} = f({x_0})$ to the final value ${y_1} = f({x_1})$.
Thus, the increment in ‘$x$
$\Delta x = {x_1} - {x_0}$
produces a corresponding increment in ‘$y$
$\Delta y = {y_1} - {y_0} = f({x_1}) - f({x_0})$
Average Rate of Change:
If $y = f(x)$ is a real valued continuous function in the interval $({x_0},{x_1})$, then the average rate of change of ‘$y$’ with respect to ‘$x$’ over this interval is
$\frac{{f({x_1}) - f({x_0})}}{{{x_1} - {x_0}}}$
But $\Delta x = {x_1} - {x_0}$
$\Rightarrow {x_1} = {x_0} + \Delta x$
$\therefore \frac{{f({x_0} + \Delta x) - f({x_0})}}{{\Delta x}}$
Instantaneous Rate of Change:
If $y = f(x)$ is a real valued continuous function in the interval $({x_0},{x_1})$, then the average rate of change of ‘$y$’ with respect to ‘$x$’ over this interval is
$\mathop {\lim }\limits_{{x_1} \to {x_0}} \frac{{f({x_1}) - f({x_0})}}{{{x_1} - {x_0}}}$
But $\Delta x = {x_1} - {x_0}$
$\Rightarrow {x_1} = {x_0} + \Delta x$
This shows that $\Delta x \to 0$, as ${x_1} \to {x_0}$
Average or Instantaneous Rate of Change of Distance
OR
Average or Instantaneous Velocity:
Suppose a particle (or an object) is moving in a straight line and its positions (from some fixed point) after times ${t_0}$ and ${t_1}$ are given by $S({t_0})$ and $S({t_1})$, then the average rate of change or the average velocity is
Also, the instantaneous rate of change of distance or instantaneous velocity is
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# Problem of the Month (August 2014)
Suppose we have n points in the interior of a square, and these points are rigid because they are constrained to be distance 1 from some subset of the sides of the square and each other. What are the possible sizes of the square, and the fewest number of points necessary to achieve them? What if the points are constrained to be distance 1 from some subset of the corners of the square and each other?
We can almost always add points to a rigid configuration without changing the side of the square, so we only show configurations containing the fewest number of points for a given side.
Here are the known configurations where points may be 1 unit from a side:
Configurations with 1 Point
2
Configurations with 2 Points
2 - 1/√2 = 1.292+ 2 + 1/√2 = 2.707+ 3
Configurations with 3 Points
2-(√6+√2)/4 = 1.034+ 2 - 2/√5 = 1.105+ 2-(√6-√2)/4 = 1.741+ 2+(√6-√2)/4 = 2.258+ 5/2 = 2.5 2 + √3/2 = 2.866+ 2 + 2/√5 = 2.894+ 2+(√6+√2)/4 = 2.965+ 4
Configurations with 4 Points
1.003+ 2 - 3/√10 = 1.051+ 7/5 = 1.4 2 - 1/√5 = 1.552+ (9 - √7)/4 = 1.588+ 1.656+ 1.874+ 2.126+ 3 - √3/2 = 2.133+ 2.274+ 2.278+ 3 - 1/√2 = 2.292+ 2.343+ (7 + √7)/4 = 2.411+ 2 + 1/√5 = 2.447+ 13/5 = 2.6 2.626+ 2.721+ 2.806+ (9 + √7)/4 = 2.911+ 2 + 3/√10 = 2.948+ 2.992+ 2.996+ 2 + √(3/2) = 3.224+ (5 + √3)/2 = 3.366+ 2 + √2 = 3.414+ 7/2 = 3.5 3 + 1/√2 = 3.707+ 2 + √3 = 3.732+ 3 + √3/2 = 3.866+ 5
Here are the known configurations where points may be 1 unit from a corner:
Configurations with 1 Point
√2 = 1.414+
Configurations with 2 Points
√(3/2) = 1.224+ (1 + √7)/2 = 1.822+ (√2 + √6)/2 = 1.931+
Configurations with 3 Points
1.782+ √(2+√2) = 1.847+ 1.95517+
Configurations with 4 Points
1.296+ 1.347+ 1.635+ 1.680+ 1.743+ 1.816+ 1.862+ √(7/2) = 1.870+ 1.95500724+ 1.95500785+ 1.95521750+ 1.95521754+ (√5 + √3)/2 = 1.984+ 1.992+ 1.999+
If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 8/1/14.
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# How do you write a polynomial in standard form given the zeros x=-4, 5. -1?
May 26, 2016
${x}^{3} - 21 x - 20 = 0$
#### Explanation:
If $\left\{\alpha , \beta , \gamma , \delta , . .\right\}$ are the zeros of a function, then the function is
$\left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right) \left(x - \delta\right) \ldots = 0$
Here zeros are $- 4$, $5$) and $- 1$, hence function is
$\left(x - \left(- 4\right)\right) \left(x - 5\right) \left(x - \left(- 1\right)\right) = 0$ or
$\left(x + 4\right) \left(x - 5\right) \left(x + 1\right) = 0$ or
$\left({x}^{2} + 4 x - 5 x - 20\right) \left(x + 1\right) = 0$ or
$\left({x}^{2} - x - 20\right) \left(x + 1\right) = 0$ or
${x}^{3} - {x}^{2} - 20 x + {x}^{2} - x - 20 = 0$ or
${x}^{3} - 21 x - 20 = 0$
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# Is -1 a Rational Number? Detailed Explanation With Sample
Yes, the number $-1$ is a rational number because we can write the number negative $1$ in $\dfrac{p}{q}$ form.
So, the question arises, “what is meant by $\dfrac{p}{q}$ form?” “What is meant by “p” and what is meant by “$q$”?” In this article, we will study in detail what makes “$-1$” a rational number and, more importantly, how we determine which number is a rational number.
At the end of this topic, you will have a firm grip on the concept of rational numbers, and you will easily differentiate between a rational and an irrational number.
## Is -1 a Rational Number?
Yes, the number “$-1$” is a rational number because it is an integer, and all integers are rational numbers. Hence, the number “$-1$” can be written as $-\dfrac{1}{1}$, so we can say that “$-1$” is a rational number.
Let us cover some examples, so that concept of rational numbers becomes crystal clear for you.
Example 1: Is the number $-1.1111$ rational number?
Solution:
Yes, the number $-1.1111$ is rational number as it can be written in $\dfrac{p}{q}$ form as $-\dfrac{11111}{10000}$.
Example 2: Is the number $1$ $\dfrac{1}{1}$ a rational number?
Solution:
Yes, the number $1$ $\dfrac{1}{1}$ is a rational number as it can be written as $\dfrac{2}{1}$ which is a fraction; hence it is a rational number.
Example 2: Is negative 2 a rational number?
Solution:
Yes, it is a rational number.
Example 2: Is negative 12 a rational number?
Solution:
Yes, it is a rational number.
Example 2: Is negative 3 a rational number?
Solution:
Yes, it is a rational number.
## Rational Numbers
The word rational is derived from the Latin word “ratio,” which in Latin means reasonable, calculate-able or having a ratio. The ratio is a comparison between 2 or more numbers given in fraction form, so we can extract that rational numbers will always be given in fraction form.
In short, the numbers which can be expressed in $\dfrac{p}{q}$ or fraction form are called rational numbers. The rational number can be a negative, positive or zero number. The only thing that should be kept in mind is that for the expression $\dfrac{p}{q}$, the value of “$q$” should be $\neq$ 0 otherwise, it will give us an indefinite answer which is not acceptable in maths.
For example, the number $\dfrac{5}{3}$ is considered to be a rational number where the integer $5$ is divided by an integer $3$ and as the value of “$q$” is not zero, hence it is a rational number.
### What Is a Number?
Numbers are used as a measurement tool in mathematics, and they are the symbols to represent the count of a thing or subject. We know numbers can be a single digit or two or more digits. To learn how to identify a rational number, it is essential that we first cover the basics related to a number itself and its types and know the difference between a number and a digit.
### Numbers vs Digits
A digit is a numerical representation of the following symbols $0,1,2,3,4,5,6,7,8$, and $9$. So all these numerical symbols are known as digits, and when we combine two or more digits together, it will give us a number. So a digit is a single numeral representation of a count or number, while a number is a numeral representation having one or more than one digits. For example, if Anna has $25$ books in her library, then $25$ is a number while “$2$” and “$5$” are digits.
Now that we know the difference between a number and a digit, let us discuss different types of numbers and their properties. There are different types of numbers, and some of them are given below.
1. Binary numbers
2. Natural numbers
3. Whole numbers
4. Integers
5. Rational numbers
6. Irrational numbers
7. Real Numbers
8. Complex numbers
Binary Numbers: In mathematics, if the numbers are only represented by 1’s and 0’s, then we call them binary numbers. This means every numerical number will be represented in the form of 1’s and 0’s. For example, “0” is represented as “$0$” in binary and similar the number “$1$” is represented as “$1$” while the number $2$ will be represented as 10 while the number $3$ is represented as $011$ and so on.
Natural Numbers: In mathematics, all the positive integers are known as natural numbers. Natural numbers start from the number $1$ up to infinity, but these are all positive numbers.
Whole Numbers: The whole numbers are basically a set of natural numbers but they also include the number “$0$” in addition to all natural numbers. So the whole numbers start from the number zero up to infinity. We can write whole numbers as $0,1,2,4$,…..
Integers: Integers consists of all the whole numbers as well as there negative counterparts, i.e., $\cdots, -4,-3,-2,-1,0,1,2,3,4,\cdots$.
Rational Numbers: The numbers which can be written as $\dfrac{p}{q}$, where both $p$ and $q$ are integers and $q\neq 0$ are called rational numbers. All natural numbers, whole numbers, and integers themselves are rational numbers. For example, we can write $-4$ as $\dfrac{-4}{1}$ and hence it is a rational number. Also, $\dfrac{5}{7}$, $\dfrac{2}{3}$ and $\dfrac{1}{8}$, etc., are examples of rational numbers.
Irrational Numbers: The number which cannot be expressed in $\dfrac{p}{q}$ form or the number which cannot be expressed in fraction/ratio form is known as an irrational number. Mathematicians initially perceived that all the numbers were rational and could be written in $\dfrac{p}{q}$ form, but later on, Greeks discovered that some roots of equations cannot be written in a fraction form, so they termed them as irrational numbers. Common irrational numbers are $\sqrt{2}$, $\pi$ etc.
Real Numbers: Real numbers consist of both rational and irrational numbers. For example, $\dfrac{1}{2}$, $0.3333$, and $\pi$ all are real numbers.
Complex Numbers: The numbers which are expressed or written in a+ix form are termed complex numbers. Here, “$a$” and “$b$” both are real numbers, while the “i” is called iota and is an imaginary number and is equal to $\sqrt{-1}$. So any real number which is written along iota will be termed an imaginary number. For example, if we are given a number “$3+4i$,” then “$3$” is called the real number while $4$ is called the imaginary number, and as a whole “$3+4i$” is called a complex number.
Types of different numbers and their definition were necessary because some of them are also types of rational numbers. Now let’s have a look at the various types of rational numbers.
### Types of Rational Numbers
Rational numbers can be classified into different types, and some of them are given below.
1. Whole Numbers
2. Natural Numbers
3. Decimal Numbers
4. Fractions
Whole Numbers: The whole numbers can be written in $\dfrac{p}{q}$ form; hence all the whole numbers are rational numbers, including the number “$0$”. For example we can write $0$ as $\dfrac{0}{1}$,$\dfrac{0}{2}$,$\dfrac{0}{3}$,$\dfrac{0}{4}$ and so on
Natural Numbers: Like whole numbers, all the natural numbers are also rational numbers as they can also be expressed in $\dfrac{p}{q}$ form. For example, $\dfrac{2}{1}$, $\dfrac{3}{1}$,$\dfrac{4}{1}$ etc
Decimal Numbers: The numbers divided into two parts which are separated by a point “.” are known as decimal numbers. The number(s) on the left side of the point are whole numbers, while the numbers on the right-hand side of the point are known as fractions. For example, the number $18.36$ is known as a decimal number where 18 is the whole number while $36$ is the decimal part or fraction part of the number.
Some of the decimal numbers are also rational numbers. There are different types of decimal numbers, for example, terminating decimal numbers, repeating decimal numbers, and non-terminating decimal numbers.
All the terminating decimals are rational numbers as they can be written in $\dfrac{p}{q}$ form; for example, $0.64$, $0.75$, and $0.67124$ all these numbers are rational numbers
All the repeating decimals are also rational numbers. Repeating decimals are the numbers where the decimal part of the number repeats itself. For example, numbers 2.1111111 and $3.121212$ are rational numbers.
Finally, the non-terminating and non-repeating decimals are not rational numbers. For example, the decimal notation of $\pi$ is $3.14159\cdots$. Note that it is a non-terminating decimal number that does not repeat itself.
Integer Numbers: All integers are rational numbers as well.
### How to Identify Rational Numbers
There are certain tricks to easily identify a rational number, and they are:
1. If the number is written in $\dfrac{p}{q}$ form such that $p$ and $q$ are integers and $q$ $\neq$ $0$, then the number is a rational number.
2. If the number is not given in fraction form but we are given a number in decimals instead, then we will check whether the fraction part is terminating or repeating. In both cases, it will be a rational number.
3. All real numbers are rational numbers, excluding those which cannot be expressed as $\dfrac{p}{q}$ form.
After learning all about numbers and how to identify rational numbers, we can develop a Venn diagram for rational and irrational numbers, which is given below.
The diagram for irrational numbers does not include any subset, and it can be drawn as:
### Practice Questions:
1. Is number $-\dfrac{1}{0}$ a rational number?
2. Is 0 a rational number?
3. Is number $\sqrt{1}$ a rational number?
4. Is number $\sqrt{-1}$ a rational number?
5. Is 1/2 a rational number?
6. -3 is a rational number, true or false.
1)
No, the number $-\dfrac{1}{0}$ is not a rational number because the value of “q” in this case is zero; hence the number is not defined, and it is not a rational number.
2)
Yes, 0 is a rational number.
3)
Yes, $\sqrt{1}$ is rational a rational number as $\sqrt{1} = 1$. Since “$1$” is a rational number, so $\sqrt{1}$ is also a rational number.
4)
No, $\sqrt{-1}$ is not a rational number. As all the rational numbers are real numbers while $\sqrt{-1}$ is an imaginary number, hence it is not a rational number.
5)
Yes, $\dfrac{1}{2}$ is a rational number.
6)
Yes, $-3$ is a rational number.
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## Precalculus (6th Edition) Blitzer
a. $a=-2, b=32, c=42$ b. 2pm,$170$ (ppm).
a. Step 1. Let $c=u, b=v, a=w$. The function becomes $y=u+vx+wx^2$. Use the data from the table to set up the system of equations: $\begin{cases} u+2v+4w=98 \\ u+4v+16w=138 \\ u+10v+100w=162 \end{cases}$ Step 2. Write the matrix and perform row operations: $\begin{bmatrix} 1 & 2 & 4 & | & 98 \\ 1 & 4 & 16 & | & 138 \\ 1 & 10 & 100 & | & 162 \end{bmatrix}\begin{array} .. \\R2-R1\to R2\\ R3-R1\to R3 \end{array}$ $\begin{bmatrix} 1 & 2 & 4 & | & 98 \\ 0 & 2 & 12 & | & 40 \\ 0 & 8 & 96 & | & 64 \end{bmatrix}\begin{array} .. \\R2/2 \to R2\\ R3/8\to R3 \end{array}$ $\begin{bmatrix} 1 & 2 & 4 & | & 98 \\ 0 & 1 & 6 & | & 20 \\ 0 & 1 & 12 & | & 8 \end{bmatrix}\begin{array} .. \\..\\ R3-R2\to R3 \end{array}$ $\begin{bmatrix} 1 & 2 & 4 & | & 98 \\ 0 & 1 & 6 & | & 20 \\ 0 & 0 & 6 & | & -12 \end{bmatrix}\begin{array} .. \\..\\ .. \end{array}$ Step 3. The last row gives $w=-2$; back-substitute to get $v=20-6(-2)=32$ and $u+2(32)+4(-2)=98, u=42$. Thus, we have $a=-2, b=32, c=42$ and the equation is $y=-2x^2+32x+42$ b. The maximum of the function can be found at $x=-\frac{b}{2a}=-\frac{32}{2(-2)}=8$ (hours after 6am, that is 2pm), which gives $y=-2(8)^2+32(8)+42=170$ (level of pollution in ppm).
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# Euclidean Algorithm
The Euclidean Algorithm, also known as Euclid’s Algorithm, is used in discrete mathematics to find the greatest common divisor of two natural numbers, namely $$a$$ and $$b$$. The greatest common divisor is typically denoted as $$gcd\left(a,b\right).$$ In general, the Euclidean Algorithm is used within numerous applications like solving Diophantine equations, constructing continued fractions, and is even used when dividing in modular arithmetic.
There are some properties that go along with finding the $$gcd\left(a,b\right)$$ in the Euclidean Algorithm:
The first two properties are very similar:
If $$a=0$$, then the $$gcd\left(a,b\right)=b$$
If $$b=0$$, then the $$gcd\left(a,b\right)=a$$
The third property is if $$a=bq+r$$ where $$b\ne0,$$ then the $$gcd\left(a,b\right)=gcd\left(b,r\right)$$, where $$q$$ is an integer and $$r$$ is an integer between $$0$$ and $$b-1$$
Notice that this property is important because it will let us take a more complex problem and break it down into a smaller problem which is not as difficult to solve.
Procedure of the Euclidean Algorithm:
We start with our $$a$$ and $$b$$ values and see how many times we can divide $$a$$ by $$b$$ then we multiply $$b$$ by $$q$$ (quotient) and add the $$r$$ (remainder). For the next line, we move our $$b$$ value and our $$r$$ value in a diagonal pattern and go through the process of dividing $$\frac{a}{b}$$. This process can be continued until we get to a $$0$$ remainder. The last non-zero remainder is our $$gcd.$$
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# How do you find a power series representation for f(x) = x / (1+x^2) and what is the radius of convergence?
Sep 29, 2015
Write out a power series that when multiplied by $1 + {x}^{2}$ gives $x$.
Find ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$ works and has radius of convergence $1$.
#### Explanation:
Consider ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} = x - {x}^{3} + {x}^{5} - {x}^{7} + \ldots$
$\left(1 + {x}^{2}\right) {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$
$= {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} + {x}^{2} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$
$= {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} - {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$
$= {\left(- 1\right)}^{0} {x}^{1} = x$
So:
${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} = \frac{x}{1 + {x}^{2}} = f \left(x\right)$
...if the sums converge.
The sum ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$ is a geometric series with common ratio $- {x}^{2}$.
To converge, the absolute value of the common ratio must be less than $1$.
That is $\left\mid - {x}^{2} \right\mid < 1$, so $\left\mid x \right\mid < 1$
That is: the radius of convergence is $1$.
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# COUNTING METHODS
0
175
1
SOLVE AND VERIFY
a. (n-2)! / (n-3)! = 5
b. (n+2)! / n! = 42
c. (n-3)! / 2!(n-4)! = 17
d. nP2 = 12
May 25, 2020
#1
+2424
+1
I will do the first two problems and then I will let you apply the techniques I demonstrated for the rest of them.
a) $$\frac{(n-2)!}{(n-3)!}=5$$
The key to this question is to manipulate the left-hand side of the equation by "expanding" one of the factorials until a major cancellation occurs. Here is the way I would approach it.
$$\frac{(n-2)(n-3)!}{(n-3)!}=5$$
This manipulation is very helpful as it is now clear that $$(n-3)!$$ is a factor of both the numerator and the denominator. We can cancel this common factor to simplify the equation significantly.
$$n-2=5\\ n=7$$
Okay, let's apply this approach to the next one, too.
b) $$\frac{(n+2)!}{n!}=42$$
The procedure is the same: Expand one of the factorial terms until a major cancellation occurs.
$$\frac{(n+2)(n+1)n!}{n!}=42$$
This time, $$n!$$ was the common factor!
$$(n+2)(n+1)=42$$
This is a fairly standard quadratic. Let's solve it now.
$$n^2+n+2n+2=42\\ n^2+3n-40=0\\$$
This quadratic happens to be factorable.
$$(n-5)(n+8)=0\\ n=5\text{ or }n=-8$$
Always quickly check to make sure that your solutions are valid. Here, $$n=-8$$ would be considered an extraneous solution because the input of a factorial function should always be nonnegative.
May 25, 2020
edited by TheXSquaredFactor May 25, 2020
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Giáo trình
# Precalculus
Mathematics and Statistics
## Geometric Sequences
Tác giả: OpenStaxCollege
Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of \$26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be \$26,520 after one year; \$27,050.40 after two years; \$27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way.
# Finding Common Ratios
The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.
Definition of a Geometric Sequence
A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If ${a}_{1}$ is the initial term of a geometric sequence and $r$ is the common ratio, the sequence will be
Given a set of numbers, determine if they represent a geometric sequence.
1. Divide each term by the previous term.
2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.
Finding Common Ratios
Is the sequence geometric? If so, find the common ratio.
1. $1\text{,}\text{\hspace{0.17em}}2\text{,}\text{\hspace{0.17em}}4\text{,}\text{\hspace{0.17em}}8\text{,}\text{\hspace{0.17em}}16\text{,}\text{\hspace{0.17em}}...$
Divide each term by the previous term to determine whether a common ratio exists.
1. $\begin{array}{llllllllll}\frac{2}{1}=2\hfill & \hfill & \hfill & \frac{4}{2}=2\hfill & \hfill & \hfill & \frac{8}{4}=2\hfill & \hfill & \hfill & \frac{16}{8}=2\hfill \end{array}$
The sequence is geometric because there is a common ratio. The common ratio is 2.
2. $\begin{array}{lllllll}\frac{12}{48}=\frac{1}{4}\hfill & \hfill & \hfill & \frac{4}{12}=\frac{1}{3}\hfill & \hfill & \hfill & \frac{2}{4}=\frac{1}{2}\hfill \end{array}$
The sequence is not geometric because there is not a common ratio.
Analysis
The graph of each sequence is shown in [link]. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.
If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?
No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.
Is the sequence geometric? If so, find the common ratio.
The sequence is not geometric because $\frac{10}{5}\ne \frac{15}{10}$.
Is the sequence geometric? If so, find the common ratio.
The sequence is geometric. The common ratio is $\frac{1}{5}$.
# Writing Terms of Geometric Sequences
Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is ${a}_{1}=-2$ and the common ratio is $r=4,$ we can find subsequent terms by multiplying $-2\cdot 4$ to get $-8$ then multiplying the result $-8\cdot 4$ to get $-32$ and so on.
The first four terms are
Given the first term and the common factor, find the first four terms of a geometric sequence.
1. Multiply the initial term, ${a}_{1},$ by the common ratio to find the next term, ${a}_{2}.$
2. Repeat the process, using ${a}_{n}={a}_{2}$ to find ${a}_{3}$ and then ${a}_{3}$ to find ${a}_{4,}$ until all four terms have been identified.
3. Write the terms separated by commons within brackets.
Writing the Terms of a Geometric Sequence
List the first four terms of the geometric sequence with ${a}_{1}=5$ and $r=–2.$
Multiply ${a}_{1}$ by $-2$ to find ${a}_{2}.$ Repeat the process, using ${a}_{2}$ to find ${a}_{3},$ and so on.
The first four terms are $\left\{5,–10,20,–40\right\}.$
List the first five terms of the geometric sequence with ${a}_{1}=18$ and $r=\frac{1}{3}.$
${ 18,6,2, 2 3 , 2 9 }$
# Using Recursive Formulas for Geometric Sequences
A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.
Recursive Formula for a Geometric Sequence
The recursive formula for a geometric sequence with common ratio $r$ and first term ${a}_{1}$ is
$a n =r a n−1 ,n≥2$
Given the first several terms of a geometric sequence, write its recursive formula.
1. State the initial term.
2. Find the common ratio by dividing any term by the preceding term.
3. Substitute the common ratio into the recursive formula for a geometric sequence.
Using Recursive Formulas for Geometric Sequences
Write a recursive formula for the following geometric sequence.
The first term is given as 6. The common ratio can be found by dividing the second term by the first term.
Substitute the common ratio into the recursive formula for geometric sequences and define ${a}_{1}.$
Analysis
The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in [link]
Do we have to divide the second term by the first term to find the common ratio?
No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.
Write a recursive formula for the following geometric sequence.
# Using Explicit Formulas for Geometric Sequences
Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.
Let’s take a look at the sequence This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is
The graph of the sequence is shown in [link].
Explicit Formula for a Geometric Sequence
The nth term of a geometric sequence is given by the explicit formula:
$a n = a 1 r n−1$
Writing Terms of Geometric Sequences Using the Explicit Formula
Given a geometric sequence with$\text{\hspace{0.17em}}{a}_{1}=3\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}{a}_{4}=24,\text{\hspace{0.17em}}$find ${a}_{2}.$
The sequence can be written in terms of the initial term and the common ratio$\text{\hspace{0.17em}}r.$
Find the common ratio using the given fourth term.
Find the second term by multiplying the first term by the common ratio.
Analysis
The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.
Given a geometric sequence with ${a}_{2}=4$ and ${a}_{3}=32$, find ${a}_{6}.$
${a}_{6}=16,384$
Writing an Explicit Formula for the nth Term of a Geometric Sequence
Write an explicit formula for the $n\text{th}$ term of the following geometric sequence.
The first term is 2. The common ratio can be found by dividing the second term by the first term.
The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.
The graph of this sequence in [link] shows an exponential pattern.
Write an explicit formula for the following geometric sequence.
${a}_{n}=-{\left(-3\right)}^{n-1}$
# Solving Application Problems with Geometric Sequences
In real-world scenarios involving arithmetic sequences, we may need to use an initial term of ${a}_{0}$ instead of ${a}_{1}.\text{\hspace{0.17em}}$In these problems, we can alter the explicit formula slightly by using the following formula:
Solving Application Problems with Geometric Sequences
In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.
1. Write a formula for the student population.
2. Estimate the student population in 2020.
1. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.
Let $P$ be the student population and $n$ be the number of years after 2013. Using the explicit formula for a geometric sequence we get
2. We can find the number of years since 2013 by subtracting.
We are looking for the population after 7 years. We can substitute 7 for $n$ to estimate the population in 2020.
The student population will be about 374 in 2020.
A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.
1. Write a formula for the number of hits.
2. Estimate the number of hits in 5 weeks.
1. The number of hits will be about 333.
Access these online resources for additional instruction and practice with geometric sequences.
# Key Equations
recursive formula for $nth$ term of a geometric sequence ${a}_{n}=r{a}_{n-1},n\ge 2$ explicit formula for$\text{\hspace{0.17em}}nth\text{\hspace{0.17em}}$term of a geometric sequence $a n = a 1 r n−1$
# Key Concepts
• A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
• The constant ratio between two consecutive terms is called the common ratio.
• The common ratio can be found by dividing any term in the sequence by the previous term. See [link].
• The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See [link] and [link].
• A recursive formula for a geometric sequence with common ratio $r$ is given by $\text{\hspace{0.17em}}{a}_{n}=r{a}_{n–1}\text{\hspace{0.17em}}$for $n\ge 2$.
• As with any recursive formula, the initial term of the sequence must be given. See [link].
• An explicit formula for a geometric sequence with common ratio $r$ is given by $\text{\hspace{0.17em}}{a}_{n}={a}_{1}{r}^{n–1}.$ See [link].
• In application problems, we sometimes alter the explicit formula slightly to $\text{\hspace{0.17em}}{a}_{n}={a}_{0}{r}^{n}.\text{\hspace{0.17em}}$See [link].
# Section Exercises
## Verbal
What is a geometric sequence?
A sequence in which the ratio between any two consecutive terms is constant.
How is the common ratio of a geometric sequence found?
What is the procedure for determining whether a sequence is geometric?
Divide each term in a sequence by the preceding term. If the resulting quotients are equal, then the sequence is geometric.
What is the difference between an arithmetic sequence and a geometric sequence?
Describe how exponential functions and geometric sequences are similar. How are they different?
Both geometric sequences and exponential functions have a constant ratio. However, their domains are not the same. Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.
## Algebraic
For the following exercises, find the common ratio for the geometric sequence.
$1,3,9,27,81,...$
$-0.125,0.25,-0.5,1,-2,...$
The common ratio is $-2$
$-2,-\frac{1}{2},-\frac{1}{8},-\frac{1}{32},-\frac{1}{128},...$
For the following exercises, determine whether the sequence is geometric. If so, find the common ratio.
$-6,-12,-24,-48,-96,...$
The sequence is geometric. The common ratio is 2.
$5,5.2,5.4,5.6,5.8,...$
$-1,\frac{1}{2},-\frac{1}{4},\frac{1}{8},-\frac{1}{16},...$
The sequence is geometric. The common ratio is $-\frac{1}{2}.$
$6,8,11,15,20,...$
$0.8,4,20,100,500,...$
The sequence is geometric. The common ratio is $5.$
For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio.
$\begin{array}{cc}{a}_{1}=8,& r=0.3\end{array}$
$\begin{array}{cc}{a}_{1}=5,& r=\frac{1}{5}\end{array}$
$5,1,\frac{1}{5},\frac{1}{25},\frac{1}{125}$
For the following exercises, write the first five terms of the geometric sequence, given any two terms.
$\begin{array}{cc}{a}_{7}=64,& {a}_{10}\end{array}=512$
$\begin{array}{cc}{a}_{6}=25,& {a}_{8}\end{array}=6.25$
$800,400,200,100,50$
For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio.
The first term is $2,$ and the common ratio is $3.$ Find the 5th term.
The first term is 16 and the common ratio is $-\frac{1}{3}.$ Find the 4th term.
${a}_{4}=-\frac{16}{27}$
For the following exercises, find the specified term for the geometric sequence, given the first four terms.
${a}_{n}=\left\{-1,2,-4,8,...\right\}.$ Find ${a}_{12}.$
${a}_{n}=\left\{-2,\frac{2}{3},-\frac{2}{9},\frac{2}{27},...\right\}.$ Find ${a}_{7}.$
${a}_{7}=-\frac{2}{729}$
For the following exercises, write the first five terms of the geometric sequence.
$\begin{array}{cc}{a}_{1}=-486,& {a}_{n}=-\frac{1}{3}\end{array}{a}_{n-1}$
$\begin{array}{cc}{a}_{1}=7,& {a}_{n}=0.2{a}_{n-1}\end{array}$
$7,1.4,0.28,0.056,0.0112$
For the following exercises, write a recursive formula for each geometric sequence.
${a}_{n}=\left\{-1,5,-25,125,...\right\}$
${a}_{n}=\left\{-32,-16,-8,-4,...\right\}$
$\begin{array}{cc}a{}_{1}=-32,& {a}_{n}=\frac{1}{2}{a}_{n-1}\end{array}$
${a}_{n}=\left\{14,56,224,896,...\right\}$
${a}_{n}=\left\{10,-3,0.9,-0.27,...\right\}$
$\begin{array}{cc}{a}_{1}=10,& {a}_{n}=-0.3{a}_{n-1}\end{array}$
${a}_{n}=\left\{0.61,1.83,5.49,16.47,...\right\}$
${a}_{n}=\left\{\frac{3}{5},\frac{1}{10},\frac{1}{60},\frac{1}{360},...\right\}$
$\begin{array}{cc}{a}_{1}=\frac{3}{5},& {a}_{n}=\frac{1}{6}{a}_{n-1}\end{array}$
${a}_{n}=\left\{-2,\frac{4}{3},-\frac{8}{9},\frac{16}{27},...\right\}$
${a}_{n}=\left\{\frac{1}{512},-\frac{1}{128},\frac{1}{32},-\frac{1}{8},...\right\}$
${a}_{1}=\frac{1}{512},{a}_{n}=-4{a}_{n-1}$
For the following exercises, write the first five terms of the geometric sequence.
${a}_{n}=-4\cdot {5}^{n-1}$
${a}_{n}=12\cdot {\left(-\frac{1}{2}\right)}^{n-1}$
$12,-6,3,-\frac{3}{2},\frac{3}{4}$
For the following exercises, write an explicit formula for each geometric sequence.
${a}_{n}=\left\{-2,-4,-8,-16,...\right\}$
${a}_{n}=\left\{1,3,9,27,...\right\}$
${a}_{n}={3}^{n-1}$
${a}_{n}=\left\{-4,-12,-36,-108,...\right\}$
${a}_{n}=\left\{0.8,-4,20,-100,...\right\}$
${a}_{n}=0.8\cdot {\left(-5\right)}^{n-1}$
${a}_{n}=\left\{-1.25,-5,-20,-80,...\right\}$
${a}_{n}=\left\{-1,-\frac{4}{5},-\frac{16}{25},-\frac{64}{125},...\right\}$
${a}_{n}=-{\left(\frac{4}{5}\right)}^{n-1}$
${a}_{n}=\left\{2,\frac{1}{3},\frac{1}{18},\frac{1}{108},...\right\}$
${a}_{n}=\left\{3,-1,\frac{1}{3},-\frac{1}{9},...\right\}$
${a}_{n}=3\cdot {\left(-\frac{1}{3}\right)}^{n-1}$
For the following exercises, find the specified term for the geometric sequence given.
Let ${a}_{1}=4,$ ${a}_{n}=-3{a}_{n-1}.$ Find ${a}_{8}.$
Let ${a}_{n}=-{\left(-\frac{1}{3}\right)}^{n-1}.$ Find ${a}_{12}.$
${a}_{12}=\frac{1}{177,147}$
For the following exercises, find the number of terms in the given finite geometric sequence.
${a}_{n}=\left\{-1,3,-9,...,2187\right\}$
${a}_{n}=\left\{2,1,\frac{1}{2},...,\frac{1}{1024}\right\}$
There are $12$ terms in the sequence.
## Graphical
For the following exercises, determine whether the graph shown represents a geometric sequence.
The graph does not represent a geometric sequence.
For the following exercises, use the information provided to graph the first five terms of the geometric sequence.
$\begin{array}{cc}{a}_{1}=1,& r=\frac{1}{2}\end{array}$
$\begin{array}{cc}{a}_{1}=3,& {a}_{n}=2{a}_{n-1}\end{array}$
${a}_{n}=27\cdot {0.3}^{n-1}$
## Extensions
Use recursive formulas to give two examples of geometric sequences whose 3rd terms are$\text{\hspace{0.17em}}200.$
Answers will vary. Examples: ${\begin{array}{cc}{a}_{1}=800,& {a}_{n}=0.5a\end{array}}_{n-1}$ and ${\begin{array}{cc}{a}_{1}=12.5,& {a}_{n}=4a\end{array}}_{n-1}$
Use explicit formulas to give two examples of geometric sequences whose 7th terms are $1024.$
Find the 5th term of the geometric sequence $\left\{b,4b,16b,...\right\}.$
${a}_{5}=256b$
Find the 7th term of the geometric sequence $\left\{64a\left(-b\right),32a\left(-3b\right),16a\left(-9b\right),...\right\}.$
At which term does the sequence exceed $100?$
The sequence exceeds $100$ at the 14th term, ${a}_{14}\approx 107.$
At which term does the sequence begin to have integer values?
For which term does the geometric sequence ${a}_{{}_{n}}=-36{\left(\frac{2}{3}\right)}^{n-1}$ first have a non-integer value?
${a}_{4}=-\frac{32}{3}\text{\hspace{0.17em}}$is the first non-integer value
Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10th term.
Use the explicit formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8th term.
Answers will vary. Example: Explicit formula with a decimal common ratio: ${a}_{n}=400\cdot {0.5}^{n-1};$ First 4 terms: $\begin{array}{cc}400,200,100,50;& {a}_{8}=3.125\end{array}$
Is it possible for a sequence to be both arithmetic and geometric? If so, give an example.
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# Foot of a $10 \mathrm{~m}$ long ladder leaning against a vertical wall is $6 \mathrm{~m}$ away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.
Given:
Foot of a $10 \mathrm{~m}$ long ladder leaning against a vertical wall is $6 \mathrm{~m}$ away from the base of the wall.
To do:
We have to find the height of the point on the wall where the top of the ladder reaches.
Solution:
Let $\mathrm{AB}$ be a vertical wall and $\mathrm{AC}$ be a ladder.
$AC=10\ m$
The top of the ladder reaches $\mathrm{A}$ and the distance of the ladder from the base of the wall $B C$ is $6 \mathrm{~m}$.
In right angled triangle $A B C$,
$A C^{2}=A B^{2}+B C^{2}$
$(10)^{2}=A B^{2}+(6)^{2}$
$100=A B^{2}+36$
$A B^{2}=100-36$
$AB^2=64$
$A B=\sqrt{64}$
$AB=8 \mathrm{~m}$
Hence, the height of the point on the wall where the top of the ladder reaches is $8 \mathrm{~m}$.
Tutorialspoint
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## Arc Length Calculator
An Arc Length Calculator is a mathematical tool used to calculate the length of an arc on a circle, given the measure of the central angle and the radius of the circle.
Desktop
Enter the radius of the circle and the central angle in degrees, then click the Calculate button.
Desktop
Desktop
# Exploring Arc Length Calculator Integral: A Comprehensive Guide
Geometry, physics, engineering, and many other disciplines depend heavily on arc length, a basic idea in mathematics. When dealing with curved or irregular shapes, calculating the length of an arc can be challenging. We will examine the arc length calculator integral in this post, which offers a sophisticated and accurate way to determine arc lengths.
We will cover various aspects of arc length calculation, including its relation to π, step-by-step instructions, the role of radius, and its applications in three-dimensional space. So, let's embark on this mathematical journey and demystify the arc length calculator integral.
## Understanding Arc Length
It's important to understand the notion itself before delving into the nuances of arc length calculation. The term "arc length" describes the length of a curve or a segment of a curve. Imagine it as the length of a string that traces the contour of the curve. There are many practical uses for arc length, which is a fundamental number in geometry and mathematics.
## The Length of an Arc Formula
The formula for the length of an arc is given by:
$L = \frac{\theta}{360^\circ} \cdot 2\pi r$
## Arc Length Calculator Integral
The formula for the arc length of a curve is:
$L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx$
For example, to calculate the arc length of the curve $$y = x^2$$ from $$x = 0$$ to $$x = 2$$, we use the following integral:
$L = \int_{0}^{2} \sqrt{1 + (2x)^2} \, dx$
## Arc Length Calculator with Steps
### Step 1: Define the Curve
$y = f(x)$
### Step 2: Determine the Interval
Determine the interval over which you want to find the arc length. Let $$a$$ and $$b$$ be the x-values that define the interval:
$a \leq x \leq b$
### Step 3: Find the Derivative
Calculate the derivative of $$f(x)$$ with respect to $$x$$, denoted as $$f'(x)$$:
$f'(x) = \frac{dy}{dx}$
### Step 4: Setup the Arc Length Integral
Use the following formula to set up the arc length integral:
$L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx$
### Step 5: Evaluate the Integral
Evaluate the integral to find the arc length. You may use analytical methods or numerical methods if necessary:
$L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx = \text{Evaluate this integral}$
### Step 6: Simplify and Calculate
Simplify the result if necessary and calculate the numerical value:
$L = \text{Simplified result}$
### Step 7: Interpret the Result
The final value $$L$$ represents the length of the arc of the curve from $$x = a$$ to $$x = b$$.
## Example Calculation
Let's calculate the arc length of the curve $$y = x^2$$ from $$x = 0$$ to $$x = 2$$.
### Step 1: Define the Curve
$y = x^2$
### Step 2: Determine the Interval
$a = 0, \quad b = 2$
### Step 3: Find the Derivative
$f'(x) = 2x$
### Step 4: Setup the Arc Length Integral
$L = \int_{0}^{2} \sqrt{1 + (2x)^2} \, dx$
### Step 5: Evaluate the Integral
$L = \int_{0}^{2} \sqrt{1 + (2x)^2} \, dx = \text{Evaluate this integral (numerically or analytically)}$
### Step 6: Simplify and Calculate
$L = \text{Simplified result}$
### Step 7: Interpret the Result
The calculated value of $$L$$ is the arc length of the curve $$y = x^2$$ from $$x = 0$$ to $$x = 2$$.
## Calculating Arc Length with Radius
The formula for calculating the arc length of a curve with a known radius is:
$L = \theta \cdot r$
Where:
• - $$L$$ is the arc length.
• - $$\theta$$ is the central angle of the arc in radians.
• - $$r$$ is the radius of the circle.
### Example Calculation:
Suppose you have a circle with a radius of 5 units, and you want to find the length of an arc that spans an angle of $$\frac{\pi}{3}$$ radians. Using the formula:
$L = \theta \cdot r$
Substitute $$\theta = \frac{\pi}{3}$$ and $$r = 5$$ into the formula:
$L = \left(\frac{\pi}{3}\right) \cdot 5$
Now, calculate the arc length:
$L = \frac{5\pi}{3}$
So, the length of the arc is $$\frac{5\pi}{3}$$ units.
## Arc Length Calculator in Terms of π
In many mathematical problems and formulas, π (pi) is an essential constant. When using the arc length calculator integral, π often appears in the calculation, especially when dealing with circles or trigonometric functions. It's a fundamental mathematical constant that relates the circumference of a circle to its diameter.
The formula for calculating the arc length of a curve in terms of π is:
$L = 2\pi r \left(\frac{\theta}{360^\circ}\right)$
Where:
• $$L$$ is the arc length.
• - $$r$$ is the radius of the circle.
• - $$\theta$$ is the central angle of the arc in degrees.
### Example Calculation:
Suppose you have a circle with a radius of 7 units, and you want to find the length of an arc that spans an angle of $$\frac{3\pi}{4}$$ radians. First, convert the angle to degrees:
$\text{degrees} = \frac{\frac{3\pi}{4} \cdot 180}{\pi} = \frac{270}{4} = 67.5^\circ$
Now, use the formula for arc length:
$L = 2\pi \cdot 7 \cdot \left(\frac{67.5}{360}\right)$
Calculate the arc length:
$L = 2\pi \cdot 7 \cdot 0.1875 = 2.355\pi \text{ units}$
So, the length of the arc is $$2.355\pi$$ units.
## Arc Length Calculator 3D
So far, we've discussed arc length in two dimensions. However, in the real world, many curves and paths exist in three-dimensional space. Calculating arc length in 3D involves extending the concepts we've covered to include an additional dimension. This can be particularly useful in physics, engineering, and computer graphics.
The formula for calculating the arc length of a parametric curve in 3D space is:
$L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt$
Where:
• $$L$$ is the arc length.
• - $$a$$ and $$b$$ are the parameter values that define the interval of interest.
• - $$x(t)$$, $$y(t)$$, and $$z(t)$$ are the parametric equations that define the curve in 3D space.
### Example:
Suppose you have a parametric curve in 3D space defined by:
\begin{align*} x(t) &= t^2 \\ y(t) &= t \\ z(t) &= 2t^3 \end{align*}
And you want to find the arc length of the curve between $$t = 0$$ and $$t = 1$$. You would use the formula above to set up the integral and evaluate it to find the arc length.
## Applications of Arc Length Calculations
Arc length calculations have wide-ranging applications. They are used in physics to describe the trajectory of particles, in engineering to design curved structures, in art to create visually pleasing curves, and in computer graphics to render realistic 3D objects. Understanding how to calculate arc length is essential for professionals in these fields.
Here are the formulas to calculate the radius of an arc:
If you know the arc length (L) and the central angle (θ):
$R = \frac{L}{\theta}$
If you know the arc length (L) and the chord length (C):
$R = \frac{C^2}{2L} + \frac{L}{2C}$
If you know the central angle (θ) and the chord length (C):
$R = \frac{C}{2\sin\left(\frac{\theta}{2}\right)}$
## Conclusion
In this comprehensive exploration of the arc length calculator integral, we've uncovered the fundamental principles of finding arc lengths. We've discussed the arc length formula, the role of radius, the use of π, and applications in both 2D and 3D spaces. Whether you're a student studying calculus or a professional applying mathematics to real-world problems, understanding arc length calculation is a valuable skill that can unlock new possibilities in your work and studies. So, next time you encounter a curve, remember the power of the arc length calculator integral to measure its length accurately and efficiently.
What is the formula used by the Arc Length Calculator?
The formula used to calculate the arc length is: Arc Length = (Central Angle / 360 degrees) × (2 × π × Radius) Here, π (pi) is a mathematical constant approximately equal to 3.14159.
Can the calculator handle angles in both degrees and radians?
Yes, the Arc Length Calculator can handle angles given in either degrees or radians. It will convert the angle to the appropriate unit for the calculation.
What are some practical applications of arc length calculations?
Arc length calculations are commonly used in architecture, engineering, physics, and other fields where circular paths or curves are involved. They are essential for determining distances along curved tracks, the length of conveyor belts, or the design of circular structures.
Is the Arc Length Calculator accurate for any circle size?
Yes, the Arc Length Calculator is accurate for any circle size as long as the provided central angle and radius values are correct.
Can I use the calculator for irregular shapes or curves that are not perfect circles?
The Arc Length Calculator is specifically designed for circles. For irregular shapes or curves, different methods may be required to calculate arc lengths.
What is an Arc Length Calculator?
An online application called an arc length calculator determines the length of an arc on a circle depending on the circle's radius and central angle. Finding arc lengths, which are sections of a circle's circumference, is made easier.
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Digital SAT Math Practice Questions – Medium : Linear inequalities in one or two variables
SAT MAth Practice questions – all topics
• Algebra Weightage: 35% Questions: 13-15
• Linear equations in one variable
• Linear equations in two variables
• Linear functions
• Systems of two linear equations in two variables
• Linear inequalities in one or two variables
SAT MAth and English – full syllabus practice tests
[calc] Question Medium
One of the two linear equations in a system is $$3 x+4 y=8$$. The system has exactly one solution.
Which of the following could be the other equation in the system?
A) $$\frac{3}{2} x+2 y=4$$
B) $$3 x+4 y=4$$
C) $$4 x+3 y=8$$
D) $$6 x+8 y=16$$
Ans: C
To determine which equation would allow the system to have exactly one solution, the slopes of the two lines must be different. This ensures that the lines intersect at exactly one point.
The given equation is:
$3x + 4y = 8$
First, we rewrite this equation in slope-intercept form ($$y = mx + b$$) to find its slope:
$3x + 4y = 8$
$4y = -3x + 8$
$y = -\frac{3}{4}x + 2$
The slope of the given equation is $$-\frac{3}{4}$$.
Now, let’s check the slopes of the other options:
Option A: $$\frac{3}{2}x + 2y = 4$$
$\frac{3}{2}x + 2y = 4$
$2y = -\frac{3}{2}x + 4$
$y = -\frac{3}{4}x + 2$
The slope here is also $$-\frac{3}{4}$$, the same as the given equation. This means the lines are parallel and would either be identical (infinitely many solutions) or parallel (no solutions), not intersecting at one point.
Option B: $$3x + 4y = 4$$
This is the same equation as the given one but with a different constant term, which makes it a parallel line. Therefore, the lines are parallel and will not intersect at one point.
Option C: $$4x + 3y = 8$$
$4x + 3y = 8$
$3y = -4x + 8$
$y = -\frac{4}{3}x + \frac{8}{3}$
The slope here is $$-\frac{4}{3}$$, which is different from $$-\frac{3}{4}$$. This means the lines have different slopes and will intersect at exactly one point.
Option D: $$6x + 8y = 16$$
This is a multiple of the given equation:
$6x + 8y = 16$
$\frac{6x + 8y}{2} = \frac{16}{2}$
$3x + 4y = 8$
This is the same equation as the given one, just scaled by a factor of 2. Therefore, they are identical and would either have infinitely many solutions or none if they were parallel lines, not intersecting at one point.
$\boxed{C}$
[Calc] Questions medium
$h(x)=3 x+3$
Which inequality represents all values of $$x$$ for which the graph of $$y=h(x)$$ in the $$x y$$-plane is above the $$x$$ axis?
A. $$x<3$$
B. $$x<-1$$
C. $$x>-1$$
D. $$x>3$$
Ans:C
The function $$h(x) = 3x + 3$$ represents a linear function with a slope of $$3$$ and a $$y$$-intercept of $$3$$. The graph of $$h(x)$$ will be above the $$x$$-axis for all values of $$x$$ greater than the $$x$$-intercept, which can be found by setting $$h(x) = 0$$:
$3x + 3 = 0$
$3x = -3$
$x = -1$
Therefore, all values of $$x$$ greater than $$-1$$ will make the graph of $$h(x)$$ above the $$x$$-axis. So, the correct inequality is $$x > -1$$.
[No calc] Question medium
$$y\leq =2x+3$$
$$y\geq 0.5x-6$$
In which graph does the shaded region represent all solutions to the given system of inequalities?
Ans: B
Given inequalities:
1. $$y \leq 2 x+3$$
2. $$y \geq 0.5 x-6$$
1. $$y \leq 2 x+3$$ :
• The boundary line is $$y=2 x+3$$.
• Since it is $$y \leq$$, the region below this line will be shaded.
2. $$y \geq 0.5 x-6$$ :
• The boundary line is $$y=0.5 x-6$$.
• Since it is $$y \geq$$, the region above this line will be shaded.
The correct solution will be the intersection of these two shaded regions.
Graph B:
• The region below the line $$y=2 x+3$$ is shaded. (Correct)
• The region above the line $$y=0.5 x-6$$ is shaded. (Correct)
• The intersection of these two regions is correctly shaded. (Correct)
[Calc] Question Medium
Sanjay works as a teacher’s assistant for $$\ 20$$ per hour and tutors privately for $$\ 25$$ per hour. Last week, he made at least $$\ 100$$ working $$x$$ hours as a teacher’s assistant and $$y$$ hours as a private tutor. Which of the following inequalities models this situation?
A) $$4 x+5 y \geq 25$$
B) $$4 x+5 y \geq 20$$
C) $$5 x+4 y \geq 25$$
D) $$5 x+4 y \geq 20$$
Ans:B
Sanjay earns $$\20$$ per hour as a teacher’s assistant and $$\25$$ per hour as a private tutor. He worked $$x$$ hours as a teacher’s assistant and $$y$$ hours as a private tutor and made at least $$\100$$.
To model this situation, we set up the following inequality:
total earnings from working $$x$$ hours as a teacher’s assistant:
$20x$
total earnings from working $$y$$ hours as a private tutor:
$25y$
3. The total earnings must be at least $$\100$$:
$20x + 25y \geq 100$
Simplify this inequality by dividing everything by 5:
$4x + 5y \geq 20$
[Calc] Question medium
According to a 2008 study, there were five known subspecies of tigers, including the Amur and Bengal, living in the wild. Scientists estimated that there were a total of 4,000 tigers in the wild. Of these, 450 were Amur tigers and x were Bengal tigers. Which inequality represents all possible numbers of tigers in the wild in 2008 that belonged to the Bengal tiger subspecies?
A. 5𝑥 ≥ 4000
B. $$\frac{x}{5}\geq 4000$$
C. 1 ≤ 𝑥 ≤ 3547
D. 3547 < 𝑥 ≤ 4000
Ans: C
To solve this problem, we need to find the range of possible values for the number of Bengal tigers (x) based on the given information.
• There were a total of 4,000 tigers in the wild.
• Out of these 4,000 tigers, 450 were Amur tigers.
• There were 5 known subspecies of tigers in the wild.
Since we know the total number of tigers and the number of Amur tigers, we can find the number of tigers belonging to the remaining 4 subspecies (including the Bengal tigers) by subtracting the number of Amur tigers from the total.
Number of tigers belonging to the remaining 4 subspecies = 4,000 – 450 = 3550
Since the Bengal tigers are one of the 4 remaining subspecies, the number of Bengal tigers (x) must be less than or equal to 3,550.
Also, the number of Bengal tigers cannot be negative or zero, as it is stated that they were one of the known subspecies living in the wild.
Therefore, the inequality that represents all possible numbers of tigers in the wild in 2008 that belonged to the Bengal tiger subspecies is:
$1 ≤ x ≤ 3550$
Among the given options, the correct answer is $C. 1 ≤ x ≤ 3547$.
[Calc] Question Medium
The function f is defined by f(x) =mx +b, where m and b are constants such that m ≥ 1 and -7 ≤b ≤7 Which of the following could be the graph of y =f(x) ?
Ans: C
[Calc] Question Medium
Teachers and students will participate in a tutoring seminar, and each teacher will lead a group of no more than 8 students. The room for the seminar can hold a maximum of 58 teachers and students. If t represents the number of teachers ands represents the number of students, which system of inequalities describes the possible numbers of teachers and students who can participate in the seminar ?
A) t + s ≥ 58
t≤ 8s
B) t + s ≤ 58
t ≥ 8s
C) t+ s ≥ 58
$$t ≥ \frac{s}{8}$$
D) t+s ≤ 58
$$t ≥ \frac{s}{8}$$
D) t+s ≤ 58
$$t ≥ \frac{s}{8}$$
Inequality 1: The total number of teachers and students cannot exceed 58. $t + s ≤ 58$
Inequality 2: The number of students for each teacher cannot exceed 8. $s ≤ 8t$
Combining the two inequalities, we get the system of inequalities: $t + s ≤ 58 s ≤ 8t$
Now, let’s check each option:
Option A: $t + s ≥ 58, t ≤ 8s$ This option is incorrect because the first inequality contradicts the given condition that the maximum number of teachers and students is 58.
Option B: $t + s ≤ 58, t ≥ 8s$ This option is incorrect because the second inequality implies that each teacher must have at least 8 students, which contradicts the given condition that each teacher will lead a group of no more than 8 students.
Option C: $t + s ≥ 58, t ≥ s/8$ This option is incorrect because the first inequality contradicts the given condition that the maximum number of teachers and students is 58.
Option D: $t + s ≤ 58, t ≥ s/8$ This option is correct because it represents the system of inequalities that satisfies both conditions:
1. The total number of teachers and students cannot exceed $58 (t + s ≤ 58)$.
2. The number of students for each teacher cannot exceed 8 $(t ≥ s/8,$ which can be rearranged as $s ≤ 8t)$.
Therefore, the correct answer is D)
[Calc] Question Medium
The shaded region shown represents the solutions to which inequality?
A. y≥-5x+3
B. y2-3x+5
C. y≤3x+5
D. y≤5x+3
Ans: D
From the graph value of y intercept is 3 so Option A and D are possible chances.
We can observe that the shaded region covers the area below the line passing through the points (0, 3) and (1, 8). The slope of this line is (8 – 3) / (1 – 0) = 5, which represents the coefficient of x in the inequality.
The y-intercept of the line is 3, which corresponds to the constant term in the inequality.
Therefore, the inequality y ≤ 5x + 3 accurately describes the shaded region, where all the points (x, y) below or on the line satisfy the inequality.
[Calc] Question medium
$$y<\frac{1}{2}x+4$$
$$y>-2x+4$$
Which ordered pair (x, y) is a solution to the given system of inequalities in the xy-plane?
A) (0,2)
B) (1,0)
C) (1,5)
D) (2,4)
D) (2,4)
Given the system of inequalities:
\begin{align*} & y < \frac{1}{2}x + 4 \\ & y > -2x + 4 \end{align*}
We need to find which ordered pair $$(x, y)$$ is a solution to this system.
Let’s analyze each inequality:
$$y < \frac{1}{2}x + 4$$ represents a region below the line $$y = \frac{1}{2}x + 4$$.
$$y > -2x + 4$$ represents a region above the line $$y = -2x + 4$$.
We need to find the intersection of these two regions.
test each ordered pair to see which one satisfies both inequalities:
• A) $$(0,2)$$:
$$2 < \frac{1}{2}(0) + 4$$ is true (below the line $$y = \frac{1}{2}x + 4$$.
$$2 > -2(0) + 4$$ is true (above the line $$y = -2x + 4$$).
Both inequalities are satisfied, so option A is a solution.
• B) $$(1,0)$$:
$$0 < \frac{1}{2}(1) + 4$$ is false (above the line $$y = \frac{1}{2}x + 4$$.
$$0 > -2(1) + 4$$ is true (above the line $$y = -2x + 4$$).
Not a solution.
• C) $$(1,5)$$:
$$5 < \frac{1}{2}(1) + 4$$ is false (above the line $$y = \frac{1}{2}x + 4$$.
$$5 > -2(1) + 4$$ is false (below the line $$y = -2x + 4$$).
Not a solution.
• D) $$(2,4)$$:
$$4 < \frac{1}{2}(2) + 4$$ is true (below the line $$y = \frac{1}{2}x + 4$$.
$$4 > -2(2) + 4$$ is true (above the line $$y = -2x + 4$$.
Both inequalities are satisfied, so option D is a solution.
Therefore, the solution is $$(2,4)$$.
[Calc] Question medium
In a forest, white pine trees between 15 and 45 years old grew 36 to 48 inches in height each year. A 15- year-old white pine tree growing in the forest was 240 inches tall. Which of the following inequalities gives all possible values for the tree’s height h, in inches, at the end of its 45th year?
A)h ≤ 540
B)h ≤ 2,160
C)240 ≤ h ≤ 1,080
D)1,320 ≤ h ≤ 1,680
D)1,320 ≤ h ≤ 1,680
To determine the possible values for the height $$h$$ of the white pine tree at the end of its 45th year, we need to consider its growth rate and current height.
Given:
The tree is currently 15 years old and 240 inches tall.
The growth rate is between 36 and 48 inches per year.
We want to find the height of the tree at the end of its 45th year (i.e., after 30 more years of growth).
Calculation:
1. Minimum Growth:
Growth per year: 36 inches
Number of years: 30
Total growth over 30 years: $$36 \, \text{inches/year} \times 30 \, \text{years} = 1080 \, \text{inches}$$
2. Maximum Growth:
Growth per year: 48 inches
Number of years: 30
Total growth over 30 years: $$48 \, \text{inches/year} \times 30 \, \text{years} = 1440 \, \text{inches}$$
3. Total Height Calculation:
Initial height at 15 years: 240 inches
Minimum possible height at 45 years: $$240 \, \text{inches} + 1080 \, \text{inches} = 1320 \, \text{inches}$$
Maximum possible height at 45 years: $$240 \, \text{inches} + 1440 \, \text{inches} = 1680 \, \text{inches}$$
The possible values for the tree’s height $$h$$ at the end of its 45th year lie between the calculated minimum and maximum heights.
Thus, the correct inequality is:D) $$1320 \leq h \leq 1680$$
[Calc] Question medium
The table gives the average speed s, in miles per hour (mph), of each lap around the track for one racing team. For how many laps was the average speed greater than or equal to 150 mph?
144
To find out how many laps had an average speed greater than or equal to $$150 \, \text{mph}$$, we need to sum up the number of laps for the speed intervals from $$150 \, \text{mph}$$ to $$165 \, \text{mph}$$.
From the table, we see that for the speed intervals $$150 \leq s < 155$$ , $$155 \leq s < 160$$ and $$160 \leq s < 165$$there are 57 and 52 laps and 35 laps respectively.
$57 + 52+35 = 144$
So, for $$150 \, \text{mph} \leq s < 165 \, \text{mph}$$, there were $$144$$ laps with an average speed greater than or equal to $$150 \, \text{mph}$$.
[No- Calc] Question Medium
$y<x-4$
Which of the following ordered pairs $$(x, y)$$ satisfies the inequality above?
A) $$(0,3)$$
B) $$(3,0)$$
C) $$(0,6)$$
D) $$(6,0)$$
Ans:D
To determine which ordered pair $$(x, y)$$ satisfies the inequality $$y < x – 4$$, we can simply substitute the $$x$$ and $$y$$ values from each option into the inequality and see which one holds true.
A) $$(0, 3)$$
$y = 3$
$x – 4 = 0 – 4 = -4$
$3 < -4$
This is false.
B) $$(3, 0)$$
$y = 0$
$x – 4 = 3 – 4 = -1$
$0 < -1$
This is false.
C) $$(0, 6)$$
$y = 6$
$x – 4 = 0 – 4 = -4$
$6 < -4$
This is false.
D) $$(6, 0)$$
$y = 0$
$x – 4 = 6 – 4 = 2$
$0 < 2$
This is true.
So, the ordered pair that satisfies the inequality is:
D) $$(6, 0)$$
[Calc] Question Medium
\begin{aligned} & y<-3 x+1 \\ & y<-\frac{1}{2} x+1 \end{aligned}
Which ordered pair $(x, y)$ is a solution to the given system of inequalities in the $x y$-plane?
A) $(-2,3)$
B) $(1,2)$
C) $(0,2)$
D) $(-1,1)$
D
[Calc] Question Medium
The coordinates of points $A, B$, and $C$ are shown in the $x y$-plane above. For which of the following inequalities will each of the points $A, B$, and $C$ be contained in the solution region?
A) $y>-x-2$
B) $y \geq-x$
C) $y<x+3$
D) $x<3$
A
[Calc] Question Medium
The function $f$ is defined for all real numbers, and the graph of $y=f(x)$ in the $x y$-plane is a line with a negative slope. Which of the following must be true?
I. If $a<b$, then $f(a)>f(b)$.
II. If $a<0$, then $f(a)>0$.
III. If $a>0$, then $f(a)<0$.
A) I only
B) II only
C) I and III only
D) II and III only
A
Question
A clothing store is having a sale on shirts and pants. During the sale, the cost of each shirt is $\$ 15$and the cost of each pair of pants is$\$25$. Geoff can spend at most $\$ 120$at the store. If Geoff buys$s$shirts and$p$pairs of pants, which of the following must be true? A.$15 s+25 p \leq 120$B.$15 s+25 p \geq 120$C.$25 s+15 p \leq 120$D.$25 s+15 p \geq 120$▶️Answer/Explanation Ans:A Question A certain elephant weighs 200 pounds at birth and gains more than 2 but less than 3 pounds per day during its first year. Which of the following inequalities represents all possible weights$w$, in pounds, for the elephant 365 days after its birth? A.$400<w<600$B.$565<w<930$C.$730<w<1,095$D.$930<w<1,295$▶️Answer/Explanation Ans: D Questions During an ice age, the average annual global temperature was at least 4 degrees Celsius lower than the modern average. If the average annual temperature of an ice age is$y$degrees Celsius and the modern average annual temperature is$x$degrees Celsius, which of the following must be true? A.$y=x-4$B.$y \leq x+4$C.$y \geq x-4$D.$y \leq x-4$▶️Answer/Explanation Ans: D Question Emma mows grass at a constant rate of 1.5 acres per hour. She mowed 2 acres before lunch and plans to spend$t$hours mowing after lunch. If Emma wants to mow at least 8 acres of grass today, which of the following inequalities best represents this situation? A.$1.5 t \geq 8$B.$1.5 t-2 \geq 8$C.$1.5 t+2 \geq 8$D.$2 t+1.5 t \geq 8$▶️Answer/Explanation Ans: C Questions A bag containing 10,000 beads of assorted colors is purchased from a craft store. To estimate the percent of red beads in the bag, a sample of beads is selected at random. The percent of red beads in the bag was estimated to be$15 \%$, with an associated margin of error of$2 \%$. If$r$is the actual number of red beads in the bag, which of the following is most plausible? A.$r>1,700$B.$1,300<r<1,700$C.$200<r<1,500$D.$r<1,300$▶️Answer/Explanation Ans: B Questions Terrence’s car contains 8 gallons of fuel. He plans to drive the car$m$miles using the fuel currently in the car. If the car can drive 20 miles per gallon of fuel, which inequality gives the possible values of$m$? A.$m \leq(8)(20)$B.$m \geq(8)(20)$C.$8 \leq 20 m$D.$8 \geq 20 m\$
Ans: A
Questions
During mineral formation, the same chemical compound can become different minerals depending on the temperature and pressure at the time of formation. A phase diagram is a graph that shows the conditions that are needed to form each mineral. The graph above is
a portion of the phase diagram for aluminosilicates, with the temperature $$T$$, in degrees Celsius (°C), on the horizontal axis, and the pressure $$P$$, in gigapascals (GPa), on the vertical axis.
Which of the following systems of inequalities best describes the region where sillimanite can form?
1. $$P \geq 0.0021T — 0.67$$
$$P \geq 0.0013T — 0.25$$
2. $$P \leq 0.0021T — 0.67$$
$$P \geq —0.0015T+ 1.13$$
3. $$P \leq 0.0013T— 0.25$$
$$P \geq —0.0015T+ 1.13$$
4. $$P \leq 0.0013T— 0.25$$
$$P \leq —0.0015T+ 1.13$$
Ans: B
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Question Video: Comparing Fractions with Negative Signs | Nagwa Question Video: Comparing Fractions with Negative Signs | Nagwa
Question Video: Comparing Fractions with Negative Signs Mathematics • First Year of Preparatory School
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Which of the following is true? [A] −3/5 = −7/9 [B] −3/5 > −7/9 [C] −3/5 < −7/9.
02:43
Video Transcript
Which of the following is true? Negative three-fifths is equal to negative seven-ninths. Negative three-fifths is greater than negative seven-ninths. Or, negative three-fifths is less than negative seven-ninths.
Here, we have two fractions we’re trying to compare. However, these fractions do not have a common denominator. Before we compare them, we’ll have to find a common denominator. I know that nine times five equals 45. And that means 45 could be used as a common denominator. To go from five to 45, we multiply by nine. And if we multiply by nine in the denominator, we have to multiply by nine in the numerator. Negative three times nine equals negative 27.
We’ll follow this same process for negative seven-ninths. Nine times five equals 45. And if we multiply by five in the denominator, we’ll need to multiply by five in the numerator. Negative seven times five equals negative 35. Now that we have a common denominator, we realize we’re comparing negative 27 over 45 and negative 35 over 45. Comparing negative numbers can sometimes be confusing. An easy way to make sure we’re comparing them correctly is to plot them on a number line.
We have zero and negative one. In our case, we could write negative one as negative 45 over 45. That’s the same thing as negative one. Negative 35 over 45 is closer to negative one than it is to zero. And negative 27 over 45 is about halfway between zero and negative one. We also want to remember the equivalent, negative seven-ninths and negative three-fifths. When we’re working with negative numbers, the value closest to zero is the largest. Negative 27 over 45, negative three-fifths, is closer to zero. And that means it’s larger than negative seven-ninths.
Negative three-fifths is larger than negative seven-ninths.
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# Difference between revisions of "2001 AMC 10 Problems/Problem 20"
## Problem
A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$. What is the length of each side of the octagon?
$\textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} {2000(\sqrt{2}-1)} \qquad \textbf{(C)} 2000(2-\sqrt{2}) \qquad \textbf{(D)} 1000 \qquad \textbf{(E)} 1000\sqrt{2}$
## Solution
$[asy] draw((0,0)--(0,10)--(10,10)--(10,0)--cycle); draw((0,7)--(3,10)); draw((7,10)--(10,7)); draw((10,3)--(7,0)); draw((3,0)--(0,3)); label("x",(0,1),W); label("x\sqrt{2}",(1.5,1.5),NE); label("2000-2x",(5,0),S);[/asy]$
$2000 - 2x = x\sqrt2$
$2000 = x(2 + \sqrt2)$
$x = \frac {2000}{2 + \sqrt2} = \frac {2000(2 - \sqrt2)}{2} = 1000(2 - \sqrt2)$
$x\sqrt2 = 1000(2\sqrt {2} - 2) = \boxed{\textbf{(B)}\ 2000(\sqrt2-1)}$.
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• Level: GCSE
• Subject: Maths
• Word count: 1675
Extracts from this document...
Introduction
Identical good tomatoes are placed in a box (see example).
Each tomato is a sphere. Each tomato just touches all the other tomatoes next to it as shown in the diagram. Tomato five goes bad. This is counted as the first hour. One hour later, all the tomatoes it touches go bad (now tomatoes 5,1,6 and 9 are bad). This continues every hour untill all the tomatoes in the box are bad.
I aim to investigate how tomatoes go bad in the above tray, and in trays of different sizes.
## How do tomatoes go bad in trays?
In this investigation, I aim to find a formula for calculating the total time required for all tomatoes to go bad in a rectangular tray of any rectangular size, and with any bad tomato starting position. To achieve this, I will first need to explain with the use of diagrams, how I arrived at a formula for the total time required for the tray to go bad. This will be done in three stages, followed by a worked example, which will answer both part 1 of the investigation, and part 2. Part 3 is an extension of the investigation concerned with the
Middle
Therefore the total time required to reach the corner is
If this were the most remote corner from the starting position, it would represent the total time needed for the whole tray to go bad.
Stage three of the analysis
Now we can calculate the time required for a rectangular tray of size M x N to go bad. A shaded cell represents the initial bad tomato.
Applying what we have learned in the previous stage of the analysis, we can firstly calculate the time required for the tomatoes in all four corners going bad, and secondly picking the longest time from these four results. The longest time, therefore will be the total time required for the whole tray to go bad. The formulas for calculating the time the bad tomatoes reach the corners are:
Corner T1 = (Q-1) + (P-1)
Corner T2 = (M-Q) + (P-1)
Corner T3 = (N-P) + (M-Q)
Corner T4 = (Q-1) + (N-P)
So the time needed for the whole tray to go rotten is
Worked example
This worked example will answer part 1 of the given investigation. The given tray size 4 by 4 can be seen in the following diagram, where the differently shaded cells are the three distinctively different starting positions.
Conclusion
T7=6
Tomato 11 is the initial bad tomato:
T11=6
Tomato 12 is the initial bad tomato:
T12=5
Tomato 13 is the initial bad tomato:
T13=4
So the average time for the tray to go bad is:
Taverage = [(8+7+6+6+5+4)/6] hours=6 hours
This can be seen easier on the following histogram:
Applying my formula for calculating the approximate average time for the tray to go bad:
Trough average = ¾(M+N) = ¾(5+5) = 7.5 hours
Again the result is that the approximate average is close to the true average different by 20%.
I predict that as the size of the tray increases, the true average time required for the tray to go bad and the approximate average time given by my formula will become closer.
Conclusion for whole investigation
I have develpoed a general formula for calculating the time for a tray of any size to become rotten, starting from one tomato. I applied the formula to the practical case of square size 4 by 4. The formula does not cover the case of several tomatoes going bad at the beginning in different places of the tray. I can only say that the tray will become rotten faster in this case. I have also devised a formula which calculates the approximate average time required for a tray of any size to go bad and tested it on trays 4 by 4 and 5 by 5 which show that it is accurate within 20%.
This student written piece of work is one of many that can be found in our GCSE Bad Tomatoes section.
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# Related GCSE Bad Tomatoes essays
equations that would cancel each other out so I was left with what A represented: 2. 4A + 2B + C = 6 - 1. A + B + C = 3 5. 3A + B = 15 4. 16A + 4B + C = 15 - 3.
2. ## GCSE Maths Bad Tomato Investigation
will equal the time for the whole tray to go bad. The reason for this expression working is using the XY rule, when a tomato in the corner goes bad, the one in the opposite corner will be the furthest one away.
4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 5 6 7
2. ## GCSE Mathematics - Bad tomatoes
I have also found out that if tomato no.6 goes bad first it would take the same amount of time to make the whole tray bad if tomato nos.7, 10 and 11 were the 1st tomatoes to go bad first.
1. ## GCSE Maths Bad Tomato Investigation
2L but there is 2 taken away from this to account for the original bad tomato and the corner tomato which the two sides share. This formula will work for any sized square were the initial tomato is on an edge and next to a corner.
2. ## In this project I am going to examine the time taken for a whole ...
The table just above this paragraph shows the total number of bad tomatoes. The columns towards the right hand side determine what the nth term will involve. If there are two differences that means the nth term will involve a 2.
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# Area of the Rectangle
The shaded rectangle is tangent to the circle at the two points indicated, and its bottom right corner lies on the circumference of that circle. What is the area of the rectangle?
#### Click to reveal solution
72
Like with many geometry problems, it is helpful to draw a few auxiliary lines:
• At the tangent point on the left, we draw a horizontal line across, parallel to the top side of the rectangle.
• At the tangent point on the top, we draw a vertical line down, parallel to the left side of the rectangle.
• From the center of the circle, draw a radius to the bottom right corner of the rectangle.
The two tangent lines and the two radii drawn from the tangent points to the center of the circle form a square. We know this is a square because:
• Its angles are right angles (the top left corner is a right angle because it is part of the rectangle, the bottom left and top right corners are right angles because radii always form right angles to the tangent lines at the tangent point), so it must be a rectangle
• The bottom and right sides are equal because they are both radii, and a rectangle with adjacent sides equal is a square
This tells us the radius of the circle is 5.
Now we know two of the three sides of the right triangle we have drawn in the bottom right. Using the Pythagorean theorem, we know the remaining side must be 3.
Therefore the sides of the rectangle are 5 + 4 = 9 and 5 + 3 = 8, which gives us an area of 72.
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# Secant of an Angle – Formulas and Examples
The secant of an angle can be calculated by relating the sides of a right triangle. The secant is defined as the reciprocal function of the cosine, so it is equal to the length of the hypotenuse over the length of the adjacent side. The secant of the most important angles are obtained by using the proportions of the known special triangles.
Here, we will learn about secant using diagrams. We will derive a formula and apply it to solve some practice problems.
##### TRIGONOMETRY
Relevant for
Learning about the secant of an angle with examples.
See examples
##### TRIGONOMETRY
Relevant for
Learning about the secant of an angle with examples.
See examples
## Definition of secant of an angle
The secant of an angle in a right triangle is defined as the length of the hypotenuse divided by the length of the side adjacent to the angle.
Additionally, we can also define the secant of an angle as the reciprocal of the cosine. This is because the cosine is defined as the adjacent side on the hypotenuse, so by taking its reciprocal, we obtain the secant.
$latex \sec (\theta)=\frac{1}{\cos}=\frac{H}{A}$
where H is the hypotenuse and A is the adjacent side.
## Formula for the secant in right triangles
We are going to use the following right triangle ABC that has a right angle at C to find the ratio of the secant of an angle.
We use lowercase letters to represent the lengths of the sides and we use capital letters to represent the angles of the triangle. For example, the letter a represents the side that is opposite angle A, the letter b represents the side that is opposite angle B, and the letter c represents the side that is opposite angle C.
The secant of an angle in a triangle rectangle is equal to the hypotenuse divided by the adjacent side:
In the triangle above, we have $latex \sec(A)=\frac{c}{b}$ and also $latex \sec(B)=\frac{c}{a}$.
## Secant for common special angles
We can use the proportions of the sides of special triangles to find the values of the secants of important angles. For example, we can consider the right isosceles triangle, which has angles 45°-45°-90°.
We find its proportions using the Pythagorean theorem: $latex {{c}^2}={{a}^2}+{{b}^2}$, but in this case, $latex a=b$, so we have $latex {{c}^2}=2{{a}^2}$. We conclude that $latex c = a \sqrt{2}$. Therefore, the secant of 45° is equal to $latex \sqrt{2}$.
Additionally, we also use the right triangle with angles 30°-60°-90°. The ratios of the sides of this triangle are 1: $latex \sqrt{3}$: 2. Using these proportions, we have $latex \sec(30^{\circ}) = \frac{2}{\sqrt{3}}$, which is equivalent to $latex \frac{2 \sqrt{3}}{3}$. We also have $latex \sec(60^{\circ})= 2$.
Finally, we can consider the angles 0 and 90°. When the angle is 0, the adjacent side like the hypotenuse is equal to 1 in the unit circle, so the secant of 0 is equal to 1. On the other hand, when the angle is 90°, the adjacent side is equal to 0 and the hypotenuse is equal to 1. However, we cannot divide by 0, so the secant of 90° is undefined.
## Secant of an angle – Examples with answers
The following examples are solved using the secant formula seen above. Each of the following examples refers to the right triangle shown above, so we use the same notation for the sides and angles.
### EXAMPLE 1
What is the value of b if we have $latex \sec(A)= 1.7$ and $latex c = 5$?
We use the right triangle above and look at the relationship $latex \sec(A) = \frac{c}{b}$. Therefore, we use this relation together with the given values to find the value of b:
$latex \sec(A)=\frac{c}{b}$
$latex 1.7=\frac{5}{b}$
$latex b=\frac{5}{1.7}$
$latex b=2.94$
The value of the b side is equal to 2.94.
### EXAMPLE 2
If we have $latex a=8$ and $latex \sec(B) = 1.44$, determine the value of c.
Using the right triangle above as a reference, we have $latex \sec(B) = \frac{c}{a}$. We use this relationship in conjunction with the given values to determine the value of c:
$latex \sec(B)=\frac{c}{a}$
$latex 1.44=\frac{c}{8}$
$latex c=1.44(8)$
$latex c=11.52$
The value of the hypotenuse is 11.52.
### EXAMPLE 3
If we have $latex c = 2$ and $latex b = \sqrt{3}$, what is the value of angle A?
Using the right triangle above as a reference, we can form the following relation $latex \sec(A) = \frac{c}{b}$. Therefore, using the given values, we have:
$latex \sec(A)=\frac{c}{b}$
$latex \sec(A)=\frac{2}{\sqrt{3}}$
$latex \sec(A)=\frac{2\sqrt{3}}{3}$
Using a calculator with the function $latex {{\sec}^{-1}}$ or using the table above, we know that we have:
$latex A=30$°
Angle A measures 30°.
## Secant of an angle – Practice problems
Use what you have learned about the secant of an angle to solve the following problems. These problems refer to the right triangle seen above, so they use the same notation for sides and angles.
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# Factors of 5
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Last updated date: 02nd Aug 2024
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Views today: 2.32k
## Factors of 5: An Introduction
A factor in mathematics is a divisor of a given number that divides it precisely with no remainder. To determine a number's components, we can adopt a number of methods, such as the division method and the multiplication approach. Only 5 and 1 are factors of 5. Both positive and negative factors are multiples of three. Both positive and negative integers are possible for the factors and pair factors of 5. The pair factor of three, for instance, is represented as (1, 5) or (-1, -5). The number 5 is the result of multiplying -1 and -5. Using the prime factorization approach and numerous solved cases, we will discover the factors of 5, pair factors of 5, and the prime factors of 5. You will learn about the three factors in this section, and for your benefit, we'll calculate the three factors' sums, identify their prime factors, and display the three factors' sums in pairs.
## What are the Factors of 5
The numbers that, when multiplied by 5, leave a remainder of zero are considered factors of a 5. Only factors 1 and 5 make up the number 5.
Keep in mind that ${-1 \times (-5)}=5$. As a product of any two negative numbers yields a positive number, (-1, -5) are also factors.
However, let's limit our discussion to positive integers.
## How to Calculate Factors of 5
Let's understand how to calculate the factors of three.
Step 1: Write down the number that needs to be factored in first. The number is 3.
Step 2: Find the two numbers whose product equals 5 in step two.
We get $1 \times 5=5$.
The factors are, therefore, 1 and 5.
## Factors of 5 Using Division Method
In the division method, the integer numbers are divided by the number 5 to determine the factors of 5. The number is a factor of 5 if it divides by 5 precisely. Divide the third number by one now, then do the same with the second and third numbers.
$5 \div 1=5$ (Factor is 1 and Remainder is 0)
$5 \div 5=1$ (Factor is 5 and Remainder is 0)
When the number 5 is divided by two, the result is one. Hence two is not a factor of three. As a result, 1 and 5 are the factors of 5.
Factors of 5
## Prime Factorization of 5
The prime factorization of 5 is expressing the number 5 as the product of its prime factors. To use the prime factorization method to determine the prime factors of 5, follow the steps below.
Take the pair factor of 5, for example (1, 5)
We cannot divide further since the number 5, in this case, is a prime number.
So, 5 is represented as $1 \times 5$.
Consequently, ($1 \times 5$) is the prime factorization of 5.
Note: If a pair factor contains a composite number, divide it into its prime factors and write the resulting numbers as the product of those prime factors.
## Factor Pairs of 5
Factor pairs of three consist of two factors combined, so their output equals three. The positive factor pairs of three are listed below.
$1 \times 5 = 5$
$5 \times 1 = 5$
Negative values are included in factors of 5, as we mentioned earlier. The list of positive factor pairs above can be changed to a list of negative factor pairs of three by simply adding a minus sign in front of each factor:
$-1 \times -5 = 5$
$-5 \times -1 = 5$
Thus, we have positive and negative factors pairs:
(1,5), (5,1), (-1,-5) and (-5,-1).
## Factor Tree
To determine a number's prime factors, a factor tree is constructed. It is constructed as a tree, with the supplied number's branches standing in for each of the number's factors. Let's learn more about the factor tree approach for determining a number's factors.
Factor Tree of 5
## Solved Examples
Example 1: To find out if 5 is a prime or a composite number, Bansi looks at the number.
Solution: Let's begin by compiling a list of all 5 factors.
Factors 1 and 5 are involved.
If a number solely consists of the number itself and one other element, such as 1. This quantity is prime.
A number is considered composite if it has more than two elements.
The number 5 is a prime number since it only has two elements.
Example 2: During a class test, Roger was asked to find the common factors between 5 and 12. Aid Rojer in figuring out the issue.
Solution: Prime factorization may be used to get the factors of 15:
We get:
12 can be factored into its primes as $2 \times 5 \times 3$
As $1 \times 5$ is the prime factorization of 5,
The common factor between 5 and 15 is 5.
## Practice Problems
1. Shri is calculating the total number of factors that equal 3. Can you locate it for him?
A. 2
B. 1
C. 13
D. 3
Ans: Option A
2. What is the greatest number that 3 and 5 have in common?
A. 5
B. 3
C. 2
D. 1
Ans: Option D
## Summary
Each factor can be combined with another to create the number 5, which is obtained by multiplying the two. In this instance, 5 is a prime number. When you reduce the factors of 5 to just the prime factors and express 5 as the product of those prime factors, you are said to have performed prime factorization. The number is a factor of 5 if it divides by 5 precisely. When the number 5 is divided by two, the result is one. Hence two is not a factor of three. As a result, 1 and 5 are the factors of 5. A number of factors, specifically its prime factorization, can be expressed as factor trees. There are factors for each branch of the tree.
## FAQs on Factors of 5
1. How many prime factors are there in 6?
Prime factorization of 5 is $1 \times 5$
Three have factors one and three. We can see that the number 5 only has two elements. As a result, there are only 2 numbers that completely divide 5 without producing a residue.
2. What is a factor in mathematics?
A number or algebraic expression that evenly divides another number or expression—i.e., leaves no remainder—is referred to as a factor in mathematics. As an illustration, 3 and 6 are factors of 12 because $12 \div 3= 4 and 12 \div 6=2$, respectively. 1, 2, 4, and 12 are the other components that make up 12. A positive integer or algebraic expression with two or fewer factors (i.e., itself and 1) greater than one is referred to as prime; one with more factors is referred to as composite.
3. What variables divide by three?
If a number's total digits equal a multiple of three or are divisible by three, it is said to be three-divisible. Take into account the multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. The sum of the digits is always divisible by 3. If a number can be divided by three, then its digit sum can also be.
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Lesson Video: Dividing Decimals by Powers of Ten | Nagwa Lesson Video: Dividing Decimals by Powers of Ten | Nagwa
# Lesson Video: Dividing Decimals by Powers of Ten Mathematics • 5th Grade
In this video, we will learn how to use patterns to investigate the effects of dividing numbers by increasing and decreasing powers of 10.
16:24
### Video Transcript
In this video, we’re going to learn how to use patterns to investigate the effects of dividing numbers by increasing and decreasing powers of 10, such as 10, 100, and 1000. Let’s remind ourselves how we multiply by various powers of 10. Let’s take the number 2.7. We know of course that 2.7 times one is equal to 2.7. This is sometimes called the multiplicative identity property. Then, we also know that 2.7 multiplied by 10 is equal to 27. What happens when we multiply by 10 is that we move every single digit in the number one place to the left. So the two that was in the ones column moves to the tens column. And the seven that was in the tens column moves to the ones column. That leaves us with two tens and seven ones, which is of course 27.
Similarly, if we were to multiply 2.7 by 100, we’d move every single digit in the number 2.7 two places to the left. And so the two moves from the ones column to the hundreds column. The seven moves from the tenths column to the tens column. And then of course we add a zero here as a placeholder. We have two hundreds, seven tens, and zero ones, giving us 270.
We continue using this pattern. This time, if we’re going to multiply 2.7 by 1000, we need to move all of the digits to the left three places. So the number two moves from the ones column to the thousands column. The seven moves from the tenths to the one hundreds. And then we add zeros as placeholders. So we now have two thousands, seven hundreds, zero tens, and zero ones. And that gives us 2700.
So let’s continue this pattern and see if we can work out how we divide by powers of 10. Now, of course, we know that multiplication and division are inverse operations to one another. That is, they are the opposite. Performing multiplication undoes a division and vice versa. And so we can say that dividing by 10 undoes the operation of multiplying by 10 such that 27 divided by 10 gives us 2.7. Similarly, dividing by 100 is the opposite to multiplying by 100. So 270 divided by 100 takes us back to 2.7, and 2700 divided by 1000 will also take us back to 2.7.
And so that should help us spot a pattern. We saw that when we multiplied by 10, we moved the digits to the left one space. And this means then that if I divide a number by 10, I’m going to move the digits to the right one space. And so our place value chart this time is going to look a little bit different. We’re going to be moving all the digits to the right. And so we need to add a few extra columns. I’ve added the hundredths, thousandths, and ten thousandths column.
Moving every single digit in our number to the right one space, and we see that the seven moves into the hundredths column. Then the two moves into the tenths column. And we’ll need a zero in front of the decimal point as a placeholder. We now have zero ones, two tenths, and seven hundredths, and so 2.7 divided by 10 is 0.27.
We’re going to now use the same sort of idea to divide 2.7 by 100. We saw that to multiply a number by 100, we moved every single digit to the left two spaces. This means given that division is the opposite operation to multiplication, we’re going to move every single digit to the right two places when we divide by 100. And so our seven moves once, twice into the thousandths column. Similarly, the two moves once, twice, and it ends up in the hundredths column. We’ll need a couple of zeros as placeholders, and we find that 2.7 divided by 100 is 0.027.
Let’s do this one more time to find the value of 2.7 divided by 1000. Remember, when we multiplied by 1000, we moved all of the digits to the left three spaces. And so we now need to do the opposite. We need to move all of the digits in the number 2.7 three places to the right. The seven moves once, twice, three times, and it ends up in the ten thousandths column. The two also moves once, twice, three times, and the two ends up in the thousandths column. We’ll add the zeros as placeholders. And so we see 2.7 divided by 1000 is 0.0027.
Now, one thing to be really aware of here is we’ve moved the digits, not the decimal point. If we consider the place value table, it doesn’t make a lot of sense to move the decimal point. If we move the decimal point to here, then we have a decimal point between the tenths column and the hundredths column, which we know doesn’t make sense at all. So we always want to consider multiplying and dividing by powers of 10 in terms of moving the digits.
Let’s recap what we’ve found so far. We’ll think about dividing by powers of 10 in terms of the place value of our digits. When we divide by 10, we move all of the digits one space to the right. When we divide by 100, we move all of the digits two spaces to the right. And when we divide by 1000, we move all of the digits three spaces to the right.
Now, we will a little bit later in this video look at the relationship between dividing by 10, 100, and 1000. But for now, if you’re struggling to remember how many places to move the digits, look at the number of zeros in the number you’re dividing by. So when we’re dividing by 10, we have one zero in that number and we move the digits one space to the right. 100 has two zeros and 1000 has three zeros. And the number of zeros tells us how many spaces we move our digits.
Let’s have a look at the application of these ideas.
Find the result of 234.5 divided by 10.
We have a process that helps us divide by powers of 10. We think about the division in terms of the place value of the numbers in the dividend. The dividend in this number of course is the number that we’re going to be dividing. It’s 234.5. And we know that to divide by 10, we move all of the digits in our dividend one space to the right. So our place value chart looks a little something like this. Our number has two hundreds, three tens, four ones, and five tenths.
We’re going to move every single one of these digits one space right. We’ll begin with the five. The five moves from the tenths column to the hundredths column. Then we take the four and we move it from the ones column to the tenths column. The three moves from the tens to the ones. And finally, the two moves from the hundreds to the tens. And now when we divide 234.5 by 10, we end up with two tens, three ones, four tenths, and five hundredths. And so we say that 234.5 divided by 10 is 23.45.
Let’s have a look at another example.
What is 668.7 divided by 1000?
We know that to divide a number by 1000, we take all of the digits in our dividend, that’s the number 668.7 here, and we move them three places to the right. When we pop the number 668.7 into a place value chart, it looks a little something like this. We have six hundreds, six tens, eight ones, and seven tenths. We’re going to move every single digit three places to the right.
So let’s begin with the seven. We move it once, twice, three times, and it goes all the way over into the ten thousandths column. So the result of 668.7 divided by 1000 will have seven ten thousandths. We then do the same with the number eight. One, two, three spaces to the right brings it to the thousandths column. We repeat this process with the remaining two digits. We’ve got the number six, which moves all the way over to the hundredths column, and we have another six. And when we move that three digits to the right, we end up in the tenths column.
Now, of course, it doesn’t make sense to just write .6687. And we know we have zero ones. So we add the zero sort of as a placeholder. And so we see that 668.7 divided by 1000 is 0.6687.
In our next example, we’ll look at the relationship between dividing by various powers of 10.
Complete the following table. And then we have a table with the number 231.72 and headings number divided by 10, number divided by 100, and number divided by 1000.
So let’s remind ourselves how we divide by each of these numbers. When we divide by 10, we move our digits to the right one space. When we divide by 100, we move them in the same direction, but this time two spaces. And when we divide by 1000, we move the digits to the right three spaces. So we’re going to pop the number 231.72 into a place value table. This number has two hundreds, three tens, one one, seven tenths, and two hundredths.
We’ll begin then by dividing it by 10. And when we do, we’re going to move every single digit one space to the right. So the two moves from the hundredths column to the thousandths column. Then we move the seven. It moves from the tenths column to the hundredths column. In a similar way, we move the one from the ones column to the tenths column. And we continue with the final two digits. And so we find that when we divide our number by 10, we end up with 23.172.
So we’re now going to divide it by 100. Now, when we divide by 100, we move every single digit two spaces to the right. The thing is when we divided by 10, we’d already moved the digits one space to the right. So we can, in fact, take every single digit in our new number and just move those one space to the right. This will result in an overall movement of two spaces.
So we take the digit two, and we move that from the thousandths column to the ten thousandths column. Then we move the seven from the hundredths column into the thousandths column. The one moves from the tenths column into the hundredths column. And then, once again, we repeat this process with the remaining two digits. So our number 231.72 divided by 100 is 2.3172.
Our last job is to divide the number by 1000. And we know that to do so, we move every digit in the original number to the right three spaces. But actually, we moved every digit right once when we divided by 10 and then once again when we divided by 100. So actually, we can take the digits in our newest number and just move those to the right ones.
So the two moves from the ten thousandths column into the hundred thousandths column. The seven moves from the thousandths column into the ten thousandths column. And then we just continue with the rest of our digits. We do need to add a placeholder in the ones column. And when we do, we find that our number divided by 1000 is 0.23172.
And so the numbers in our table are 23.172, 2.3172, and 0.23172.
Now, we’re actually able to find a relationship between dividing by 10, 100, and 1000. We saved ourselves a little bit of time when dividing by 100 by taking the digits that we had after dividing by 10 and just moving them to the right one space. Now, this makes a lot of sense because 100 is the same as 10 squared. So when we divide by 100, we can say that that is the same as dividing by 10 and then by dividing by 10 again. Similarly, when we divided by 1000, we took the digits that we’d already moved twice to the right and then just moved them once again. And this also makes a lot of sense because 10 cubed is the same as 1000. So dividing by 1000 is just like dividing by 10, then 10 again, and then 10 a third time.
And so we see that there are, in fact, two different ways that we can perform the same calculation. Dividing by 100 is the same as dividing by 10 once and then twice. And dividing by 1000 is the same as dividing by 10 three times in a row.
Let’s consider one final example.
Complete the figure shown.
Our figure contains three series of numbers. We have 348.9, which becomes 3.489 and then becomes 0.3489. We then have 8.5, which goes along in a similar pattern, and 21.5, and we’re given 0.215. But we’re not told what the final number in this list is. We are, however, told that the first thing that happens to each of our numbers is that they’re divided by 100. And this makes a lot of sense since when we divide by 100, we move all of the digits in our dividend two spaces to the right.
If we look at 348.9, for instance, the three has moved from the hundreds column to the ones column. The four has moved from the tens column to the tenths column, and so on. And so what has happened between column two and column three? Well, let’s take the three in the number 3.489. That has moved one space to the right from the ones column into the tenths column. Then if we look at the four, that’s moved one space to the right too, from the tenths column into the hundredths column.
A similar pattern happens in this number down here. The eight moves exactly one space to the right, as does the number five. And, of course, we know that when we divide by 10, we do actually move all the digits right one space. And so the operation that’s occurring between column two and column three is a division by 10.
Now, if we look at the bottom of our screen, we see that this is the same as dividing by 1000. And we know that one way to divide by 1000 is to move all of the digits in our dividend right three spaces. Alternatively, we could just take all of the digits in the number 0.215 and move them right one space, since dividing by 1000 is the same, as we see, as dividing by 100 and then dividing by 10.
When we move each digit to the right one place, the two moves from the tenths column into the hundredths column. The one and the five both move down with it. And we add a zero in the tenths column as a placeholder. So 0.215 divided by 10 is the same as 21.5 divided by 1000. It’s 0.0215.
We’ll now recap the key points from our lesson. In this lesson, we recalled that multiplication and division are inverse operations to one another. This helped us to find rules for dividing by various powers of 10. We said that to divide by 10 itself, we move all of the digits right exactly one space. To divide by 100, we move those very same digits right two spaces. But also we saw that we could consider this as dividing by 10 and then dividing by 10 again and that if we wanted to divide by 1000 of course, we move all of the digits to the right three spaces. And this is in fact the same as dividing by 10, dividing by 10, and then dividing by 10 a third time.
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# Integral of Tan x
Last Updated : 06 Mar, 2024
Integral of tan x is ln |sec x| + C. Integral of tan x refers to finding the integration of the trigonometric function tan x with respect to x which can be mathematically formulated as ∫tan x dx. The tangent function, tan x, is an integrable trigonometric function that is defined as the ratio of the sine and cosine functions.
This article discusses the formula for the integral of tan x along with derivation, definite integral of tan x, and integral of tan inverse x. We will also discuss some solved examples based on the integral of tan x along with Practice Questions and FAQs.
## What is Integral of Tan x?
Integration of tan x is ln |sec x| + C or -ln |cos x| + C. There can be both indefinite and definite integrals of tan x with respect to x. The Indefinite Integral is represented as ∫tan x dx and the definite integral of tan x is given as [Tex]\int_a^btan\ x\ dx [/Tex] Tan x is an integrable function. Tan is one of the six trigonometric functions in maths. The domain of tan x is all real numbers except odd multiples of π/2 and the range of tan x is all real numbers. Hence, tan x is continuous at all values of Real numbers except π/2. Hence, the trigonometric function tan x is integrable in its domain only which consists of all the Real Numbers except odd Multiples of π/2.
## Integral of Tan x Formula
Let’s have a look at the formula for the integral of tan x with respect to x is shown below.
∫ tan x dx = ln |sec x| + C
or
∫ tan x dx = -ln |cos x| + C
In order to understand the process of Integration of tan x dx we need to have some requisite knowledge which can be gained from the following article: Integration By Parts Method.
## How to do Integration of tan x dx?
Let’s have a look at the proof for the above formula for the indefinite integration of tan x. Here, the Integration by Substitution Method and Logarithmic Properties are used.
I = ∫ tan x dx
We know that tan X = sin X / cos X
Thus, ∫ tan x dx = ∫ (sin x /cos x) dx
I = ∫ (1/cos x) sin x dx
Let’s apply the substitution method of integration.
Let t = cos x
now differentiating above equation with respect to x.
⇒ dt/dx = – sin x
⇒ sin x dx = -dt
So, ∫ tan x dx = ∫(1 /cos x) sin x dx
= ∫ (1/t) (-dt) = – ∫ (1/t) dt
I = – ln |t| + C ∴ (C is added due to indefinite integral)
Substituting the value of t back in the equation.
I = -ln |cos x| + C {Using Logarithmic Properties}
I = ln |1/cos x| + CI = ln |sec x| + C
Therefore, ∫ tan x dx = -ln |cos x| + C = ln |sec x| + C
So, by using above steps, we have proved the formula for the Indefinite Integration of tan x with respect to x.
## Definite Integral of Tan x
We know that, by the fundamentals properties of definite integrals, we can calculate the definite integral of tan x within any two time intervals limits.
Definite Integral of f(x) = [Tex]\int_a^b f(x)\ dx~=~F(b) – F(a)[/Tex]
Definite Integral of f(x) within ‘a’ to ‘b’ interval = [Tex]\int_a^b tan\ x\ dx~=~|ln\ sec \ x|_a^b~=~|ln\ sec \ b|~-~|ln\ sec \ a| [/Tex]
Using the above property to calculate the definite integral of tan x. Let’s take an example of definite integral of tan x between the time interval Ï€/4 and Ï€/3.
we know that, [Tex]\int tan x dx = ln |sec x| + C[/Tex]
Then ,[Tex]\int_{\pi/4}^{\pi/3} tan\ x\ dx = ln|sec \ \pi/3| – ln|sec\\pi /4|[/Tex]
we know that,
• sec Ï€/ 3 = 2
• sec Ï€/ 4 = √(2)
so, [Tex]\int_{\pi /4}^{\pi / 3} tan\ x\ dx~=~ln |2| – ln|\sqrt{2|}[/Tex]
we know that, ln|a| – ln|b| = ln|a/b|
so, [Tex]\int_{\pi /4}^{\pi /3} tan\ x\ dx~=~ln |2/\sqrt{2}|~=~0.3465735902799727[/Tex]
## Integration of Tan Inverse x
The indefinite integration of tan inverse x with respect to x can be calculated using the integration by parts method. The formula for the integration of tan inverse x with respect to x is given as
∫ tan-1x dx = x tan-1x – (1/2) ln |1 + x2| + C
(Note: Derivation for the above formula can be seen further in the Solved Example 4.)
Also, Check
## Solved Examples on Integral of Tan x
Example 1: Calculate the integral of cot x with respect to x.
Solution:
∫ cot x dx = ∫ cos x / sin x dx
Substitute sin x = t.
Differentiating with respect to x.
Then cos x dx = dt.
Then the above integral becomes
= ∫ (1/t) dt
= ln |t| + C (Because ∫ 1/x dx = ln|x| + C)
Substitute back t = sin x back here,
= ln |sin x| + C
Thus, ∫ cot x dx = ln |sin x| + C.
Example 2: Calculate the integral of sec x tan x with respect to x.
Solution:
We can write sec x = 1/cos x and tan x = sin x/cos x.
We have, ∫sec x tan x dx = ∫(1/cos x)(sin x/cos x) dx
= ∫(sin x/cos2x) dx
Now, assume cos x = t.
Differentiating both sides
-sin x dx = dt
⇒ sin x dx = dt
Substituting back these values
We have, ∫sec x tan x dx = ∫(-1/t2) dt
= (1/t) + C
Substitute back t = sin x back here,
= 1/cos x + C
= sec x + C
Thus, ∫ sec x tan x dx = sec x + C
Example 3: Calculate the integral of tan2x with respect to x.
Solution:
Let us find the integral of (tan2 x) with respect to x.
= ∫ tan2 x dx
Using the identity sec2 X – tan2 X = 1,
∫ tan2 x dx = ∫ (sec2 x – 1) dx
= ∫ sec2 x dx – ∫ 1 dx
Using the standard integration formulas, ∫sec2 x dx = tan x + C and ∫ 1 dx = x + C,
we get, tan x – x + C
Hence, ∫ tan2 x dx = tan x – x + C.
Example 4: Calculate the integral of tan-1 x with respect to x.
Solution:
We know that,
Formula for integration by parts is ∫f(x)g(x)dx = f(x) ∫g(x)dx – ∫[d(f(x))/dx × ∫g(x) dx] dx.
∫tan-1x dx = ∫tan-1x.1 dx
= tan-1x ∫1dx – ∫[d(tan-1x)/dx × ∫1 dx] dx
= x tan-1x – ∫[1/(1 + x2) × x] dx
= x tan-1x – ∫x/(1 + x2) dx
[Multiplying and dividing by 2]
= x tan-1x – (1/2) ∫2x/(1 + x2) dx
{Using formula ∫f'(x)/f(x) dx = ln |f(x)| + C}
= x tan-1x – (1/2) ln |1 + x2| + C
Hence, the integral of tan inverse x is x tan-1x – (1/2) ln |1 + x2| + C.
Example 5: Calculate the integral of sec x with respect to x.
Solution:
Firstly, we multiply and divide the integrand with (sec x + tan x).
∫ sec x dx = ∫ sec x · (sec x + tan x) / (sec x + tan x) dx
= ∫ (sec2x + sec x tan x) / (sec x + tan x) dx
Now assume that sec x + tan x = t.
Differentiating with respect to x.
Then (sec x tan x + sec2x) dx = dt.
Substituting these values in the above integral,
∫ sec x dx = ∫ dt / t = ln |t| + C
Substituting t = sec x + tan x back here,
Hence, ∫ sec x dx = ln |sec x + tan x| + C.
## Practice Questions on Integral of Tan x
Q1. Find the integral of cosec x with respect to x.
Q2. Find the integral of tan(√x) with respect to x.
Q3. Find the definite integral of (tan2 x – 1) between the interval Ï€/3 to Ï€/2.
Q4. Calculate the integral of 1/(1 + tan x) with respect to x.
Q5. Calculate the integral of tan2 (2x – 7) with respect to x.
## FAQs on Integral of Tan x
### What is Integration of tan x?
The integral of tan x is : ln |sec x| + C or -ln |cos x| + C
### What is Integration? Name its types.
Integration is the inverse process of differentiation. We are given with the derivative of a function and we are asked to find the original function, such process is called as Integration. In integrals, we find the area enclosed between the curve and the axis.
There are two types of integrals in calculus:
• Definite Integral (Integral within limits)
• Indefinite Integral (Integral without limits)
### Why do we add an Constant C in Indefinite Integrals?
We know that the derivative of any constant is 0, so while integration when we are willing to get the original function, we are unable to obtain the constant term the original function might had. Hence, we add a constant C to represent the constant term of the original function, which could not be obtained through this anti-derivative process.
### What is the Integral of sec x tan x?
The integral of sec x tan x is sec x + C
### What is the Derivative of tan x?
The derivative of tan x is sec2x
### What is the Integral of arc tan x?
The integral of arc tan x is ∫ tan-1x dx = x tan-1x – (1/2) ln |1 + x2| + C
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# How do you find the integral for (x)/(sqrt(1+2x))dx?
Apr 1, 2015
I would probably use parts, but since this is asked in Integration by substitution, I'll do it that way.
$\int \frac{x}{\sqrt{1 + 2 x}} \mathrm{dx}$
Let $u = 1 + 2 x$. This makes $\mathrm{du} = 2 \mathrm{dx}$ and $\mathrm{dx} = \frac{1}{2} \mathrm{du}$.
The substitution is not complete until we also note that
$u = 1 + 2 x$ makes $x = \frac{u - 1}{2}$
$\int \frac{x}{\sqrt{1 + 2 x}} \mathrm{dx} = \int x \frac{1}{\sqrt{1 + 2 x}} \cdot \mathrm{dx}$
$= \int \frac{u - 1}{2} \cdot \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \mathrm{du} = \frac{1}{4} \int \left(u - 1\right) {u}^{- \frac{1}{2}} \mathrm{du}$
$= \frac{1}{4} \int \left({u}^{\frac{1}{2}} - {u}^{- \frac{1}{2}}\right) \mathrm{du}$
You can probably finish this yourself.
$= \frac{1}{4} \left[\frac{2}{3} {u}^{\frac{3}{2}} - \frac{2}{1} {u}^{\frac{1}{2}}\right] + C = \frac{1}{6} \left[{u}^{\frac{3}{2}} - 3 {u}^{\frac{1}{2}}\right] + C$
$\int \frac{x}{\sqrt{1 + 2 x}} \mathrm{dx} = \frac{1}{6} \left[{\sqrt{1 + 2 x}}^{3} - 3 \sqrt{1 + 2 x}\right] + C$
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# High School: Number and Quantity
### The Complex Number System HSN-CN.A.1
1. Know there is a complex number i such that i2 = -1, and every complex number as the form a + bi with a and b real.
It's a little-known fact that in Disney World, in the Journey into the Imagination Pavilion, lives a purple dragon named Figment. (No, Figment is not Barney. They're not even related. Figment is a dragon, not a dinosaur, and he doesn't have that annoying voice or theme song.) Figment is quite a rule breaker—he does things that others tell him he simply can't.
Before the ride was rehabbed, there was a wall toward the end of the ride. It pictured all sorts of things imaginary—pigs that flew and three headed cows and the expression "i2= -1."
What on Earth are we talking about? Well, what the Imagineers at Disney remembered from high school is that there is a field of numbers based on something imaginary. We call this field of numbers "complex numbers" (since "imaginary" sounds a tad too mythical) and its most basic unit is the number i. Yes, the number. Not the letter.
What's the big deal about i? Well, the big deal is that i = .
Yeah, we know. Square roots and negative numbers just don't go together. Well that was then, and this is now.
Once your students get past the idea that -1 can have a square root, they can have lots of fun with imaginary numbers. The complex number system is composed of numbers in the form "a + bi," where both a and b are real numbers. (That means we can have numbers like 2 + 5i or 7 – 12i.) Eventually, they can even do all sorts of operations with complex numbers.
We'll take it one step at a time, though.
#### Drills
1. What's the big deal about ? Why does is need to be imaginary?
Because a negative squared is positive and a positive squared is positive
Think about it. There's no "normal" way to find the square root of -1. Using -1 won't work (since -1 × -1 = 1) and using 1 won't work (since 1 × 1 = 1). The only way to make the problem work is to invent a number, i. While all the other choices are true, they don't explain the need for the number i.
2. If i = , what does equal?
2i
Our first venture into the world of imaginary numbers! We can rewrite as . The first radical simplifies to 2 and the second to i. Since they're products, we just multiply the two together. So the answer is 2i.
3. What does equal?
We can rewrite as , and then simplify it. The first radical simplifies to i, although "simplify" might not be the best word to use since we're making things imaginary and all. The second doesn't simplify. So the right answer is (C).
4. What does 4 + 2i equal?
4 + 2i
We should treat imaginary numbers as you would treat any other radical. (Imaginary friends get the same treatment as friends do. Why shouldn't imaginary numbers get the same treatment numbers do?) Unfortunately, they're imaginary, so we'll treat them like they're irrational and only combine like terms.
5. What does i2 equal?
-1
We said that , right? If all we do is square both sides of that equation, we'll have our answer. The square root on the cancels out when we square it, and we're left with i2 = -1. That's (B), and it's kind of the whole idea behind the complex number system.
6. What does 3 + 7 + 8i equal?
10 + 8i
Here, we have two real numbers (3 and 7) and one imaginary number (8i). We can add real numbers, but real numbers and imaginary ones can only take the form a + bi. We can't do more than that. So in this case, we can add 3 and 7, but that 8i is all by itself. A match.com profile might help it be less lonely, though. In any case, our answer is (B).
7. What is ?
7i
If we split up what's under the radical, we can have , which will give us 7 × . Since we know that , we should have no problem arriving at (A) as the correct answer.
8. What is ?
10 + 5i
This question still takes the format a + bi, as it should. In this case, our a is , or 10 and our bi is , or 5i. The two are added together, and since a + bi is the final formatting on complex numbers, the most simplified answer we can give is (C).
9. What does equal?
Before we even look at imaginary numbers, we have two of the exact same number. Instead of writing , we can simply write . Now, let's take a look at what i has to offer us. If we pull out an i from under the radical, we get . Since that's the simplest we can make it, our answer is (D).
10. Are there any uses for imaginary numbers (aside from rides at Disney World)?
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# Difference between revisions of "2012 AMC 12A Problems/Problem 24"
## Problem
Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$, and in general,
$$a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}$$
Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$. What is the sum of all integers $k$, $1\le k \le 2011$, such that $a_k=b_k?$
$\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\2012$ (Error compiling LaTeX. )
## Solution
### Solution 1
We begin our solution by understanding two important functions: $f(x) = b^x$ for $0 < b < 1$, and $g(x) = x^k$ for $k > 0$. The first function is a decreasing exponential function. This means that for numbers $m > n$, $f(m) < f(n)$. The second function is an increasing function on the interval $[0, \infty]$. This means that for numbers $m > n$, $g(m) > g(n)$. $f(x)$ is used to establish inequalities when we change the exponent keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant.
We will now begin by examining the first few terms.
Comparing $a_1$ and $a_2$, $0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$.
Comparing $a_2$ and $a_3$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$.
Comparing $a_1$ and $a_3$, $0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 0$.
Therefore, $0 < a_1 < a_3 < a_2 < 1$.
Continuing in this manner, it is easy to see a pattern.
We claim that $0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1$.
We will now use induction to prove this statement. (Note that this is not necessary on the AMC):
Rearranging in decreasing order gives $1 > b_1 = a_2 > b_2 = a_4 > ... > b_{1005} = a_{2010} > b{1006} = a_{2011} > ... > b_{2010} = a_3 > b_{2011} = a_1 > 0$.
Therefore, the only $k$ when $a_k = b_k$ is when $2(k-1006) = 2011 - k$. Solving gives $\boxed{textbf{(C)} 1341}$.
2012 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
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# How do you solve abs(8x-10)>=6?
May 8, 2017
See a solution process below:
#### Explanation:
The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
$- 6 \ge 8 x - 10 \ge 6$
First, add $\textcolor{red}{10}$ to each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:
$- 6 + \textcolor{red}{10} \ge 8 x - 10 + \textcolor{red}{10} \ge 6 + \textcolor{red}{10}$
$4 \ge 8 x + 0 \ge 16$
$4 \ge 8 x \ge 16$
Now, divide each segment of the system by $\textcolor{red}{8}$ to solve for $x$ while keeping the system balanced:
$\frac{4}{\textcolor{red}{8}} \ge \frac{8 x}{\textcolor{red}{8}} \ge \frac{16}{\textcolor{red}{8}}$
$\frac{1}{2} \ge \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} x}{\cancel{\textcolor{red}{8}}} \ge 2$
$\frac{1}{2} \ge x \ge 2$
Or
$x \le \frac{1}{2}$ and $x \ge 2$
Or, in interval notation:
$\left(- \infty , \frac{1}{2}\right]$ and $\left[2 , \infty\right)$
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Collection of recommendations and tips
# What is the greatest common factor of 33 and 43?
## What is the greatest common factor of 33 and 43?
As you can see when you list out the factors of each number, 1 is the greatest number that 33 and 43 divides into.
What is the greatest common factor of 46 and 33?
As you can see when you list out the factors of each number, 1 is the greatest number that 33 and 46 divides into.
### What is the greatest common factor of 46 23 and 46?
The greatest common factor of 46 and 23 is 23.
How do you find the greatest common factor of three numbers?
To find the greatest common factor (GCF) between numbers, take each number and write its prime factorization. Then, identify the factors common to each number and multiply those common factors together. Bam! The GCF!
## What is a factor of 46?
The factors of 46 are 1, 2, 23, 46 and its negative factors are -1, -2, -23, -46.
What are the factors for 33?
Factors of 33
• Factors of 33: 1, 3, 11 and 33.
• Negative Factors of 33: -1, -3, -11 and -33.
• Prime Factors of 33: 3, 11.
• Prime Factorization of 33: 3 × 11 = 3 × 11.
• Sum of Factors of 33: 48.
### What are the greatest common factors of 46?
Factors of 46 are integers that can be divided evenly into 46. It has a total of 4 factors of which 46 is the biggest factor and the positive factors of 46 are 1, 2, 23 and 46.
What is the greatest common factor of 30 and 45?
15
The GCF of 30 and 45 is 15. To calculate the greatest common factor of 30 and 45, we need to factor each number (factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30; factors of 45 = 1, 3, 5, 9, 15, 45) and choose the greatest factor that exactly divides both 30 and 45, i.e., 15.
## How do u find the greatest common factor?
To find the GCF of a set of numbers, list all the factors of each number. The greatest factor appearing on every list is the GCF. For example, to find the GCF of 6 and 15, first list all the factors of each number. Because 3 is the greatest factor that appears on both lists, 3 is the GCF of 6 and 15.
How do I find the highest common factor?
The highest common factor is found by multiplying all the factors which appear in both lists: So the HCF of 60 and 72 is 2 × 2 × 3 which is 12. The lowest common multiple is found by multiplying all the factors which appear in either list: So the LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.
### What are the factors of 43?
43 has only two factors and they are 1 and 43. The number 43 is a prime number.
What is the greatest common factor of 46?
Factors of 46 are integers that can be divided evenly into 46. It has a total of 4 factors of which 46 is the biggest factor and the positive factors of 46 are 1, 2, 23 and 46. The sum of all factors of 46 is 72.
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Question
# A radio can tune over the frequency range of a portion of the MW broadcast band: (from $800\mathrm{kHz}$ to $1200\mathrm{kHz}$). If its LC circuit has an effective inductance of $200\mathrm{\mu H}$, what must be the range of its variable capacitor?
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Solution
## Step 1: Given dataThe lower tuning frequency of the radio is ${f}_{1}=800KHz=800×{10}^{3}Hz.$The upper tuning frequency of the radio is ${f}_{2}=1200KHz=1200×{10}^{3}Hz.$The effective inductance of the radio is $L=200\mu H=200×{10}^{-6}H.$Step 2: The capacitance of the LCR circuitWe know at the resonance of the LCR circuit the angular frequency is $\omega =\frac{1}{\sqrt{\mathrm{LC}}}$, where, L is the inductance, C is the capacitance, $\omega$ and is the angular frequency of the wave.Angular momentum, $\omega =2\mathrm{\pi f}$, where, f is the frequency of a wave.Step 3: Finding the range of the capacitorAs we know the angular frequency of an LCR circuit is $\omega =\frac{1}{\sqrt{\mathrm{LC}}}$.So, $C=\frac{1}{{\omega }^{2}L}.................\left(1\right)$ Now, the capacitance of the variable capacitor for frequency ${f}_{1}$ and angular frequency ${\mathrm{\omega }}_{1}$ is ${C}_{1}=\frac{1}{{{\omega }_{1}}^{2}L}=\frac{1}{{2{\mathrm{\pi f}}_{1}}^{2}×L}=\frac{1}{\left(2×3.14\right)×{\left(800×{10}^{3}\right)}^{2}×\left(200×{10}^{-6}\right)}\phantom{\rule{0ex}{0ex}}or{C}_{1}=1.98×{10}^{-10}F................\left(1\right)$And, the capacitance of the variable capacitor for frequency ${f}_{2}$ and angular frequency ${\mathrm{\omega }}_{2}$ is ${C}_{2}=\frac{1}{{{\omega }_{2}}^{2}L}=\frac{1}{{2{\mathrm{\pi f}}_{2}}^{2}×L}=\frac{1}{\left(2×3.14\right)×{\left(1200×{10}^{3}\right)}^{2}×\left(200×{10}^{-6}\right)}\phantom{\rule{0ex}{0ex}}or{C}_{2}=0.880×{10}^{-10}F..............\left(2\right)$From equations 1 and 2 we get, $C=\left[{C}_{1},{C}_{2}\right]=\left[\left(1.98×{10}^{-10}\right),\left(0.880×{10}^{-10}\right)\right]$Therefore, the range of its variable capacitor is $C=\left[\left(1.98×{10}^{-10}\right)F,\left(0.880×{10}^{-10}\right)F\right]$.
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Lesson Explainer: Graphing Speed | Nagwa Lesson Explainer: Graphing Speed | Nagwa
# Lesson Explainer: Graphing Speed Physics • First Year of Secondary School
In this explainer, we will learn how to interpret graphs of distance and time and graphs of speed and time that represent the motion of objects.
A graph is a convenient way to show how two quantities change together. One of the quantities in a graph can be time that passes, so that the graph shows how the other quantity on the graph, whatever it is, changes as time passes. In this explainer, we will look at graphs showing either of the following:
• how the distance traveled by an object changes with time,
• how the speed of an object changes with time.
A graph of how the distance traveled by an object changes with time is called a distance–time graph, or a graph.
A graph of how the speed of an object changes with time is called a speed–time graph, or a graph.
Let us begin by looking at the distance–time graph and speed–time graph for an object that is at rest when time equals zero and does not change speed. This is shown in the following figure.
For both graphs, all the points plotted are on the time axis. This is because the time axis corresponds to zero distance on the distance–time graph and to zero speed on the speed–time graph.
Let us now look at the distance–time graph and speed–time graph for an object that is moving at a constant speed when time equals zero and does not change speed. This is shown in the following figure.
We can draw lines of best fit for both these graphs, as shown in the following figure.
We see that the line for the speed–time graph is horizontal and that the line for the distance–time graph is straight and slopes upward.
The value of on the speed axis of the speed–time graph equals the constant speed of the object.
The speed of an object is related to the distance that the object travels and the time that it travels for by the formula where is the speed of the object, is the distance the object travels, and is the time for which the object travels.
As the speed of the object for these graphs is constant, for equal changes in time there are equal changes in distance, as shown in the following figure.
The gradient of the line of the distance–time graph is equal to the speed of the object.
The following figure shows distance–time and speed–time graphs for two objects, where the speed of the faster object (shown by the green line) is twice the speed of the slower object (shown by the blue line).
We can see that on the distance–time graph the gradient of the green line is twice that of the blue line.
Let us now look at an example of interpreting a distance–time graph.
### Example 1: Identifying the Time Interval in Which an Object Is at Rest
The changes in the distance covered by an object during a time interval are shown in the graph. The graph is split into three sections: I, II, and III. In which section of the graph does the object have a speed of zero?
An object has a speed of zero when the change in the distance it travels, , is zero in a time interval . This happens on the graph in a section where the distance does not change throughout the section.
In sections I and III, the distances traveled by the object are greater at the end of the section than at the start of the section. Only in section II are the distances at the start and the end of the section equal. The line of best fit in section II is horizontal, so there is no change in distance in this section. This must mean that the speed in section II is zero.
Let us now look at another similar example.
### Example 2: Identifying the Time Interval in Which the Speed of an Object Is Greatest
The changes in the distance moved by an object during a time interval are shown in the graph. The graph is split into three sections: I, II, and III. In which section of the graph is the speed of the object greatest?
An object has its greatest speed when the distance it travels, , in a time interval is greatest.
The following figure shows the line in each section of the graph in a different color. Below this, the figure shows, in the same colors, the distance traveled in the same time interval for each section, directly compared.
We see that is the greatest for the blue line, and so in section I the speed is greatest.
Let us now look at another example.
### Example 3: Analyzing the Distance–Time Graph for an Object That Changes Speed
The graph shows the changes in the distance walked by a dog in a time interval of 8 seconds.
1. At what time did the dog change its speed?
2. Was the dog’s speed higher or lower before the point when its speed changed?
3. What is the difference between the speed of the dog before and after it changed speed?
Part 1
The gradient of the line of best fit of the graph changes at a time value of 4 seconds. This is when the speed changes.
Part 2
The gradient of the line is less steep after 4 seconds. The speed of the dog must have decreased.
Part 3
In the first 4 seconds that the dog walks, it walks a distance of 12 metres. The speed of the dog in this time interval is given by where is the speed of the dog, is the distance the dog travels, and is the time for which the dog travels.
We see that
If the motion of the dog over less than full 12 metres is used, we would get the same answer.
For example, we could take the motion of the dog between 6 metres and 12 metres.
We see for the motion of the dog over this distance that
After 4 seconds, the dog walks for a further 4 seconds. In this time, the dog walks a distance of 8 metres. In this time interval, the speed of the dog is given by
We see then that the speed of the dog changes by
Consider the speed–time graph for an object that is shown in the following figure.
We can see that the speed of the object increases as time passes. We can compare this to the speed–time graph for an object that is shown in the following figure.
We can see that the speed of the object decreases as time passes.
Let us look at an example involving speed–time graphs for objects with different speeds.
### Example 4: Comparing the Speeds of Multiple Objects Using a Speed–Time Graph
The changes in the speeds of three objects during the same time interval are shown in the graph.
1. Which object has the greatest initial speed?
2. Which object has the greatest final speed?
3. Which object has the greatest average speed?
4. Which object was not moving?
Part 1
The initial speed of an object shown by the graph is the speed of the object at the origin of the time axis. We see that at this time, the yellow line has the greatest speed, so object III has the greatest initial speed.
Part 2
The final speed of an object shown by the graph is the speed of the object at the furthest point along the time axis from its origin. We see that at this time, the blue line has the greatest speed, so object II has the greatest final speed.
Part 3
The average speed of an object with constant speed is the same as the constant speed of the object. Only object I has a constant speed, as the other objects change speeds.
The speeds of object II and object III change uniformly, so the average speed for each object is given by
No numbers are given in the question, but the average speeds for objects II and III can be estimated from the graph by inspection, as shown in the following figure.
We see from this that object I has the greatest average speed.
If the initial and final speed values were given, the average speeds could have been calculated using the equation for average speed.
In this question, no values are stated for the speeds, as the question is intended to require visually interpreting lines on a graph in terms of how their gradients and intercepts compare.
It is important to understand that the scale drawing method for obtaining the answer shown here is included to give a very clear explanation of how to identify the line corresponding to the greatest average speed.
A much simpler explanation might have simply stated that by inspection it can be seen that the red line corresponds to the greatest average speed. This would be correct but would not clarify how to use inspection to compare the speeds.
Part 4
The line for object I is horizontal. A horizontal line on a distance–time graph indicates zero speed, but this graph is a speed–time graph, and the speed of object I is constantly greater than zero. Object I, and the other objects, move. None of the objects do not move.
We have seen that the speed of an object can remain constant, increase, or decrease. The distance moved by an object can only ever remain constant or increase, however.
The following figure shows speed–time and distance–time graphs for an object that increases in speed and an object that decreases in speed. It is important to note that the distance moved by the object increases as time passes for both objects, but the increases do not have equal rates.
The lines for both distance–time graphs are curved.
Let us look at an example involving a distance–time graph with both a straight and a curved line.
### Example 5: Comparing the Motion of Two Objects Using a Distance–Time Graph
The red and blue lines show the change in distance moved with time for two objects.
1. Which color line corresponds to the object that moves the greater distance?
2. Which color line corresponds to the object that moves at the greater average speed?
3. Which color line corresponds to the object that has the greater maximum speed?
Part 1
It is important to remember that a distance–time graph should not be confused with a diagram showing the path traveled by an object. The red line is clearly longer than the blue line. If the red and blue lines were in a diagram showing the paths traveled by objects, then the red line would indicate a greater distance.
However, on a distance–time graph, the distance traveled is indicated not by the length of a line but by the greatest value on the distance axis for the line. As distance traveled can never decrease with time, the maximum value on the distance for the line equals the value at the end of the line corresponding to the end of the motion of the object. This is shown in the following figure.
We can see that both the red and blue lines end at the same point, and so the lines show equal distances traveled by objects.
Part 2
The average speed of an object in a time interval is the distance that the object moves in the time interval divided by the time interval. We have already seen that the objects travel the same distance, as the red and blue lines end at the same point. This fact tells us that the objects move for the same amount of time. Two objects that travel the same distance in the same time have the same average speed in that time.
Part 3
The average speeds of the objects are equal, but the speeds of the objects are not equal to each other throughout the time interval in which the objects move. The speed of an object is given by where is the speed of the object, is the distance the object travels, and is the time for which the object travels.
For the blue line, the value of is constant. For the red line, though, the object travels greater distance in a time near the end point of its motion than near the start point of its motion. This is shown in the following figure.
The speeds of the objects are equal only at the instant where the red line and blue line are parallel. After that time, the object corresponding to the red line has greater speed than the object corresponding to the blue line. The red line therefore represents the object that has the greater maximum speed.
Alternatively, instead of answering the question this way, it is possible to consider defining a section of the red line corresponding to an arbitrary time interval, less than the full time for which both objects move.
The section of the red line would start and end at arbitrarily selected points on the -axis that were within the full time interval of the motion of the objects.
This section could then be translated in the positive -direction such that the left-hand end of the section was at the same -axis position as the point on the blue line corresponding to the start of the time interval.
Any such section for which the right-hand end of the red line corresponded to a greater increase in distance than a section of the blue line of the same length starting from the same time would demonstrate that within the time interval for the section, the red line must correspond to a greater average speed than the blue line.
Different time intervals could be chosen by trial and error. For any such interval, the speed represented by the blue line would be the same.
We see then that finding even one time interval for which the red line corresponded to a greater average speed than the blue line would show that the red line represented the object with the greater maximum speed.
Let us now summarize what has been learned in these examples.
### Key Points
• The line of a distance–time graph for an object at rest is a horizontal line along the time axis.
• The line of a speed–time graph for an object with a constant speed is a horizontal line that intersects the speed axis of the graph at the value of the speed of the object.
• The line of a distance–time graph for an object with a constant speed is a straight line with a gradient equal to the speed of the object.
• The line of a speed–time graph for an object with a changing speed is a straight line that has a positive gradient for an increasing speed and a negative gradient for a decreasing speed.
• The line of a distance–time graph for an object with a changing speed is a curved line.
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# What is 31/40 as a percentage?
All through our schooling years, we encounter fractions. Many questions involve solving a fraction as a percentage. For example, we may need to calculate 31/40 as a percentage.
Sometimes we need to express the given fraction in terms of percentage and you need to learn the steps for the way you can convert the given fraction to its percentage form.
Here we will take up the simple example: 31/40 where we will learn to convert it to its percentage form. The percentage form can be expressed through different methods and here we are going to learn about both of them.
What is 31/40 as a percent? The answer is 31/40 as a percentage is 77.5%
## Basic concepts of converting a fraction to a percentage
1) As we all know that a fraction is divided into two parts, they are called the numerator and the denominator.
2) The numerator is present above the line of division and the denominators are digits that are present below the line of division.
3) Here, in this case, the numerator is 31 and the denominator is 40.
4) Hence when we are conducting division with fraction, the numerator becomes the dividend and the denominator becomes the divisor which divides the given dividend in the given process.
5) What do you mean by percentage – the process of percentage consists of expressing all the numerical values in terms of hundred. We can also say that during the process of percentage we compare the two values based on 100. In other words, the percentage is defined as a fraction of a hundred.
6) So if we write 33%, we mean 33/100, and if we say 73% we mean 73/100.
Also read: What is 24/50 as a percentage?
## Calculation to express 31/40 as a Percentage
### Method #1
Step 1
In the first step, we are going to change the denominator value to a hundred and to do so we will have to divide 100 by the denominator.
#### 100÷40 = 2.5
$\frac{100}{40}=2.5$
Step 2
In the next step, you need to use the value that was acquired in the first step to multiply with both the numerator and the denominator.
#### (31×2.5)÷(40×2.5) = 77.5/100
$\frac{31\times2.5}{40\times2.5}=\frac{77.5}{100}$
Thus, we see that the results of the change in fraction 31/40 to percentage form will be 77.5%
### Method #2
Step 1
In the first step, you will have to divide the given numerator by the denominator where 31 is the numerator that acts as the dividend now and 40 is the denominator that acts as the divisor now.
#### 31÷40= 0.775
Step 2
Finally, in the last step, you have to multiply the obtained results in decimals by 100 to yield the results in their percentage form.
#### 0.775×100 = 77.5%
Hence we see that when we apply both the steps to find out the results, the answer is the same in both cases. Thus, it is clear that both the processes are valid and you can use any one of them, whichever is suitable for you to yield the results.
However, it is also important for you to understand both the methods carefully so that you can solve your sum either way if presenting, in exams.
If you practice the method every day with different fractions then you will surely be able to exercise expertise over the sums and find your answer without any trouble.
In this case,
#### The answer to the conversion of fraction 31/40 to percentage is 77.5%
Also read: What is 37/50 as a percentage?
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# How do you factor 2v^2-4-7v ?
Mar 1, 2017
$2 {v}^{2} - 4 - 7 v = \left(2 v + 1\right) \left(v - 4\right)$
#### Explanation:
First put in standard order of decreasing powers of $v$:
$2 {v}^{2} - 7 v - 4$
Next, use an AC method:
Find a pair of factors of $A C = 2 \cdot 4 = 8$ which differ by $B = 7$.
The pair $8 , 1$ works.
Use this pair to split the middle term and factor by grouping:
$2 {v}^{2} - 7 v - 4 = 2 {v}^{2} - 8 v + v - 4$
$\textcolor{w h i t e}{2 {v}^{2} - 7 v - 4} = \left(2 {v}^{2} - 8 v\right) + \left(v - 4\right)$
$\textcolor{w h i t e}{2 {v}^{2} - 7 v - 4} = 2 v \left(v - 4\right) + 1 \left(v - 4\right)$
$\textcolor{w h i t e}{2 {v}^{2} - 7 v - 4} = \left(2 v + 1\right) \left(v - 4\right)$
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# Probability and Statistics (Test 3)
## Problem Solving And Reasoning : Mathematics Or Quantitative Aptitude
| Home | | Problem Solving And Reasoning | | Mathematics Or Quantitative Aptitude | | Probability and Statistics |
Probability and Statistics
| Probability and Statistics |
Q.1
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that at most one is blue?
A. 15/91
B. 85/91
C. 25/29
D. 75/91
E. None of these
Answer : Option B
Explaination / Solution:
Probabilities if at most one is blue =
[(4C0*10C24C1*10C1)/14C2] = (1*45 + 4*10)/( 91) = 85/91
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Q.2
Direction: Study the following information carefully and answer the questions that follow:
A committee of 10 persons is to be formed from 7 men and 6 women.
In how many ways this can be done if at least 5 men to have to be included in a committee.
A. 251
B. 265
C. 167
D. 340
E. None of these
Answer : Option A
Explaination / Solution:
Number of Ways when if at least 5 men include in committee = 7C5*6C5+7C6*6C4+7C7*6C3
= 21*6+7*15+1*20
= 251
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Q.3
Direction: Study the following information carefully and answer the questions that follow:
A committee of 10 persons is to be formed from 7 men and 6 women.
In how many ways of these committee the women are in majority
A. 45
B. 35
C. 110
D. 56
E. None of these
Answer : Option B
Explaination / Solution:
Number of ways when women are majority in committee = 6C6*7C4 = 1*35 = 35
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Q.4
Direction: Study the following information carefully and answer the questions that follow:
A committee of 10 persons is to be formed from 7 men and 6 women.
In how many of these committee the men are in majority
A. 118
B. 135
C. 140
D. 125
E. None of these
Answer : Option D
Explaination / Solution:
Number of ways when men are majority in committee = 7C6*6C4+7C7*6C3
7*15+1*20 = 125
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Q.5
Direction: Study the following information carefully and answer the questions that follow:
A committee of 10 persons is to be formed from 7 men and 6 women.
In how many ways this can be done if 6 men and 4 women be included in a committee
A. 105
B. 114
C. 81
D. 210
E. None of these
Answer : Option A
Explaination / Solution:
If 6 men and 4 women include in committee = 7C6*6C4 --> 7*15 --> 105
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Q.6
Direction: Study the following information carefully and answer the questions that follow:
A committee of 10 persons is to be formed from 7 men and 6 women.
In how many ways this can be done if 5 men and 5 women be included in a committee
A. 102
B. 140
C. 136
D. 126
E. None of these
Answer : Option D
Explaination / Solution:
If 5 men and 5 women include in committee = 7C5*6C5 --> 21*6 = 126
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Q.7
In a basket there are 7 apples and 8 oranges. 4 fruits are picked at random. What is the probability that two fruits are apples and 2 are oranges?
A. 72/455
B. 82/455
C. 89/455
D. 84/455
E. None of these
Answer : Option E
Explaination / Solution:
Total number of fruits n(s) =7 + 8=15
Probability = (7C2*8C2)/15C4
= (21*28)/1365 = 28/65
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Q.8
In how many ways the letter of the word IMMEDIATELY can be arranged so that vowels always come together ?
A. 56800
B. 75600
C. 64800
D. 84560
E. None of these
Answer : Option C
Explaination / Solution:
There are 11 letters .out of which 5 are vowels and 6 are consonants.
So taking vowels together as a single letter , we have 7 letters
So no of arrangements =( 7! * 5!)/ (2!)3 [there are 2 i,m e]
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Q.9
In how many ways can the word ENGINEER be arranged so that ‘G’ and ‘R’ are never together?
A. 3120
B. 2830
C. 2520
D. 3220
E. None of these
Answer : Option C
Explaination / Solution:
Number of ways of rearranging the word ENGINEER = (8!)/(3!*2!) = 3360
Finding the number of ways of arranging the word ENGINEER such that G and R are always together is done by taking GR as a single alphabet and then finding the permutation.
Number of ways of arranging the word ENGINEER such that G and R are always together = (7!)/(3!*2!) = 420*2 = 840
∴Number of ways of arranging the word ENGINEER such that G and R are never together = Number of ways of rearranging the word ENGINEER - Number of ways of arranging the word ENGINEER such that G and R are always together
⇒Number of ways of arranging the word ENGINEER such that G and R are never together
= 3360 – 840
= 2520
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Q.10
Read the following information to answer the following questions :
Two unbiased dice are thrown simultaneously
What is the probability of getting a doublet?
A. 1/3
B. 1/6
C. 1/4
D. 2/3
E. None of these
Answer : Option B
Explaination / Solution:
Total possible outcomes = 6 × 6 = 36
E = Events of getting a doublet
= (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6
PE = 6/36 = 1/6
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# 4100 in Words
4100 in words can be written as Four Thousand One Hundred. Students will be able to learn the conversion of 4100 in words which will help them understand the applications of numbers in our daily lives. If you buy a grinder for Rs. 4100, then you can say that “I have bought a grinder for Four Thousand One Hundred Rupees”. The number 4100 can be written in words using the English alphabet. The numbers in words can be grasped easily by the students using the resources given at BYJU’S. 4100 in English can be read as “Four Thousand One Hundred”.
4100 in words Four Thousand One Hundred Four Thousand One Hundred in Numbers 4100
## How to Write 4100 in Words?
Students will learn about the conversion of 4100 into words from place value charts. The number 4100 has four digits. For 4100, the place value chart is prepared in a table form to help students understand it effectively.
Thousands Hundreds Tens Ones 4 1 0 0
4100 in expanded form is explained in brief here:
4 × Thousand + 1 × Hundred + 0 × Ten + 0 × One
= 4 × 1000 + 1 × 100 + 0 × 10 + 0 × 1
= 4000 + 100
= 4100
= Four Thousand One Hundred
Therefore, 4100 in words is written as Four Thousand One Hundred.
4100 is a natural number that precedes 4101 and succeeds 4099.
4100 in words – Four Thousand One Hundred
Is 4100 an odd number? – No
Is 4100 an even number? – Yes
Is 4100 a perfect square number? – No
Is 4100 a perfect cube number? – No
Is 4100 a prime number? – No
Is 4100 a composite number? – Yes
## Frequently Asked Questions on 4100 in Words
### How do you write 4100 in words?
4100 can be written as “Four Thousand One Hundred” in words.
### Is 4100 an even number?
4100 is an even number because it is divisible by 2.
4100/2 = 2050
### How can Four Thousand One Hundred be written in numbers?
Four Thousand One Hundred can be written in numbers as 4100.
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# Into Math Grade 3 Module 9 Lesson 4 Answer Key Use Mental Math to Assess Reasonableness
We included HMH Into Math Grade 3 Answer Key PDF Module 9 Lesson 4 Use Mental Math to Assess Reasonableness to make students experts in learning maths.
## HMH Into Math Grade 3 Module 9 Lesson 4 Answer Key Use Mental Math to Assess Reasonableness
I Can use mental math to determine the reasonableness of statements and answers.
A store has 212 shirts for sale. The store sells 77 shirts. Luke says that there are about 130 shirts left to sell. Micah says that there are about 300 shirts left to sell. Whose statement is reasonable? Explain your thinking.
Luke statement is reasonable than Micah.
Explanation:
A store has 212 shirts for sale.
Store sells 77 shirts.
The difference between these two is 212 – 77 = 135 and Luke said there are about 130 shirts to sell.
Micah said that there are about 300 shirts left to sell. Hence Luke’s statement is reasonble than Micah.
Turn and Talk For the statement that is not reasonable, what scenario would make the statement reasonable?
Build Understanding
1. A market has 53 cans of fruit and 44 cans of vegetables for sale. Rachel says that there are about 140 cans of food for sale. Is Rachel’s statement reasonable?
Rachel’s statement is not reasonable.
Explanation:
In a market sale has 53 cans of fruit and 44 cans of vegetables.
And rachel says there are about 140 cans of food for sale.
In market a total of 97 cans of food are there but rachel said there are 140 cans of food. Therefore rachel’s statement is wrong.
A. How can you use a number line to show your thinking?
B. How does your thinking compare to Rachel’s statement?
The market has a total of 97 cans of food and rachel said there are about 140 cans of food for sale. By comparing both rachel said more number of can food than in the market.
C. Is Rachel’s statement reasonable? How do you know?
No it is not reasonable.
Explanation:
As we are told rachel said more number of can food than in the market. Hence the given statement is wrong.
Think: When you decide if an answer is reasonable, the quantities and the operations have to make sense.
Turn and Talk How many fruit cans and vegetable cans could there be in order for Rachel’s statement to be reasonable?
70 cans of fruit and 70 cans of vegetables.
Explanation:
There should be 70 cans of fruit and 70 cans of vegetables could there be in order for Rachel’s statement to be reasonable.
Step It Out
2. A toy store has 283 toys for sale. There are 69 board games. Hector says there are about 215 toys that are not board games. Is Hector’s statement reasonable? Use mental math to decide.
A. About how many toys are for sale?
• Make 283 a friendly number. 283 – __ = ___
283 – 215 = 69.
Explanation:
The number of toys are for sale is 280 and it is a friendly number.
B. About how many board games are there?
• Make 69 a friendly number. 69 + ___ = ___
69 + 1 = 70
Explanation:
There are 70 board games in the toy store.
C. Using friendly numbers, what equation can you write to solve the problem?
_______________
215 + 69 = 283
Explanation:
The equation we can write to solve the problem is 215 + 69 = 283.
D. About how many toys are not board games? about ___ toys
215 toys.
Explanation:
There are about 215 toys that are not board games and there are 69 board games.
E. Is Hector’s statement reasonable? Explain.
_______________
Yes the Hector’s statment is reasonable.
Explanation:
A toy store has 283 toys for sale and 69 board games.
Hector says there are about 215 toys that are not board games.
By doing the difference 283 – 69 = 215.
Check Understanding Math Board
Question 1.
Mrs. Bice spends $93 on groceries and$47 on the heating bill. She says that she spends about $100. Is Mrs. Bice’s statement reasonable? Show your thinking. Answer: Mrs. Bice’s statement is not reasonable. Explanation: Mrs. Bice spends$93 on groceries and $47 on the heating bill. Mrs. Bice said she spent about$100.
The total sum of groceries is $110. By comparing both we can know that she spent more than$100. Hence the given statement is not reasonable.
Question 2.
STEM Cellular phone memory, which is stored on a Secure Digital card, or SD card, is measured in units of information called gigabytes or GB. The SD card in Ms. Patel’s phone has 128 GB. If she buys a new phone with 256 GB, she says she will have about 25 GB more. Is this reasonable? Explain.
The statement Ms. patel’s said is not reasonable.
Explanation:
Ms. Patel’s phone has 128 GB and she bought a new phone with 256 GB. She also said that she have 25 GB more. So by doing the difference 256 – 128 GB the answer we get is 128 GB. Hence the given statment is wrong.
Question 3.
Critique Reasoning David has $50 in his pocket. He spends$23 on meat and $18 on fruits and vegetables at the market. He says he has about$40 left in his pocket. Is David’s statement reasonable? Explain.
Question 4.
Critique Reasoning Ina says 185 – 79 is about 105. Is Ina’s statement reasonable? Explain your thinking.
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# Notes | EduRev
## Class 10 : Notes | EduRev
The document Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
Q.1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2x2 – 3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
(iii) 2x2 – 6x + 3 = 0
Solutions:
(i) Given,
2x2 – 3x + 5 = 0
Comparing the equation with ax2 + bx + c = 0, we get
a = 2, b = -3 and c = 5
We know, Discriminant = b2 – 4ac
( – 3)2 – 4 (2) (5) = 9 – 40
= – 31
As you can see, b2 – 4ac < 0
Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0.
(ii) 3x2 – 4√3x + 4 = 0
Comparing the equation with ax2 + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
We know, Discriminant = b2 – 4ac
= (-4√3)– 4(3)(4)
= 48 – 48 = 0
As b2 – 4ac = 0,
Real roots exist for the given equation and they are equal to each other.
Hence the roots will be –b/2a and –b/2a.
b/2= -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and 2/√3.
(iii) 2x2 – 6x + 3 = 0
Comparing the equation with ax2 + bx + c = 0, we get
a = 2, b = -6, c = 3
As we know, Discriminant = b2 – 4ac
= (-6)2 – 4 (2) (3)
= 36 – 24 = 12
As b2 – 4ac > 0,
Therefore, there are distinct real roots exist for this equation, 2x2 – 6x + 3 = 0.
= (-(-6) ± √(-62-4(2)(3)) )/ 2(2)
= (6±2√3 )/4
= (3±√3)/2
Therefore the roots for the given equation are (3+√3)/2 and (3-√3)/2
Q.2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
Solutions:
(i) 2x2 + kx + 3 = 0
Comparing the given equation with ax2 + bx + c = 0, we get,
a = 2, b = k and c = 3
As we know, Discriminant = b2 – 4ac
= (k)2 – 4(2) (3)
= k2 – 24
For equal roots, we know,
Discriminant = 0
k2 – 24 = 0
k2 = 24
k = ±√24 = ±2√6
(ii) kx(x – 2) + 6 = 0
or kx2 – 2kx + 6 = 0
Comparing the given equation with ax2 + bx + c = 0, we get
a = k, b = – 2k and c = 6
We know, Discriminant = b2 – 4ac
= ( – 2k)2 – 4 (k) (6)
= 4k2 – 24k
For equal roots, we know,
b2 – 4ac = 0
4k2 – 24k = 0
4k (k – 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘.
Therefore, if this equation has two equal roots, k should be 6 only.
Q.3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Solution:
Let the breadth of mango grove be l.
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)= 2l2
2l= 800
l= 800/2 = 400
l– 400 =0
Comparing the given equation with ax2 + bx + c = 0, we get
a = 1, b = 0, c = 400
As we know, Discriminant = b2 – 4ac
=> (0)2 – 4 × (1) × ( – 400) = 1600
Here, b2 – 4ac > 0
Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.
= ±20
As we know, the value of length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m
Q.4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let’s say, the age of one friend be x years.
Then, the age of the other friend will be (20 – x) years.
Four years ago,
Age of First friend = (x – 4) years
Age of Second friend = (20 – x – 4) = (16 – x) years
As per the given question, we can write,
(x – 4) (16 – x) = 48
16x – x2 – 64 + 4x = 48
– x2 + 20x – 112 = 0
x2 – 20x + 112 = 0
Comparing the equation with ax2 + bx + c = 0, we get
a = 1, b = -20 and c = 112
Discriminant = b2 – 4ac
= (-20)2 – 4 × 112
= 400 – 448 = -48
b2 – 4ac < 0
Therefore, there will be no real solution possible for the equations. Hence, condition doesn’t exist.
Q.5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.
Solution:
Let the length and breadth of the park be and b.
Perimeter of the rectangular park = 2 (l + b) = 80
So, l + b = 40
Or, b = 40 – l
Area of the rectangular park = l×b = l(40 – l) = 40l= 400
l2 40l + 400 = 0, which is a quadratic equation.
Comparing the equation with ax2 + bx + c = 0, we get
a = 1, b = -40, c = 400
Since, Discriminant = b2 – 4ac
=(-40)2 – 4 × 400
= 1600 – 1600 = 0
Thus, b2 – 4ac = 0
Therefore, this equation has equal real roots. Hence, the situation is possible.
Root of the equation,
l = –b/2a
l = (40)/2(1) = 40/2 = 20
Therefore, length of rectangular park, = 20 m
And breadth of the park, = 40 – = 40 – 20 = 20 m.
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## Mathematics (Maths) Class 10
51 videos|346 docs|103 tests
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Time for Class 2 | Maths Olympiad
# Time
## Time - Class 2
• What is Time?
• Units of Time
• In this article, we will take a journey into the world of time, learn about clocks and discover how we measure moments.
## What is Time?
Time is an exciting concept that helps in our daily routine from waking up in the morning to going to bed at night. It helps us to organize our day and understand. Time can be defined as the ongoing sequence of events.
Measurement of Time
Time is measured by a clock. On a clock face, there are 12 numbers to know about the time. A clock consists of three hands:
1. The big hand represents minutes.
2. The small hand represents the hours.
3. The thin hand represents seconds.
In a digital clock, the numbers are related to hours, minutes and seconds, shown as:
This is how the minute hand's movement displays various times within one hour.
Quarters of Time
Understanding a "quarter of time" is a simple concept related to fractions of time. When we talk about a quarter of the time, we are referring to dividing time into four equal parts or segments.Each of these segments represents one quarter which is described using the figure:
Example: Which of the following options states the correct time for the clock shown below?
a) Quarter to 12
b) Quarter to 1
c) Quarter past 12
d) Quarter past 1
Explanation: The time is 12 : 45 which means quarter to 1.
## Units of Time
Years, months, weeks, days, hours, minutes and seconds
Arrangement of the units of time in increasing order:
seconds < minutes < days < weeks < months < years
### Conversion of One Unit to Another Unit
Time conversion includes:
• 1 year = 12 months
Note:
a) 1 month can have 28, 29, 30 or 31 days depending upon the month.
b) 1 year = 365 days
c) 1 leap year = 366 days
d) 1 year has 52 weeks and one day whereas a leap year has 52 weeks and two days.
e) 1 week = 7 days
• 1 day = 24 hours
• 1 hour = 60 minutes
• 1 minute = 60 seconds
• 1 hour = 3600 seconds
Understanding time in both the 12-hour and 24-hour formats is necessary. Here's a simple explanation of both formats:
12-Hour Format:
In the 12-hour format, we divide the day into two parts:
Ante Meridiem (AM) and Post Meridiem (PM).
• AM is from midnight (12:00 AM) to midday (12:00 PM).
• PM is from midday (12:00 PM) to midnight (12:00 AM).
The cycle of the day is shown as:
This format is often used to tell time on regular clocks and watches. To tell time in the 12-hour format, we use numbers from 1 to 12 followed by "AM" or "PM" to show whether it's morning, afternoon, evening or night.
24-Hour Format:
The 24-hour format is also known as the international time format. In this format, the entire day is divided into 24 hours and we don't use "AM" or "PM".
• The hours run from 00:00 (midnight) to 23:59 (just before the next midnight).
• It's a continuous count of time so there is no need to switch between AM and PM.
For example, 6:10 AM in the 12-hour format is equivalent to 06:10 in the 24-hour format and 8:45 PM in the 12-hour format is equivalent to 20:45 in the 24-hour format.
This clock shows both 12-hour and 24-hour formats.
Let's learn about terms related to days:
• Today means the day we are in right now.
• Yesterday means the day that came before today.
• Tomorrow means the day that comes after today.
Let’s understand by taking an example:
Example: If a new baby was born on April 10, 2020, then what was his age on January 10, 2021?
a) 6 months
b) 8 months
c) 9 months
d) 12 months
Explanation: The months of the new baby are shown as:
The age of the baby was 9 months.
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# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate geometry is also called analytical or Cartesian geometry. It is a branch in mathematics where we can study geometry using coordinate points. This is mainly useful in locating a given point with the ordered pair of numbers. It helps us find the distance between two points using coordinates. The NCERT Maths Chapter 7 coordinate geometry deals with three major sections such as
1) Introduction to the cartesian system and distance formula
2) Section formula
3) Area of triangle
Introduction to the Cartesian System and Distance Formula
Apart from what we have seen above, coordinate geometry is also very useful in the study of graphs. These are the visual representation of our data. Some of the most useful terms to be remembered while studying coordinate geometry are;
Coordinate axes – In a standard graph, there must be two axes. One is the horizontal one, which is X-axis, and the other one is the vertical one, which is Y-axis.
Origin – The centre or the intersecting point of the X and Y axes is called the origin. It is denoted as (0,0).
Quadrant – The axes divide the plane into four equal parts. These are called quadrants.
Meanwhile, the distance formula for identifying the distance between the two points is as follows.
Section formula
This formula is focused on identifying the coordinates of the point dividing the line into m1:m2 ratio.
The following is the section formula,
m1x2+m2x1/m2+ m1 , m1y2+m2y1/m2+ m1
Area of triangle
This helps us in finding the area of any triangle in terms of coordinates and its vertices. The below-mentioned formula is used in identifying the area of the quadrilaterals,
Area of a Triangle = 1 / 2 * base * altitude sq.units.
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# Is the equation 13x² + 13y² - 26x + 52y = -78 a line, parabola, ellipse, hyperbola, or circle?
Jul 25, 2018
No - it contains no real points.
#### Explanation:
Given:
$13 {x}^{2} + 13 {y}^{2} - 26 x + 52 y = - 78$
First note that all of the coefficients are divisible by $13$, so let's divide the whole equation by $13$ to get:
${x}^{2} + {y}^{2} - 2 x + 4 y = - 6$
Adding $1 + 4 = 5$ to both sides of the equation and rearranging slightly, this becomes:
${x}^{2} - 2 x + 1 + {y}^{2} + 4 y + 4 = - 1$
That is:
${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = - 1$
Note that this is always false for real values of $x$ and $y$, so this equation describes an empty set of values in the $x$ $y$ plane.
Imaginary circle
#### Explanation:
The given equation:
$13 {x}^{2} + 13 {y}^{2} - 26 x + 52 y = - 78$
$13 \left({x}^{2} + {y}^{2} - 2 x + 4 y\right) = - 78$
${x}^{2} + {y}^{2} - 2 x + 4 y = - 6$
$\left({x}^{2} - 2 x + 1\right) + \left({y}^{2} + 4 y + 4\right) - 5 = - 6$
${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = - 1$
${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = {i}^{2}$
Above equation shows a circle with a radius $r = i$ i.e. an imaginary circle with unit radius.
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1.
Introduction
2.
How to Find Number of Divisors of a Number?
3.
The sum of divisors of a number
4.
4.1.
How can you find the total no. of divisors of a number?
4.2.
What is the multiplicative identity?
4.3.
How do you find the number of odd divisors of a number?
4.4.
How do you find the number of odd divisors of a number?
4.5.
What are the proper divisors of a number?
5.
Conclusion
Last Updated: Mar 27, 2024
# Sum of all proper divisors of a natural number
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## Introduction
Finding the number of divisors and the sum of divisors of a number is a very important concept for programming. Even though it is easy to find the number of divisors and the sum of divisors of a number if it is small, finding the number and sum of divisors of a large number will be difficult.
So let's first begin with finding the number of divisors of a number, then we will learn about finding the sum of divisors of a number.
## How to Find Number of Divisors of a Number?
To find the number of divisors of a number, we start by finding the prime factorization of the number, as all the divisors of a number will be a subset of the number's prime factorization. In simple words, if we can devise a way to find all the subset of prime factorizations of a number, we can find the number of divisors of that number.
Let's say we have a number with x terms in its prime factorization. Then to find the total number of divisors of this number, we will have to find the total number of ways to select any number of terms from these x terms or the total number of subsets of x. It would have been easier because the total number of subsets of a set is 2no. Of elements in the set, but here the x might have some repeated terms. So we first find the prime factorization of a number, then represent the number in terms of distinct prime factors. For example- we can represent 36 as 22.32
Now, suppose we have a number F which appears p times in the prime factorization of a number X, then it can appear in subsets in p+1 ways, i.e.,
So, for a number, We can say that each distinct factor Fhas Pi+1 ways to appear in subsets. So the total number of divisors of a number Y=
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## The sum of divisors of a number
To find the sum of all the divisors of a number, we will start with the formulae we used to find the number of divisors.
As we know, if a number F appears p times in the prime factorization of a number X, it can appear in subsets in p+1 ways, i.e., Now, if we multiply the set of choices of each factor, we will get a set of all possible divisors.
For example, if we have a number with two factors, then all the terms in the product will form the set containing all the divisors.
Since we need the sum of all the divisors of a number, we will find the product of the sum of all the choices for each factor. For the above example- the will represent the sum of all the divisors of a number.
So, for any number, the sum of all the divisors of a number will be
In this expression, we can observe that each term in the product is a GP. and for any GP, the sum of all the terms is
Using this for every term in the formula
We will get the result as-
Check out this problem - Maximum Product Subarray
### How can you find the total no. of divisors of a number?
To find the total number of divisors of a number, we first find the prime factorization of the number and then find each distinct prime number's exponents. Then we add one to all the exponents and then find their product. The product is the number of divisors of a number.
### What is the multiplicative identity?
Multiplicative identity states that the product of any number with 1 is the number itself. Thus, we can also say that one is a divisor of every number.
### How do you find the number of odd divisors of a number?
To find the number of odd divisors of a number, we first find the prime factorization of the number and then find each distinct prime number's exponents. Now, we remove even prime numbers in it. Then we add one to all the exponents of odd prime numbers and then find their product. The product is the number of odd divisors of a number.
### How do you find the number of odd divisors of a number?
Since we know how to find the total number of divisors of a number and odd divisors of a number, the difference of both will be the even divisors of a number.
### What are the proper divisors of a number?
The proper divisors of a number refer to the divisors, which are smaller than the number itself.
## Conclusion
In this blog, we have learned the formula for the number of divisors of a number and the sum of divisors of a number-
• We learned about the formula to find the number of divisors of a number because any divisor of a number will be formed by selecting some factors from the prime factorization of the number. Then we found that the number of choices for each factor in forming another divisor is one more than the number of times it appears in the prime factorization of the number. Thus, the total number of choices for each factor gives us the total number of divisors of a number.
• Then we learned about the formula to find the sum of divisors of a number using the similar concept we used for finding the number of divisors of a number. Since the product of all the choices for each factor for forming a divisor gives the total number of divisors. So, we find the product of the sum of the total number of choices for each distinct factor in the prime factorization of the number, which gives us the sum of all the number's divisors.
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# m18#24
Author Message
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30 Jan 2011, 13:55
For every point $$(a, b)$$ lying on line 1, point $$(b, -a)$$ lies on line 2. If the equation of line 1 is $$y = 2x + 1$$ , what is the equation of line 2 ?
(C) 2008 GMAT Club - m18#24
* $$y = \frac{1}{2} + \frac{x}{2}$$
* $$2y = 1 - x$$
* $$\frac{x + y}{2} = -1$$
* $$y = \frac{x}{2} - 1$$
* $$x = 2y + 1$$
Find two points on line 2 and use their coordinates to build the line's equation. Points $$(0, 1)$$ and $$(-\frac{1}{2}, 0)$$ on line 1 correspond to points $$(1, 0)$$ and $$(0, \frac{1}{2})$$ on line 2. The equation of line 2 is $$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$ .
i cant understand what the formula used here for building equation line 2. Can somebody explain? what is the Formula ?
thank you
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### Show Tags
30 Jan 2011, 16:19
The formula for finding the slope of a line is (change in y)/(change in x)
In your case, two coordinates (1,0) and (0,1/2) where (x,y)
(0-1/2)/(1-0) = -1/2 which is the slope of the line. The common formula for a line is y = kx + m where k is the slope and m is the y-intercept.
The second coordinate (0,1/2) tells us that when x = 0 y = 1/2, which is the y-intercept or 'm' in the formula y = kx+m where m is the y-intercept and 'k' is the slope. Plugging in the slope (-1/2) and the intercept (1/2) into the formula gives us y = 1/2-(x/2)
multiplying by 2 on both sides:
2y = 1-x
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Last edited by Mackieman on 31 Jan 2011, 04:34, edited 3 times in total.
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31 Jan 2011, 01:56
thanks . yet im actually wondering what is the formula used here that comes to direct equation. ???
The equation of line 2 is \frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1} or 2y = 1 - x
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31 Jan 2011, 01:58
sorry, here is the equation of my interest :
The equation of line 2 is $$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$ .
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31 Jan 2011, 02:16
Mackieman wrote:
The formula for finding the slope of a line is (change in y)/(change in x)
In your case, two coordinates (1,0) and (0,1/2) where (x,y)
(0-1/2)/(1-0) = -1/2 which is the slope of the line. The common formula for a line is y = kx + m where k is the slope and m is the y-intercept.
The first coordinate (1,0) tells us that when y = 0 x = 1, which is the y-intercept or 'm' in the formula y = kx+m where m is the intercept and 'k' is the slope. Pluggin in the slope (-1/2) and the intercept (1) into the formula gives ut y = 1-(x/2)
y- intercept is at x=0 as i know. so y= 1/2 when x=0 Am i missing something?
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31 Jan 2011, 04:33
tinki wrote:
Mackieman wrote:
The formula for finding the slope of a line is (change in y)/(change in x)
In your case, two coordinates (1,0) and (0,1/2) where (x,y)
(0-1/2)/(1-0) = -1/2 which is the slope of the line. The common formula for a line is y = kx + m where k is the slope and m is the y-intercept.
The first coordinate (1,0) tells us that when y = 0 x = 1, which is the y-intercept or 'm' in the formula y = kx+m where m is the intercept and 'k' is the slope. Pluggin in the slope (-1/2) and the intercept (1) into the formula gives ut y = 1-(x/2)
y- intercept is at x=0 as i know. so y= 1/2 when x=0 Am i missing something?
Sorry, you are correct, I had a rough day yesterday I hope it makes sense now.
(edited my post above)
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31 Jan 2011, 11:50
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tinki wrote:
For every point $$(a, b)$$ lying on line 1, point $$(b, -a)$$ lies on line 2. If the equation of line 1 is $$y = 2x + 1$$ , what is the equation of line 2 ?
(C) 2008 GMAT Club - m18#24
* $$y = \frac{1}{2} + \frac{x}{2}$$
* $$2y = 1 - x$$
* $$\frac{x + y}{2} = -1$$
* $$y = \frac{x}{2} - 1$$
* $$x = 2y + 1$$
Find two points on line 2 and use their coordinates to build the line's equation. Points $$(0, 1)$$ and $$(-\frac{1}{2}, 0)$$ on line 1 correspond to points $$(1, 0)$$ and $$(0, \frac{1}{2})$$ on line 2. The equation of line 2 is $$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$ .
i cant understand what the formula used here for building equation line 2. Can somebody explain? what is the Formula ?
thank you
Check this: math-coordinate-geometry-87652.html (chapter "Lines in Coordinate Geometry").
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31 Jan 2011, 11:57
I saw the link. VERY IMPRESSIVE!!!! + kudo from me
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13 Feb 2013, 01:31
Bunuel wrote:
tinki wrote:
For every point $$(a, b)$$ lying on line 1, point $$(b, -a)$$ lies on line 2. If the equation of line 1 is $$y = 2x + 1$$ , what is the equation of line 2 ?
(C) 2008 GMAT Club - m18#24
* $$y = \frac{1}{2} + \frac{x}{2}$$
* $$2y = 1 - x$$
* $$\frac{x + y}{2} = -1$$
* $$y = \frac{x}{2} - 1$$
* $$x = 2y + 1$$
Find two points on line 2 and use their coordinates to build the line's equation. Points $$(0, 1)$$ and $$(-\frac{1}{2}, 0)$$ on line 1 correspond to points $$(1, 0)$$ and $$(0, \frac{1}{2})$$ on line 2. The equation of line 2 is $$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$ .
i cant understand what the formula used here for building equation line 2. Can somebody explain? what is the Formula ?
thank you
Check this: math-coordinate-geometry-87652.html (chapter "Lines in Coordinate Geometry").
bunuel/Karishma
This is how I solved and it worked but I don't know why it worked :D Please enligthen me about why it worked...
I plugged in (a,b) in line 1 equeation to get b=2a+1..
then I started plugging in (b,-a) in the answer choices.. the 2nd answer choice resulted in b=2a+1.. and hence I marked B..
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14 Feb 2013, 21:21
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Sachin9 wrote:
I plugged in (a,b) in line 1 equeation to get b=2a+1..
then I started plugging in (b,-a) in the answer choices.. the 2nd answer choice resulted in b=2a+1.. and hence I marked B..
Responding to a pm:
a and b stand for two numbers which define a co-ordinate on the plane. When you say that (a, b) lies on y = 2x+1, it means the relation between a and b is b = 2a + 1. e.g. if a = 0, b = 1; if a = 1, b = 3... At the end of the day, a line is nothing but a depiction of how one variable changes with another. A line just shows you the relation between 2 variables.
If (b, -a) lies on a line 2y = 1-x, this is just a different way of expressing the same relation between the two numbers a and b.
a and b are the same set of numbers (i.e. if a = 0, b = 1; if a = 1, b = 3...)
So after manipulating the equation a little, you are bound to get b = 2a + 1 only.
As you figured out, the approach is a little un-intuitive. When I looked at the problem, I actually solved it exactly the same way except that I took numbers rather than a and b.
I said, if (a, b) lies on y = 2x + 1, if a = 1, b = 3.
So (3, -1) must lie on the new equation of the line. When I put (3, -1) in the options, I see that only (B) satisfies.
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For every point (a,b) lying on line 1, point (b,-a) lies on line [#permalink]
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28 Dec 2013, 23:57
it took me a while to get this
so what i did was create a value for x then sub it into line 1 to find out what y is. now that you have (a,b) create the point (b,-a).
now what you do is sub is points of line 2 into various equations until you find one that makes sense.
For every point (a,b) lying on line 1, point (b,-a) lies on line [#permalink] 28 Dec 2013, 23:57
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# m18#24
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1. # What Is The First Step When Constructing An Angle Bisector Using Only A Compass And A Straightedge?
Constructing an angle bisector using only a compass and a straightedge is possible, and it is one of the most commonly used constructions in mathematics. Constructing an angle bisector involves drawing two line segments of equal length, at right angles to each other, with one line segment forming the angle bisector. This construction can be used for many different purposes, such as determining the midpoint of a line segment or constructing a perpendicular line from an existing point. But what’s the first step when constructing an angle bisector using only a compass and a straightedge? In this article, we will discuss the steps involved in constructing an angle bisector with a compass and straightedge.
## Constructing an Angle Bisector
There are a few different ways to go about constructing an angle bisector, but the most common method is using a compass and straightedge. Here are the steps you need to take:
1. Place the point of your compass on one endpoint of the line segment whose angle you want to bisect.
2. Swing the compass around until the other endpoint of the line segment is inside the circle created by the compass.
3. Draw an arc that intersects the other endpoint of the line segment.
4. Place your compass point on one of the intersection points and swing the compass around until it intersects with the arc you just drew at another point.
5. Draw a line through this second intersection point and extend it until it meets up with the first endpoint of the line segment (the one you started with in Step 1). This new line is your angle bisector!
## The First Step
If you want to construct an angle bisector using only a compass and a straightedge, the first step is to draw a line segment. This line segment will be the base of your angle bisector. Next, use your compass to draw two arcs that intersect at two points on the line segment. These arcs should be equal in length. Finally, use your straightedge to connect the two points of intersection.
## Other Steps
There are a few other steps that are worth mentioning when constructing an angle bisector using only a compass and a straightedge. First, make sure the point you’re bisecting is on the line segment between the two points that define the angle. Next, draw a circle with the compass centered at one endpoint of the line segment. Finally, use the straightedge to draw a line through the point you’re bisecting and the point where the circle intersects the line segment.
## Conclusion
Constructing an angle bisector using only a compass and a straightedge is an important topic to understand when learning Geometry. The first step in constructing an angle bisector is to draw two rays that make up the given angle, and then use the compass to draw two arcs intersecting them. Finally, connect these discrete points with a straight line to form your desired angle bisector. With practice and knowledge of the steps involved, anyone can construct angles quickly and accurately by following this method.
2. When constructing an angle bisector using only a compass and a straightedge, the first step is to draw a line segment. This line segment will be the base of the angle bisector.
Next, set the compass to the length of the line segment and draw an arc that intersects each end of the line segment. Then, draw a line through the two points of intersection. This line will bisect the original angle.
The next step is to draw an arc that is the same length as the original line segment. This arc should intersect both of the points of the original line segment, but it should also intersect the line that was just drawn.
Finally, draw a line from the point of intersection between the two arcs to the midpoint of the original line segment. This line is the angle bisector.
By following these steps, you can construct an angle bisector using only a compass and a straightedge. It’s an easy and precise way to measure the angles of any shape. So, the next time you need to bisect an angle, give it a try!
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# NCERT solution class 12 chapter 5 Three Dimensional Geometry exercise 5.2 mathematics part 2
## EXERCISE 5.2
#### Question 1:
Show that the three lines with direction cosines
are mutually perpendicular.
Two lines with direction cosines, l1m1n1 and l2m2n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0
(i) For the lines with direction cosines, and , we obtain
Therefore, the lines are perpendicular.
(ii) For the lines with direction cosines, and , we obtain
Therefore, the lines are perpendicular.
(iii) For the lines with direction cosines, and , we obtain
Therefore, the lines are perpendicular.
Thus, all the lines are mutually perpendicular.
#### Question 2:
Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6).
The direction ratios, a1b1c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4.
The direction ratios, a2b2c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.
AB and CD will be perpendicular to each other, if a1a2 + b1b2c1c2 = 0
a1a2 + b1b2c1c2 = 2 × 3 + 5 × 2 + (− 4) × 4
= 6 + 10 − 16
= 0
Therefore, AB and CD are perpendicular to each other.
#### Question 3:
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).
Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5).
The directions ratios, a1b1c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4.
The direction ratios, a2b2c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4.
AB will be parallel to CD, if
Thus, AB is parallel to CD.
#### Question 4:
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector.
It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is
It is known that the line which passes through point A and parallel to is given by is a constant.
This is the required equation of the line.
#### Question 5:
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector and is in the direction .
It is given that the line passes through the point with position vector
It is known that a line through a point with position vector and parallel to is given by the equation,
This is the required equation of the line in vector form.
Eliminating λ, we obtain the Cartesian form equation as
This is the required equation of the given line in Cartesian form.
#### Question 6:
Find the Cartesian equation of the line which passes through the point
(−2, 4, −5) and parallel to the line given by
It is given that the line passes through the point (−2, 4, −5) and is parallel to
The direction ratios of the line, are 3, 5, and 6.
The required line is parallel to
Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0
It is known that the equation of the line through the point (x1y1z1) and with direction ratios, abc, is given by
Therefore the equation of the required line is
#### Question 7:
The Cartesian equation of a line is . Write its vector form.
The Cartesian equation of the line is
The given line passes through the point (5, −4, 6). The position vector of this point is
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of vector,
It is known that the line through position vector and in the direction of the vector is given by the equation,
This is the required equation of the given line in vector form.
#### Question 8:
Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
The required line passes through the origin. Therefore, its position vector is given by,
The direction ratios of the line through origin and (5, −2, 3) are
(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3
The line is parallel to the vector given by the equation,
The equation of the line in vector form through a point with position vector and parallel to is,
The equation of the line through the point (x1y1z1) and direction ratios abc is given by,
Therefore, the equation of the required line in the Cartesian form is
#### Question 9:
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.
Since PQ passes through P (3, −2, −5), its position vector is given by,
The direction ratios of PQ are given by,
(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11
The equation of the vector in the direction of PQ is
The equation of PQ in vector form is given by,
The equation of PQ in Cartesian form is
i.e.,
#### Question 10:
Find the angle between the following pairs of lines:
(i)
(ii) and
(i) Let Q be the angle between the given lines.
The angle between the given pairs of lines is given by,
The given lines are parallel to the vectors, and , respectively.
(ii) The given lines are parallel to the vectors, and , respectively.
#### Question 11:
Find the angle between the following pairs of lines:
(i)
(ii)
1. Let and be the vectors parallel to the pair of lines, respectively.
and
The angle, Q, between the given pair of lines is given by the relation,
(ii) Let be the vectors parallel to the given pair of lines, and , respectively.
If Q is the angle between the given pair of lines, then
#### Question 12:
Find the values of p so the line and
are at right angles.
The given equations can be written in the standard form as
and
The direction ratios of the lines are −3,, 2 and respectively.
Two lines with direction ratios, a1b1c1 and a2b2c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0
Thus, the value of p is .
#### Question 13:
Show that the lines and are perpendicular to each other.
The equations of the given lines areand
The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.
Two lines with direction ratios, a1b1c1 and a2b2c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0
∴ 7 × 1 + (−5) × 2 + 1 × 3
= 7 − 10 + 3
= 0
Therefore, the given lines are perpendicular to each other.
#### Question 14:
Find the shortest distance between the lines
The equations of the given lines are
It is known that the shortest distance between the lines, and , is given by,
d = b1→×b2→.a2→-a1→b1→×b2→
Comparing the given equations, we obtain
Substituting all the values in equation (1), we obtain
Therefore, the shortest distance between the two lines is units.
#### Question 15:
Find the shortest distance between the lines and
The given lines are and
It is known that the shortest distance between the two lines, is given by,
Comparing the given equations, we obtain
Substituting all the values in equation (1), we obtain
Since distance is always non-negative, the distance between the given lines is units.
#### Question 16:
Find the shortest distance between the lines whose vector equations are
The given lines are and
It is known that the shortest distance between the lines, and , is given by,
Comparing the given equations with and , we obtain
Substituting all the values in equation (1), we obtain
Therefore, the shortest distance between the two given lines is units.
#### Question 17:
Find the shortest distance between the lines whose vector equations are
The given lines are
r→=(s+1)i^+(2s-1)j^-(2s+1)k^⇒r→=(i^-j^-k^)+s(i^+2j^-2k^) …(2)It is known that the shortest distance between the lines, and , is given by,
For the given equations,
Substituting all the values in equation (3), we obtain
Therefore, the shortest distance between the lines isunits.
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# Lesson 1
Angles and Steepness
## 1.1: Ratios Galore (5 minutes)
### Warm-up
In the previous unit, students studied similar triangles and used the properties of rigid transformations and dilations to establish that in similar triangles:
• All pairs of corresponding angles are congruent.
• Lengths of all pairs of corresponding sides are proportional.
This warm-up helps students to recall that prior learning and to practice expressing ratios of side lengths in similar triangles and noticing they are equal. Monitor for students who use ratios comparing lengths within each triangle ($$\frac{AB}{AC} = \frac{DE}{DF}$$) and students who use ratios comparing lengths of corresponding sides ($$\frac{AB}{DE} = \frac{AC}{DF}$$).
### Student Facing
Triangle $$ABC$$ is similar to triangle $$DEF$$. Write as many equations as you can to describe the relationships between the sides and angles of the 2 triangles.
### Activity Synthesis
Invite students to share their equations using angles. Then invite students to share their equations using side lengths. Continue collecting examples until there are some ratios comparing lengths within each triangle and some ratios comparing lengths of corresponding sides.
## 1.2: Can You Calculate? (10 minutes)
### Activity
As students complete this activity, they ask themselves whether they have enough information to find unknown side lengths in right triangles, and if not, what information they might need. Students may notice that without two side lengths given, they can’t use the Pythagorean Theorem to find the length of the third side. In the synthesis, there is an opportunity to connect back to what students know about triangle similarity and congruence, and to preview the fact that while we don’t yet have enough information to find the length $$z$$, the length $$z$$ is fixed once we know the measures of angles $$G$$ and $$H$$ and the length of $$GH$$.
Monitor for students who:
• attempt to measure or estimate the length of $$z$$
• assume that angles $$H$$ and $$E$$ are congruent and use similarity to find the length of $$z$$ (the angles are not congruent, angle $$E$$ actually measures $$48^{\circ}$$)
### Student Facing
Find the values of $$x, y, \text{ and } z$$. If there is not enough information, what else do you need to know?
### Anticipated Misconceptions
If students are struggling to find $$x$$ or $$y$$ ask them what they notice about the triangles. (They are all right triangles.) If necessary, prompt them to check their reference chart for information about right triangles. (The Pythagorean Theorem applies.)
### Activity Synthesis
The purpose of this synthesis is to discuss whether it is possible to calculate $$z$$. Invite the previously selected students to share their attempts to calculate $$z$$.
Ask, “If we don’t have enough information to find the exact length of $$z$$, does that mean that $$z$$ could be any length?” (No, by the Angle-Side-Angle Triangle Congruence Theorem, any right triangle with a 50 degree angle and a side with length 3 between the known angles has to be congruent to this one. You can’t draw any other triangle with those measurements.)
If similarity is not mentioned by students, ask, “If we had a triangle with the same angles as triangle $$GHJ$$ but different side lengths, would that be helpful for finding $$z$$?” (Yes, then we could use the scale factor to find $$z$$.)
Speaking: MLR8 Discussion Supports. Use this routine to support whole-class discussion. For each student that shares, ask peers to restate what they heard using precise mathematical language. Consider providing students time to restate what they hear to a partner before selecting one or two students to share with the class. Ask the original speaker if their peer was accurately able to restate their thinking. Call students' attention to any words or phrases that helped to clarify the original statement. This provides more students with an opportunity to produce language as they interpret the reasoning of others.
Design Principle(s): Support sense-making
## 1.3: Is it Accessible? (20 minutes)
### Activity
In this activity, students are building skills that will help them in mathematical modeling (MP4). Students formulate a model by designing a ramp that they think will be accessible. Then they validate their results by comparing them to the Americans with Disabilities Act (ADA) guidelines. Students then use measurement and computation to determine if their ramp is acceptable and redesign it if necessary.
Monitor for the methods students choose to validate their design:
• Use the angle to check their ramps by measuring the angle with a protractor.
• Use the slope information by measuring to make sure that for every inch of vertical length, there are 12 inches of horizontal length.
• Construct one or more slope triangles along their ramp to demonstrate that it meets the guidelines.
Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).
### Launch
You may want to do an image search for “bad wheelchair ramp” to show some examples that are not safe to help students come up with good characteristics.
Arrange students in groups. After students make their design, distribute one copy of the ADA guidelines, cut from the blackline master, per group.
Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer interactions. Prior to the whole-class discussion, invite students to share their work with a partner. Display sentence frames to support student conversation such as: “First, I _____ because . . .”, “I noticed _____ so I . . .”, “Why did you . . .?”, “How did you get. . .?”
Supports accessibility for: Language; Social-emotional skills
### Student Facing
1. Some buildings offer ramps in addition to stairs so people in wheelchairs have access to the building. What characteristics make a ramp safe?
2. A school has 4 steps to the front door. Each step is 7 inches tall. Design a ramp for the school.
3. Your teacher will give you the Americans with Disabilities Act (ADA) guidelines. Does your design follow the rules of this law? If not, draw a new design that does.
### Student Facing
#### Are you ready for more?
A ramp with a length to height ratio of $$12:1$$ forms a right triangle with a 4.8 degree angle.
1. What is the angle measurement if the base is only 6 units long and the height is 1 unit tall?
2. When the length is half as long does that make the angle half as big?
3. What is the angle measurement if the base is 6 units long and the height is increased to 2 units tall?
4. When the height is twice as tall does that make the angle twice as big?
### Anticipated Misconceptions
Some students may be struggling to design a ramp. Ask them what shape is a good model for the side view of a ramp. (A right triangle.)
### Activity Synthesis
The goal of this discussion is to see that having a 4.8 degree angle and a ratio of vertical to horizontal length of $$1:12$$ are equivalent.
Display previously selected drawings of student ramp designs. Tell students, “Simplified diagrams of right triangles are a mathematical model of the cross section of a ramp.” This previews the work with cross sections that students will do in a subsequent unit but there’s no need to define cross section now.
Invite students who used the ratio of $$1:12$$ to check their ramps to share their process. Make sure that all students understand how to use the ratio to generate the horizontal length of the ramp given the vertical length, as students will need to do this in the cool-down.
Next, invite students who used the angle of 4.8 degrees to share.
Ask students to confirm that ramps in which students checked for a ratio of $$1:12$$ also have an angle of 4.8, and vice versa. Students should end convinced that having a 4.8 degree angle and a ratio of vertical to horizontal length of $$1:12$$ are equivalent. In the lesson synthesis, they will connect this concept to triangle similarity.
Representing, Conversing: MLR7 Compare and Connect. Use this routine to prepare students for the whole-class discussion. At the appropriate time, invite students to create a visual display that includes their ramp design and a justification for whether it meets ADA guidelines. Invite students to quietly circulate and read at least 2 other displays in the room. Next, ask students to find a new partner to discuss what is the same and what is different about the different methods groups chose to validate their designs. Listen for and amplify observations that include mathematical language and reasoning about angles and ratios of side lengths.
Design Principle(s): Cultivate conversation; Optimize output (for justification)
## Lesson Synthesis
### Lesson Synthesis
Ask students, “Will any ramp with one angle of 4.8 degrees have a slope ratio of $$1:12$$? Will any ramp with a slope ratio of $$1:12$$ have an angle of 4.8 degrees?” (The ramp makes a right triangle. If we know the measure of one angle is 4.8 degrees, then we know any ramps with those measurements will be similar right triangles by the Angle-Angle Triangle Similarity Theorem. So they will all have the same ratio of the vertical side to the horizontal side. If we know the ratio of the vertical side to the horizontal sides in the ramps are $$1:12$$, then we know that two pairs of sides are scaled copies of each other and the corresponding angles between them are congruent, so the triangles are similar by the Side-Angle-Side Triangle Similarity Theorem. So they will all have the same angles, because similar triangles have congruent corresponding angles.)
Ask students, “How do you think the people who wrote the Americans with Disabilities Act found the ratio of vertical length to horizontal length that is true for every 4.8 degree angle?” (They built ramps and measured them. They used math that we’re about to learn. They looked it up.)
“In the cool-down, you’ll see an example of another guideline for a different type of ramp with a different angle and ratio. Does every angle in a right triangle give you a fixed ratio of vertical length to horizontal length?” (Yes, because when you’re dealing with right triangles, all the triangles with the same angle measure for one of the other angles are similar, so they have the same ratios.)
## Student Lesson Summary
### Student Facing
Because of the Pythagorean Theorem, if we know any 2 sides of a right triangle, we can calculate the length of the third side. But what if we know a side and an angle rather than 2 sides?
All right triangles with one pair of congruent acute angles are similar by the Angle-Angle Triangle Similarity Theorem. Knowing just one side length in addition to those angle measures is enough to uniquely define the triangle.
The Americans with Disabilities Act includes guidelines for safe and accessible wheelchair ramps. Ramps must form a maximum 4.8 degree angle with the ground, which creates a maximum $$1:12$$ ratio for the legs of the right triangle.
Let's assume we are building a ramp for a 3 inch threshold.
To find length $$x$$ we can use similarity. By corresponding sides, $$\frac{1}{12}=\frac{3}{x}$$ so $$x$$ is 36 units. To find length $$y$$ we can use the Pythagorean Theorem, $$3^2+36^2=y^2$$. So $$y=\sqrt{1,\!305}$$ or about 36.1 units. To build a ramp that goes up 3 inches we need to start 36 inches, or 3 feet, out from the edge of the threshold and use a board that's about 36.1 inches long.
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# (a) A person is suffering from hypermetropia. The near point of the person is $1.5m$. Calculate the focal length of the convex lens used in his spectacles. Assume the near point of the normal eye is $25cm$.(b)What do you mean by the term range of vision? Also write its value.(c)What is the limit of accommodation of a healthy human eye?
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Hint: As we know Focal length, we can find by using lens maker formula. After finding the focal length we can easily calculate the value of power.People with farsightedness have powers with a positive sign. For example $+1.50$ or $+2.50$ as their corrective powers.
Formula used:
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
Here $f$ is the focal length, $u$ is the object distance and $v$ is the image distance.
Complete step by step answer:
(a) As we know that Hypermetropia is a condition of the eye where the person is not able to see the near things clearly. That person can see the farther objects very clearly. Also the normal eye point of the eye is$25cm$. For the normal view point, an object must be at a distance of at least $25cm$. Now, apply the formula,
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
Here f is the focal length, u is the object distance and v is the image distance.
Here virtual image is formed, so formula will become-
$\dfrac{1}{f} = \dfrac{1}{u} - \dfrac{1}{v}$
$\Rightarrow\dfrac{1}{f} = \dfrac{1}{{25}} - \dfrac{1}{{100}}$
By solving this, we get value of focal length, that is$= 33.3cm$
Now, for finding the power of the eye,
$\therefore P = \dfrac{{100}}{f}$
So, power comes out to be $3D$.
(b) The range of vision of a person is defined as the distance between the near point and far point from the eyes of that person.
Near point:The point which is at closer distance, by which we can see things clearly.
Far Point: The farther distance at which the object is seen clearly.
For a normal eye, the far point lies at infinity. So, a person with normal vision has an infinite range of vision.
(C) For a healthy human eye, power of accommodation is approximately $4D$.
Note:Hyperopia is when a person can see nearby objects very clearly and cannot see farther objects. In this question, a virtual image is formed, remembering to substitute the value of $v$ by using sign convention.It is important to remember that Hyperopia is an eye defect and not an eye disease. Therefore, it can be corrected.
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# How to Write a C Program to Check if Three Values Can Form a Triangle
write a C program that reads three floating-point values and checks if is possible to make a triangle with them. determine the perimeter of triangle
In this program, you will learn how to write a C program that reads three floating-point values and checks if it is possible to make a triangle with them. You will also learn how to determine the perimeter of the triangle if the given values are valid.
#### What is a Triangle?
A triangle is a polygon with three sides and three angles. The sum of the angles of a triangle is always 180 degrees. The length of each side of a triangle can vary, but there is a condition that must be satisfied for any three values to form a valid triangle. This condition is called the triangle inequality theorem.
#### What is the Triangle Inequality Theorem?
The triangle inequality theorem states that for any three values a, b, and c, they can form a valid triangle only if the following inequalities hold:
• a + b > c
• b + c > a
• c + a > b
These inequalities mean that the sum of any two sides of a triangle must be greater than the third side. This ensures that the sides can meet at the vertices and form a closed shape. If any of these inequalities is violated, then the values cannot form a triangle.
For example, suppose we have three values 5, 7, and 10. We can check if they can form a triangle by applying the triangle inequality theorem:
• 5 + 7 > 10 (True)
• 7 + 10 > 5 (True)
• 10 + 5 > 7 (True)
Since all the inequalities are true, we can conclude that these values can form a valid triangle.
However, suppose we have three values 3, 4, and 9. We can check if they can form a triangle by applying the triangle inequality theorem:
• 3 + 4 > 9 (False)
• 4 + 9 > 3 (True)
• 9 + 3 > 4 (True)
Since one of the inequalities is false, we can conclude that these values cannot form a valid triangle.
#### C Program
Now that we understand the concept of triangles and the triangle inequality theorem, we can write the C program to implement it. The program will do the following steps:
• Declare three floating-point variables to store the input values
• Prompt the user to enter three values
• Read the input values and store them in the variables
• Check if the values are valid for a triangle using the triangle inequality theorem
• If valid, calculate the perimeter of the triangle and print it
• If not valid, print an error message
The code for the program is shown below:
#include <stdio.h>
int main()
{
float a, b, c;
printf("Enter three floating-point values:\n");
scanf("%f %f %f", &a, &b, &c);
if (a + b > c && b + c > a && c + a > b)
{
float perimeter = a + b + c;
printf("The perimeter of the triangle is %.2f\n", perimeter);
}
else
{
printf("The given values cannot form a triangle.\n");
}
return 0;
}
#### Output
To run and test the program, you need to compile it using a C compiler and execute it on your terminal or command prompt. You can use any online C compiler or IDE to run the program as well.
Here are some sample inputs and outputs of the program:
Enter three floating-point values:
5.6 7.8 9.2
The perimeter of the triangle is 22.60
Enter three floating-point values:
3.4 4.5 9.8
The given values cannot form a triangle.
Enter three floating-point values:
6.7 8.9 10.1
The perimeter of the triangle is 25.70
#### Explanation
The explanation of the above program is as follows:
1. The program starts by including the standard input/output header file 'stdio.h' which provides the functions 'printf' and 'scanf' for printing and reading data.
2. The program then defines the main function which is the entry point of the program. The main function returns an integer value which indicates the status of the program execution.
3. Inside the main function, the program declares three floating-point variables 'a', 'b', and 'c' to store the input values. Floating-point variables can store decimal numbers with fractional parts.
4. The program then prints a message to prompt the user to enter three values using the 'printf' function. The '\n' character is used to move to a new line after printing the message.
5. The program then reads the input values from the user using the 'scanf' function. The 'scanf' function takes a format string and a list of pointers to variables as arguments. The format string specifies the type and format of the input data. The '%f' specifier is used to indicate that the input data is a floating-point value. The '&' operator is used to get the address of the variables where the input data will be stored. The 'scanf' function will read three values from the user and store them in the variables 'a', 'b', and 'c'.
6. The program then checks if the values are valid for a triangle using the triangle inequality theorem. The triangle inequality theorem states that for any three values a, b, and c, they can form a valid triangle only if a + b > c, b + c > a, and c + a > b. The program uses the logical AND operator '&&' to combine these three inequalities into one condition. If this condition is true, then the program executes the statements inside the if block. Otherwise, it executes the statements inside the else block.
7. If the condition is true, then the program calculates the perimeter of the triangle by adding up the values of 'a', 'b', and 'c'. The program stores this value in another floating-point variable called 'perimeter'. The program then prints this value using the 'printf' function. The '%f' specifier is used to indicate that the value is a floating-point number. The '.2' modifier is used to limit the number of decimal places to two. The '\n' character is used to move to a new line after printing the value.
8. If the condition is false, then the program prints an error message using the 'printf' function. The error message informs the user that the given values cannot form a triangle.
9. Finally, the program returns 0 from the main function which indicates that it has executed successfully.
This is how the above program works and what it does.
#### Conclusion
In this program, you learned how to write a C program that reads three floating-point values and checks if it is possible to make a triangle with them. You also learned how to determine the perimeter of the triangle if the given values are valid. You also learned about the concept of triangles and the triangle inequality theorem. You can use this program to practice your C programming skills and learn more about geometry and logic. You can also modify the program to calculate other properties of triangles, such as area, angles, or types. Have fun coding!
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# CBSE Class 8 Math, Mensuration
Class VIII Math
Notes for Mensuration
STANDARD UMTS OF VOLUME
1 dm = 10 cm
1 dm3 or litre = 1 dm × 1 dm × 1 dm = (10 × 10 × 10) cm3 = 1000 cm3
1 cm = 10 mm
1m =100 cm
1m3 = (100 × 100 × 100) cm3 = 1000000 cm3
1 m3 = 1000 × 1000 cm3 = 1000 litre
1 kilolitre = 1m3 = 1000 litre
(mm) millimeter, ( cm) centimetre, (dm) decimetre, (m) metre
Plans figures
The geometrical figures which have only two dimensions are called as the plane figures.
Booster 1
A square with sides of 1 cm has an area of 1 cm2.
Find the area of the shaded shape.
Explanation
The shape covers 11 squares, so its area is 11 cm2.
Booster 2
Find the area of the shaded triangle.
Explanation
The triangle covers 6 full squares marked F, and 4 half squares marked H. Area = 6 + 2 = 8 cm2.
Booster 3
Estimate the area of the shape shaded in the diagram.
Explanation
This is a much more complicated problem as there are only 9 full squares marked F, but many other part square. You need to combine part squares that approximately make a whole square. For example, the squares marked make about 1 full square; the squares marked � make about 1 full square; the squares marked + make about 1 full square; the squares marked • make about 1 full square. Thus the total area is approximately
9 + 4 = 13 cm2.
Triangle
(i)
Area of triangle
(ii) Area of an equilateral triangle
(iii) Area of an isosceles triangle :- base = b, equal side = a
Booster 4
The parallel sides of a trapezium are 20 cm and 10 cm. Its non-parallel sides are both equal, each being 13 cm. Find the area of the trapezium.
Explanation
Let ABCD be a trapezium such that,
AB = 20 cm, CD = 10 cm and AD = BC = 13 cm
Draw CL || AD and CM || AB.
Now, CL || AD and CD || AB.
∴ ALCD is a parallogram.
⇒ AL = CD = 10 cm and CL = AD = 13 cm
In ΔCLB, we have CL = CB = 13 cm
∴ΔCLB is an isosceles triangle.
[∵BL= AB – AL = (20 – 10) cm = 10 cm]
Applying pythagoras theorem in δCML, we have
CL2 = CM2 + LM2
132 = CM2 + 52
CM2 = 169 – 25 = 144
Area of parallelogram ALCD = AL × CM = (10 × 12) = 120 cm2
Hence, Area of trapezium ABCD = Area of parallelogram ALCD + Area of δCLB = (120 + 60) cm2 = 180 cm2
Booster 5
If the area of a rhombus be 24 cm2 and one of its diagonals be 4 cm, find the perimeter of the rhombus.
Explanation
Let ABCD be a rhombus such that its one diagonal AC = 4 cm. Suppose the diagonals AC and BD intersect at O.
Area of rhombus ABCD = 24 cm2
Thus, we have AC = 4 cm and BD = 12 cm
Since the diagonals of a rhombus bisect each other at right angle. Therefore, δOAB is right triangle, right angled at 0.
Using pythagoras theorem in δOAB, we have
AB2 = OA2 + OB2
AB2 = 22 + 62 = 40
Hence, perimeter of rhombus ABCD
Booster 6
Find the area of a regular octagon each of whose sides measures 4 cm.
Explanation
Area of the octagon
Booster 7
The diagram shows a lorry.
Find the volume of the load-carrying part of the lorry.
Explanation
The load-carrying part of the lorry is represented by a cuboid, so its volume is given by
V = 2 × 2.5 × 4 = 20m3.
Cube
A cuboid whose length, breadth and height are all equal is called a cube.
(i) Surface area of a cube . Since alI the faces of a cube are squares of the same size i.e., for a cube we have l = b = h. Thus if l cm is the length of the edge of side of a cube, then
Total Surface area of the cube = 2(l × l + l × l + l × l)
= 2 × 3l2 = 6l2 = 6(Edge)2
(ii) Lateral surface area of the cube = 2(l × l + l × l)
= 2(l2 + l2) = 4l2 = 4(Fdge)2
(iii) Volume of a cube = l × l × l = l3
Booster 8
Find the total surface of a hollow cylinder open at ends, if the length is 12 cm, the external diameter 10 cm and thickness 2 cm.
Explanation
The outer radius (R) = 5 cm, Thickness = 2 cm
∴. Inner radius (r) = 5 cm – 2 cm = 3 cm
Outer curved surface of the cylinder
= 2πrh = 2 × π × 5 × 12 = 120πcm2
Inner curved surface of the cylinder 27πrh
= 2 × π × 3 × 12 = 72 π cm2.
Both ends of the cylinder will be of the shape, as shown in figure (ii).
∴. Area of one of end of the cylinder = πR2 – πr2
= π × 52 – π × 32 = 25π – 9π = 16π cm2.
Area of both ends = (2 × 16π) cm2 = 32π cm2
∴Total surface of the cylinder = External curved surface + Internal curved surf ace + 2 (Area of the base of the ring)
Remark : It is advisible to put the value of π is the end in such calculations.
Circular cone
There are many objects around us which are conical in shape like an ice-cream cone, a conical, tent a birthday cap etc. These objects are of the shape of a right circular cone.
δAOC is right angled at O.
By Pythagoras theorem,
AC2 = AO2 + OC2
i.e. l2 = h2 + r2
(ii) Total surface area of cone = Curved surface area of cone + Area of the circular base
= πrl + πr2
= πr (r + l) sq. units.
(iii) Volume of a cone (where ‘r’ r is the radius and ‘h’ is the height of cone).
Sphere and hemisphere
Sphere: The set of all the points in space which are equidistant from a fixed point is called a sphere.
Hemisphere: A lane through the centre of a sphere divides the sphere into two equal parts and each part is called a hemisphere.
(i) Curved surface area of hemisphere = 2πr2
(ii) Total surface area of hemisphere 2πr2 + πr2 = 3πr2
Booster 9
If radius of the base of a cone is 140 dm and its slant height is 9 m. Find the
(i) curved surface area
(ii) total surface area
Explanation
Radius of the base of the cone (r) = 140 dm = 14 m
Slant height (l) = 9 m
(i) Curved surf ace area of the cone
(ii) Total surface area of cone = πrl + πr2
Question 12
The surface area of a sphere is 2464 dm2. Find its diameter.
Solution
Surface area of a sphere = 2464 dm2
Therefore, 4πr2 = 2464
Diameter of the given sphere = 2 × 14 = 28 dm.
Booster 10
The dome of a building is in the form of a hemisphere of radius 63 dm.
Find the cost if it is to be painted at the rate of Rs 5 per m2.
Explanation
Radius of the hemisphere = 63 dm
Surface area of hemisphere = 2πr2
Booster 11
The earth taken out while digging a pit, is evenly sprea over a rectangular field of length 90 m, width 60 m.
If the volume of the earth dug is 3078 m3, find the height of the field raised.
Explanation
3078 m3 = 90 m × 60m × h
height of field raised = 0.57 m
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# Commutative property
In mathematics, a binary operation is commutative if changing the order of the operands does not change the result. It is a fundamental property of many binary operations, and many mathematical proofs depend on it. Perhaps most familiar as a property of arithmetic, e.g. "3 + 4 = 4 + 3" or "2 × 5 = 5 × 2", the property can also be used in more advanced settings. The name is needed because there are operations, such as division and subtraction, that do not have it (for example, "3 − 5 ≠ 5 − 3"); such operations are not commutative, and so are referred to as noncommutative operations. The idea that simple operations, such as the multiplication and addition of numbers, are commutative was for many years implicitly assumed. Thus, this property was not named until the 19th century, when mathematics started to become formalized.[1][2] A similar property exists for binary relations; a binary relation is said to be symmetric if the relation applies regardless of the order of its operands; for example, equality is symmetric as two equal mathematical objects are equal regardless of their order.[3]
Type Property Algebra A binary operation is commutative if changing the order of the operands does not change the result. ${\displaystyle x*y=y*x\quad \forall x,y\in S.}$
## Mathematical definitions
A binary operation ${\displaystyle *}$ on a set S is called commutative if[4][5]
${\displaystyle x*y=y*x\qquad {\mbox{for all }}x,y\in S.}$
In other words, an operation is commutative if every two elements commute. An operation that does not satisfy the above property is called noncommutative.
One says that x commutes with y or that x and y commute under ${\displaystyle *}$ if
${\displaystyle x*y=y*x.}$
That is, a specific pair of elements may commute even if the operation is (strictly) noncommutative.
## Examples
### Noncommutative operations
Some noncommutative binary operations:[6]
#### Division, subtraction, and exponentiation
Division is noncommutative, since ${\displaystyle 1\div 2\neq 2\div 1}$ .
Subtraction is noncommutative, since ${\displaystyle 0-1\neq 1-0}$ . However it is classified more precisely as anti-commutative, since ${\displaystyle 0-1=-(1-0)}$ .
Exponentiation is noncommutative, since ${\displaystyle 2^{3}\neq 3^{2}}$ . This property leads to two different "inverse" operations of exponentiation (namely, the nth-root operation and the logarithm operation), which is unlike the multiplication. [7]
#### Truth functions
Some truth functions are noncommutative, since the truth tables for the functions are different when one changes the order of the operands. For example, the truth tables for (A ⇒ B) = (¬A ∨ B) and (B ⇒ A) = (A ∨ ¬B) are
A B A ⇒ B B ⇒ A
F F T T
F T T F
T F F T
T T T T
#### Function composition of linear functions
Function composition of linear functions from the real numbers to the real numbers is almost always noncommutative. For example, let ${\displaystyle f(x)=2x+1}$ and ${\displaystyle g(x)=3x+7}$ . Then
${\displaystyle (f\circ g)(x)=f(g(x))=2(3x+7)+1=6x+15}$
and
${\displaystyle (g\circ f)(x)=g(f(x))=3(2x+1)+7=6x+10}$
This also applies more generally for linear and affine transformations from a vector space to itself (see below for the Matrix representation).
#### Matrix multiplication
Matrix multiplication of square matrices is almost always noncommutative, for example:
${\displaystyle {\begin{bmatrix}0&2\\0&1\end{bmatrix}}={\begin{bmatrix}1&1\\0&1\end{bmatrix}}{\begin{bmatrix}0&1\\0&1\end{bmatrix}}\neq {\begin{bmatrix}0&1\\0&1\end{bmatrix}}{\begin{bmatrix}1&1\\0&1\end{bmatrix}}={\begin{bmatrix}0&1\\0&1\end{bmatrix}}}$
#### Vector product
The vector product (or cross product) of two vectors in three dimensions is anti-commutative; i.e., b × a = −(a × b).
## History and etymology
Records of the implicit use of the commutative property go back to ancient times. The Egyptians used the commutative property of multiplication to simplify computing products.[8][9] Euclid is known to have assumed the commutative property of multiplication in his book Elements.[10] Formal uses of the commutative property arose in the late 18th and early 19th centuries, when mathematicians began to work on a theory of functions. Today the commutative property is a well-known and basic property used in most branches of mathematics.
The first recorded use of the term commutative was in a memoir by François Servois in 1814,[1][11] which used the word commutatives when describing functions that have what is now called the commutative property. Commutative is the feminine form of the French adjective commutatif, which is derived from the French noun commutation and the French verb commuter, meaning "to exchange" or "to switch", a cognate of to commute. The term then appeared in English in 1838.[2] in Duncan Gregory's article entitled "On the real nature of symbolical algebra" published in 1840 in the Transactions of the Royal Society of Edinburgh.[12]
## Propositional logic
### Rule of replacement
In truth-functional propositional logic, commutation,[13][14] or commutativity[15] refer to two valid rules of replacement. The rules allow one to transpose propositional variables within logical expressions in logical proofs. The rules are:
${\displaystyle (P\lor Q)\Leftrightarrow (Q\lor P)}$
and
${\displaystyle (P\land Q)\Leftrightarrow (Q\land P)}$
where "${\displaystyle \Leftrightarrow }$ " is a metalogical symbol representing "can be replaced in a proof with".
### Truth functional connectives
Commutativity is a property of some logical connectives of truth functional propositional logic. The following logical equivalences demonstrate that commutativity is a property of particular connectives. The following are truth-functional tautologies.
Commutativity of conjunction
${\displaystyle (P\land Q)\leftrightarrow (Q\land P)}$
Commutativity of disjunction
${\displaystyle (P\lor Q)\leftrightarrow (Q\lor P)}$
Commutativity of implication (also called the law of permutation)
${\displaystyle {\big (}P\to (Q\to R){\big )}\leftrightarrow {\big (}Q\to (P\to R){\big )}}$
Commutativity of equivalence (also called the complete commutative law of equivalence)
${\displaystyle (P\leftrightarrow Q)\leftrightarrow (Q\leftrightarrow P)}$
## Set theory
In group and set theory, many algebraic structures are called commutative when certain operands satisfy the commutative property. In higher branches of mathematics, such as analysis and linear algebra the commutativity of well-known operations (such as addition and multiplication on real and complex numbers) is often used (or implicitly assumed) in proofs.[16][17][18]
## Mathematical structures and commutativity
### Associativity
The associative property is closely related to the commutative property. The associative property of an expression containing two or more occurrences of the same operator states that the order operations are performed in does not affect the final result, as long as the order of terms does not change. In contrast, the commutative property states that the order of the terms does not affect the final result.
Most commutative operations encountered in practice are also associative. However, commutativity does not imply associativity. A counterexample is the function
${\displaystyle f(x,y)={\frac {x+y}{2}},}$
which is clearly commutative (interchanging x and y does not affect the result), but it is not associative (since, for example, ${\displaystyle f(-4,f(0,+4))=-1}$ but ${\displaystyle f(f(-4,0),+4)=+1}$ ). More such examples may be found in commutative non-associative magmas. Furthermore, associativity does not imply commutativity either – for example multiplication of quaternions or of matrices is always associative but not always commutative.
### Symmetry
Some forms of symmetry can be directly linked to commutativity. When a commutative operation is written as a binary function ${\displaystyle z=f(x,y),}$ then this function is called a symmetric function, and its graph in three-dimensional space is symmetric across the plane ${\displaystyle y=x}$ . For example, if the function f is defined as ${\displaystyle f(x,y)=x+y}$ then ${\displaystyle f}$ is a symmetric function.
For relations, a symmetric relation is analogous to a commutative operation, in that if a relation R is symmetric, then ${\displaystyle aRb\Leftrightarrow bRa}$ .
## Non-commuting operators in quantum mechanics
In quantum mechanics as formulated by Schrödinger, physical variables are represented by linear operators such as ${\displaystyle x}$ (meaning multiply by ${\displaystyle x}$ ), and ${\textstyle {\frac {d}{dx}}}$ . These two operators do not commute as may be seen by considering the effect of their compositions ${\textstyle x{\frac {d}{dx}}}$ and ${\textstyle {\frac {d}{dx}}x}$ (also called products of operators) on a one-dimensional wave function ${\displaystyle \psi (x)}$ :
${\displaystyle x\cdot {\mathrm {d} \over \mathrm {d} x}\psi =x\cdot \psi '\ \neq \ \psi +x\cdot \psi '={\mathrm {d} \over \mathrm {d} x}\left(x\cdot \psi \right)}$
According to the uncertainty principle of Heisenberg, if the two operators representing a pair of variables do not commute, then that pair of variables are mutually complementary, which means they cannot be simultaneously measured or known precisely. For example, the position and the linear momentum in the ${\displaystyle x}$ -direction of a particle are represented by the operators ${\displaystyle x}$ and ${\displaystyle -i\hbar {\frac {\partial }{\partial x}}}$ , respectively (where ${\displaystyle \hbar }$ is the reduced Planck constant). This is the same example except for the constant ${\displaystyle -i\hbar }$ , so again the operators do not commute and the physical meaning is that the position and linear momentum in a given direction are complementary.
## Notes
1. ^ a b Cabillón & Miller, Commutative and Distributive
2. ^ a b Flood, Raymond; Rice, Adrian; Wilson, Robin, eds. (2011). Mathematics in Victorian Britain. Oxford University Press. p. 4. ISBN 9780191627941.
3. ^
4. ^ Krowne, p. 1
5. ^ Weisstein, Commute, p. 1
6. ^ Yark, p. 1
7. ^ "User MathematicalOrchid". Mathematics Stack Exchange. Retrieved 20 January 2024.
8. ^ Lumpkin 1997, p. 11
9. ^ Gay & Shute 1987
10. ^ O'Conner & Robertson Real Numbers
11. ^ O'Conner & Robertson, Servois
12. ^ Gregory, D. F. (1840). "On the real nature of symbolical algebra". Transactions of the Royal Society of Edinburgh. 14: 208–216.
13. ^ Moore and Parker
14. ^ Copi & Cohen 2005
15. ^ Hurley & Watson 2016
16. ^ Axler 1997, p. 2
17. ^ a b Gallian 2006, p. 34
18. ^ Gallian 2006, pp. 26, 87
19. ^ Gallian 2006, p. 236
20. ^ Gallian 2006, p. 250
## References
### Books
• Axler, Sheldon (1997). Linear Algebra Done Right, 2e. Springer. ISBN 0-387-98258-2.
Abstract algebra theory. Covers commutativity in that context. Uses property throughout book.
• Copi, Irving M.; Cohen, Carl (2005). Introduction to Logic (12th ed.). Prentice Hall. ISBN 9780131898349.
• Gallian, Joseph (2006). Contemporary Abstract Algebra (6e ed.). Houghton Mifflin. ISBN 0-618-51471-6.
Linear algebra theory. Explains commutativity in chapter 1, uses it throughout.
• Goodman, Frederick (2003). Algebra: Abstract and Concrete, Stressing Symmetry (2e ed.). Prentice Hall. ISBN 0-13-067342-0.
Abstract algebra theory. Uses commutativity property throughout book.
• Hurley, Patrick J.; Watson, Lori (2016). A Concise Introduction to Logic (12th ed.). Cengage Learning. ISBN 978-1-337-51478-1.
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Construct a 2 x 2 matrix $$A = [ a_{ij} ]$$, whose elements are given by $$a_{ij}$$ is $$\large\frac{( \hat i + \hat j )^2}{2}$$
Toolbox:
• The number or functions occurring in the rectangular array are called the elements of the matrix or entries of the matrix. The element of the matrix is given by ith row and jth column.
• 2 x 2 matrix = $A=\begin{bmatrix}a{11} & a{12}\\a{21} & a{22}\end{bmatrix}$
Step1:
Given:
$a_{ij}=\frac{(i+j)^2}{2}$
$a_{11}$ where i=1,j=1
$a_{11}=\frac{(1+1)^2}{2}=\frac{2^2}{2}=\frac{4}{2}=2.$
$a_{12}$ where i=1,j=2
$a_{11}=\frac{(1+2)^2}{2}=\frac{3^2}{2}=\frac{9}{2}.$
Step 2:
$a_{21}$ where i=2,j=1
$a_{21}=\frac{(2+1)^2}{2}=\frac{3^2}{2}=\frac{9}{2}.$
$a_{22}$ where i=2,j=2
$a_{22}=\frac{(2+2)^2}{2}=\frac{4^2}{2}=\frac{16}{2}=8.$
Step3:
Hence $A=\begin{bmatrix}2 & \frac{9}{2}\\\frac{9}{2} & 8\end{bmatrix}$
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Edit Article
# How to Multiply Two Digit by Two Digit Numbers (Using the Count to 99 on Your Fingers Method)
Community Q&A
This method will allow you to multiply two digit numbers without using more than one line. If you do not know how to count to 99 on your fingers, use the wikihow link below.
## Steps
1. 1
Write down this practice problem: 23 times 57 *23 x57
2. 2
Multiply 3 x 7 = 21. Place your hands in the "21" position, putting down your right index finger and your index and middle fingers on your left hand.
3. 3
Write down "1" (from your right hand) in the right -hand-most position under the problem. Clear your right hand.
4. 4
Carry the "2" by moving the 20 (from the left hand) to the "2" position on the right. Clear your left hand.
5. 5
Multiply 7 x 2 = 14. ADD 14 to the 2 already in your hands. Your hands should be in the "16" position.
6. 6
Multiply 3 x 5 = 15. ADD 15 to the 16 already in your hands. Your hands should now be in the "31" position.
7. 7
Write down the "1" (from your right hand) in the position left of the "1" you wrote down earlier.
8. 8
Carry the "30" in your left hand to the "3" in your right hand. Clear your left hand.
9. 9
Multiply 2 x 5 = 10. ADD "10" to the "3". Your hands should now be in the "13" position.
10. 10
Write down "13" in the position to the left of the number "11" you have already written. Clear your hands! You should have 1311. It's OK if you don't get it right away. Chisanbop takes practice.
## How to deal with numbers greater than 100
1. Consider the problem 79 x 86:
*79
x86
2. Multiply 9 x 6 = 54. Write down the "4" and you have already moved the "5" to the right hand.
3. Multiply 8 x 9 = 72. You add 72 to the 5 already in your hands. Your hands are in the "77" position.
• Uh Oh! We have multiplied 6 x 7 = 42, but we do not have a way to exceed 99!
5. Write down the "9" and clear the right hand.
6. Carry the "7" in the left hand to the right hand.
8. Multiply 7 x 8 = 56. Add this to the "11" already in your hands. Your hands should be in the "67" position.
9. Write down the 67. The answer is 6794!
## Article Info
Categories: Multiplication and Division
In other languages:
Thanks to all authors for creating a page that has been read 62,547 times.
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## Probability
A measure of the likelihood that a specific event will occur is the probability of that event. Before determining the probability of an event, students must be able to find the total number of possible outcomes. The total number of possible outcomes can be found by making an organized list, by making a tree diagram, or by using multiplication. If there are m possible choices, or outcomes, for one experiment or problem and n possible choices or outcomes for a second experiment or problem, then the total number of possible choices or outcomes is m × n. If you have 3 sizes of shirts and 2 choices of color, there are two possible colors for each of the three sizes, which means there are six possible outcomes because 3 × 2 = 6. Multiplication can be used when you only need to find the total number of possible outcomes. This is known as the fundamental counting principle.
The result of an experiment is called an outcome. If the experiment is to toss a 1−6 number cube, then there are six possible outcomes, one for each face of the cube. An event is any collection of outcomes. Examples of events for tossing a number cube are that the number tossed is even, that the number is 1 or 2, or that the number is 3. The probability of an event is a measure of the likelihood that an event will occur. The probability is always a number between 0 and 1. A probability of 0 means that an event is impossible while a probability of 1 means that an event is certain.
When a coin is tossed, there are two outcomes, heads or tails. Either outcome is equally likely. When a 1−6 number cube is tossed, each face is equally likely to turn up. When a marble is chosen from a bag of thoroughly mixed marbles of the same size, without looking, each marble has the same chance of being chosen. Such outcomes are said to be equally likely outcomes. Sometimes the word fair is used when the outcomes are equally likely, as in “fair coin” or “fair number cube.” To indicate that each outcome is equally likely, the word random is used, such as in saying “The object is chosen at random.”
If each outcome is equally likely, the theoretical probability of an event is the ratio of the number of outcomes in the event to the total number of possible outcomes. At this grade level, students will determine simple probability by finding the ratio of favorable outcomes to all outcomes.
Teaching Model 23.4: Problem-Solving Strategy: Make an Organized List
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NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6
# NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6 are the part of NCERT Solutions for Class 6 Maths (Rationalised Contents). Here you can find the NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6.
## NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6 (Rationalised Contents)
### Ex 3.6 Class 6 Maths Question 1.
Find the HCF of the following numbers:
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75
Solution:
(a) The given numbers are 18 and 48.
Prime factorisations of 18 and 48 are:
Here, the common factors are 2 and 3.
Hence, the HCF = 2 × 3 = 6
(b) The given numbers are 30 and 42.
Prime factorisations of 30 and 42 are:
Here, the common factors are 2 and 3.
Hence, the HCF = 2 × 3 = 6
(c) The given numbers are 18 and 60.
Prime factorisations of 18 and 60 are:
Here, the common factors are 2 and 3.
Hence, the HCF of 18 and 60 = 2 × 3 = 6
(d) The given numbers are 27 and 63.
Prime factorisations of 27 and 63 are:
Here, the common factor is 3 × 3.
Hence, the HCF = 3 × 3 = 9
(e) The given numbers are 36 and 84.
Prime factorisations of 36 and 84 are:
Here, the common factors are 2, 2 and 3.
Hence, the HCF = 2 × 2 × 3 = 12
(f) The given numbers are 34 and 102.
Prime factorisations of 34 and 102 are:
Here, the common factors are 2 and 17.
Thus, the HCF = 2 × 17 = 34
(g) The given numbers are 70, 105 and 175.
Prime factorisations of 70, 105 and 175 are:
Here, common factors are 5 and 7.
Hence, the HCF of 70, 105 and 175 is 5 × 7 = 35
(h) The given numbers are 91, 112 and 49.
Prime factorisations of 91, 112 and 49 are:
Here, the common factor is 7.
Hence, the HCF = 7
(i) The given numbers are 18, 54 and 81.
Prime factorisations of 18, 54 and 81 are:
Here, the common factor is 3 × 3.
Thus, the HCF = 3 × 3 = 9
(j) The given numbers are 12, 45 and 75.
Prime factorisations of 12, 45 and 75 are:
Here, the common factor is 3.
Hence, the HCF = 3
### Ex 3.6 Class 6 Maths Question 2.
What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution:
(a) The common factor of two consecutive numbers is always 1.
Hence, the HCF of two consecutive numbers = 1
(b) The common factors of two consecutive even numbers are 1 and 2.
Hence, the HCF of two consecutive even numbers = 1 × 2 = 2
(c) The common factor of two consecutive odd numbers is 1.
Hence, the HCF of two consecutive odd numbers = 1
### Ex 3.6 Class 6 Maths Question 3.
HCF of co-prime numbers 4 and 15 was found as follows by factorisation:
4 = 2 × 2 and 15 = 3 × 5. Since there is no common prime factors, so HCF of 4 and 15 is 0.
Is the answer correct? If not, what is the correct HCF?
Solution:
No, the answer is not correct.
Reason: 0 is not the prime factor of any number.
1 is always the prime factor of co-prime numbers.
Hence, the correct HCF of 4 and 15 is 1.
NCERT Solutions for Maths Class 7
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12
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# wikiHow to Find the Determinant of a 3X3 Matrix
The determinant of a matrix is frequently used in calculus, linear algebra, and higher level geometry. Outside the academic world, engineers and computer graphics programmers use matrices and their determinants all the time.[1] To find the determinant of a 3x3 matrix, read this wikiHow.
## 10 Second Summary
1. Write your 3 x 3 matrix. More ↓
2. Choose a single row or column.
3. Cross out the row and column of your first element.
4. Find the determinant of the 2 x 2 matrix.
7. Repeat this process for the second element in your reference row or column.
8. Repeat with the third element.
### Part 1 Finding the Determinant
1. 1
Write your 3 x 3 matrix. We'll start with a 3 x 3 matrix A, and try to find its determinant |A|. Here's the general matrix notation we'll be using, and our example matrix:
• ${\displaystyle M={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}={\begin{pmatrix}1&5&3\\2&4&7\\4&6&2\end{pmatrix}}}$
2. 2
Choose a single row or column. This will be your reference row or column. You'll get the same answer no matter which one you choose. For now, just pick the first row. Later, we'll give some advice on how to choose the easiest option to calculate.
• Let's choose the first row of our example matrix A. Circle the 1 5 3. In general terms, circle a11 a12 a13.
3. 3
Cross out the row and column of your first element. Look at the row or column you circled and select the first element. Draw a line through its row and column. You should be left with four numbers. We'll treat these as a 2 x 2 matrix.
• In our example, our reference row is 1 5 3. The first element is in row 1 and column 1. Cross out all of row 1 and column 1. Write the remaining elements as a 2 x 2 matrix:
• 1 5 3
2 4 7
4 6 2
4. 4
Find the determinant of the 2 x 2 matrix. Remember, the matrix ${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}}$ has a determinant of ad - bc.[2] You may have learned this by drawing an X across the 2 x 2 matrix. Multiply the two numbers connected by the \ of the X. Then subtract the product of the two numbers connected by the /. Use this formula to calculate the determinate of the matrix you just found.
• In our example, the determinant of the matrix ${\displaystyle {\begin{pmatrix}4&7\\6&2\end{pmatrix}}}$ = 4 * 2 - 7 * 6 = -34.
• This determinant is called the minor of the element we chose in our original matrix.[3] In this case, we just found the minor of a11.
5. 5
Multiply the answer by your chosen element. Remember, you selected an element from your reference row (or column) when you decided which row and column to cross out. Multiply this element by the determinant you just calculated for the 2x2 matrix.
• In our example, we selected a11, which had a value of 1. Multiply this by -34 (the determinant of the 2x2) to get 1*-34 = -34.
6. 6
Determine the sign of your answer. Next, you'll multiply your answer either by 1 or by -1 to get the cofactor of your chosen element. Which you use depends on where the element was placed in the 3x3 matrix. Memorize this simple sign chart to track which element causes which:
• + - +
- + -
+ - +
• Since we chose a11, marked with a +, we multiply the number by +1. (In other words, leave it alone.) The answer is still -34.
• Alternatively, you can find the sign with the formula (-1)i+j, where i and j are the element's row and column.[4]
7. 7
Repeat this process for the second element in your reference row or column. Return to the original 3x3 matrix, with the row or column you circled earlier. Repeat the same process with this element:
• Cross out the row and column of that element. In our case, select element a12 (with a value of 5). Cross out row one (1 5 3) and column two ${\displaystyle {\begin{pmatrix}5\\4\\6\end{pmatrix}}}$.
• Treat the remaining elements as a 2x2 matrix. In our example, the matrix is ${\displaystyle {\begin{pmatrix}2&7\\4&2\end{pmatrix}}}$
• Find the determinant of this 2x2 matrix. Use the ad - bc formula. (2*2 - 7*4 = -24)
• Multiply by the chosen element of the 3x3 matrix. -24 * 5 = -120
• Determine whether to multiply by -1. Use the sign chart or the (-1)ij formula. We chose element a12, which is - on the sign chart. We must change the sign of our answer: (-1)*(-120) = 120.
8. 8
Repeat with the third element. You have one more cofactor to find. Calculate i for the third term in your reference row or column. Here's a quick rundown of how you'd calculate the cofactor of a13 in our example:
• Cross out row 1 and column 3 to get ${\displaystyle {\begin{pmatrix}2&4\\4&6\end{pmatrix}}}$
• Its determinant is 2*6 - 4*4 = -4.
• Multiply by element a13: -4 * 3 = -12.
• Element a13 is + on the sign chart, so the answer is -12.
9. 9
Add your three results together. This is the final step. You've calculated three cofactors, one for each element in a single row or column. Add these together and you've found the determinant of the 3x3 matrix.
• In our example the determinant is -34 + 120 + -12 = 74.
### Part 2 Making the Problem Easier
1. 1
Pick the reference with the most zeroes. Remember, you can pick any row or column as your reference. You'll get the same answer no matter which you pick. If you pick a row or column with zeros, you only need to calculate the cofactor for the nonzero elements. Here's why:
• Let's say you pick row 2, with elements a21, a22, and a23. To solve this problem, we'll be looking at three different 2x2 matrices. Let's call them A21, A22, and A23.
• The determinant of the 3x3 matrix is a21|A21| - a22|A22| + a23|A23|.
• If terms a22 and a23 are both 0, our formula becomes a21|A21| - 0*|A22| + 0*|A23| = a21|A21| - 0 + 0 = a21|A21|. Now we only have to calculate the cofactor of a single element.
2. 2
Use row addition to make the matrix easier. If you take the values of one row and add them to a different row, the determinant of the matrix does not change. The same is true of columns. You can do this repeatedly — or multiply the values by a constant before adding — to get as many zeroes in the matrix as possible. This can save you a lot of time.
• For example, say you have a 3 x 3 matrix: ${\displaystyle {\begin{pmatrix}9&-1&2\\3&1&0\\7&5&-2\end{pmatrix}}}$
• In order to cancel out the 9 in position a11, we can multiply the second row by -3 and add the result to the first. The new first row is [9 -1 2] + [-9 -3 0] = [0 -4 2].
• The new matrix is ${\displaystyle {\begin{pmatrix}0&-4&2\\3&1&0\\7&5&-2\end{pmatrix}}}$ Try to use the same trick with columns to turn a12 into a 0 as well.
3. 3
Learn the shortcut for triangular matrices. In these special cases, the determinant is simply the product of the elements along the main diagonal, from a11 in the top left to a33 in the lower right. We're still talking about 3x3 matrices, but "triangular" ones have special patterns of nonzero values:[5]
• Upper triangular matrix: All the non-zero elements are on or above the main diagonal. Everything below is a zero.
• Lower triangular matrix: All the non-zero elements are on or below the main diagonal.
• Diagonal matrix: All the non-zero elements are on the main diagonal. (A subset of the above.)
## Community Q&A
Search
• What is the formula for the determinant?
Prem Shah
The formula to find the determinant for a quadratic formula is (b^2-4ac), which is all in a square root.
• How do I adjoin a matrix?
wikiHow Contributor
The adjoint of a square matrix is the transpose of the matrix Cij (cofactor of the original matrix).
• What is the formula of A' in matrix?
• How do I find a determinant of 3*3 matrices, where the first row has 1, 1, 2, second row has 2,-1,3, and third row has 3.2 -1?
200 characters left
## Tips
• This method extends to square matrices of any size. For example, if using this for a 4x4 matrix, your "crossing out" leaves you with a 3x3 matrix, for which you calculate the determinate as described above. Be warned, this gets very tedious by hand!
• If all elements of a row or column are 0, the determinant of that matrix is 0.
## Article Info
Categories: Linear Algebra
In other languages:
Español: encontrar el determinante de una matriz 3x3, Deutsch: Die Determinante einer 3x3 Matrix ermitteln, Português: Achar a Determinante de uma Matriz 3X3, Italiano: Calcolare il Determinante di una Matrice 3 x 3, Русский: найти определитель матрицы 3Х3, Français: calculer le déterminant d'une matrice 3 x 3, Bahasa Indonesia: Menentukan Determinan Matriks 3X3, Nederlands: De determinant van een 3x3 matrix bepalen
Thanks to all authors for creating a page that has been read 924,601 times.
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# 4.06 Graphs and characteristics of cube root functions
Lesson
The basic cube root function has the form $y=\sqrt[3]{x}$y=3x
Originally the cube root was defined as the side length of a cube whose volume was $x$x, as shown in the diagram.
Thus we have $\sqrt[3]{x}\times\sqrt[3]{x}\times\sqrt[3]{x}=x$3x×3x×3x=x.
Today we define the cube root to include negative numbers, so that for example $\sqrt[3]{-8}=-2$38=2 and $\sqrt[3]{8}=2$38=2
### The graph of $f\left(x\right)=\sqrt[3]{x}$f(x)=3√x
Positive cube roots greater than $1$1 are smaller than their argument, for example $\sqrt[3]{27}=3$327=3. Negative cube roots less than $-1$1 behave similarly, so that $\sqrt[3]{-27}=-3$327=3
If we let $f\left(x\right)=\sqrt[3]{x}$f(x)=3x, then $f\left(-x\right)=-\sqrt[3]{x}$f(x)=3x, and so the function is an odd function. It exhibits rotational symmetry about the origin.
The table of values can help us to see this:
$x$x $-27$27 $-8$8 $-1$1 $0$0 $1$1 $8$8 $27$27
$f\left(x\right)=\sqrt[3]{x}$f(x)=3x $-3$3 $-2$2 $-1$1 $0$0 $1$1 $2$2 $3$3
Putting these facts together, we should be able to understand why the cubic graph has the shape shown here:
Using the graph we can identify the key characteristics of the square root function, $f\left(x\right)=\sqrt[3]{x}$f(x)=3x.
Characteristic $f\left(x\right)=\sqrt[3]{x}$f(x)=3x
Domain
Words: $x$x can be any real number
Interval form: $\left(-\infty,\infty\right)$(,)
Range Words: $y$y can be any real number
Interval form: $\left(-\infty,\infty\right)$(,)
Extrema None
$x$x-intercept $\left(0,0\right)$(0,0)
$y$y-intercept $\left(0,0\right)$(0,0)
Increasing/decreasing Increasing over its domain
End behavior
As $x\to\infty$x, $y\to\infty$y
As $x\to-\infty$x$y\to-\infty$y
### Comparing $\sqrt{x}$√x, $\sqrt[3]{x}$3√x and $\sqrt[4]{x}$4√x
As a comparison, this second graph compares the positions of the graphs of $y=\sqrt{x}$y=x$y=\sqrt[3]{x}$y=3x and $y=\sqrt[4]{x}$y=4x within the interval $-1\le x\le1$1x1. Note that the square root and fourth root functions exist only in the first quadrant.
You can see that in this region $y=\sqrt[4]{x}$y=4x rises at a faster rate inside this unit square interval than either of $y=\sqrt{x}$y=x and $y=\sqrt[3]{x}$y=3x. The graph of $y=\sqrt[3]{x}$y=3x rises vertically through the origin.
Use this applet below see the different shapes for different powers, as well as how to impose transformations.
What we have already learned about transformations holds true for the cube root function.
#### Practice questions
##### question 1
Consider the function $y=\sqrt[3]{x}$y=3x.
1. Complete the table of values.
Round any values to two decimal places if necessary.
$x$x $y$y $-100$−100 $-10$−10 $-8$−8 $-3$−3 $-1$−1 $0$0 $1$1 $3$3 $8$8 $10$10 $100$100 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
2. The graph of $y=\sqrt[3]{x}$y=3x is given.
Is $y=\sqrt[3]{x}$y=3x an increasing function or a decreasing function?
Increasing
A
Decreasing
B
Increasing
A
Decreasing
B
##### question 2
Consider the graph of the function $y=-\sqrt[3]{x}$y=3x.
1. Is $y=-\sqrt[3]{x}$y=3x an increasing function or a decreasing function?
Increasing function
A
Decreasing function
B
Increasing function
A
Decreasing function
B
2. How would you describe the rate of decrease of the function?
As $x$x increases, the function decreases at a faster and faster rate.
A
As $x$x increases, the function decreases at a slower and slower rate.
B
As $x$x increases, the function decreases more and more rapidly up to $x=0$x=0, and from $x=0$x=0 onwards, the rate of decrease slows down.
C
As $x$x increases, the function decreases at a constant rate.
D
As $x$x increases, the function decreases at a faster and faster rate.
A
As $x$x increases, the function decreases at a slower and slower rate.
B
As $x$x increases, the function decreases more and more rapidly up to $x=0$x=0, and from $x=0$x=0 onwards, the rate of decrease slows down.
C
As $x$x increases, the function decreases at a constant rate.
D
##### question 3
Graph the function $f\left(x\right)=\sqrt[3]{-x}-5$f(x)=3x5.
2. What is the domain?
All real numbers.
A
$x\ge-5$x5
B
$x\le0$x0
C
$x\ge0$x0
D
All real numbers.
A
$x\ge-5$x5
B
$x\le0$x0
C
$x\ge0$x0
D
3. What is the range?
$y\ge-5$y5
A
All real numbers.
B
$y\ge0$y0
C
$y\le-5$y5
D
$y\ge-5$y5
A
All real numbers.
B
$y\ge0$y0
C
$y\le-5$y5
D
### Outcomes
#### F.IF.B.4'''
For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. '''Include rational, square root and cube root; emphasize selection of appropriate models.
#### F.IF.B.5'''
Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. '''Include rational, square root and cube root; emphasize selection of appropriate models.
#### F.IF.C.7'''
Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. '''Focus on using key features to guide selection of appropriate type of model function
#### F.IF.C.7.B'''
Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. '''Focus on using key features to guide selection of appropriate type of model function
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# Give me the answers to my math problems
Give me the answers to my math problems can be found online or in mathematical textbooks. We can help me with math work.
## The Best Give me the answers to my math problems
One tool that can be used is Give me the answers to my math problems. This method can be used to solve any quadratic formula calculator. We use our knowledge about quadratic formula calculator in this step. We know that if we have any linear equation like x + 2 = y where x 0, then we need to subtract 2 from both sides of this equation (this will give us a linear equation). We also know that if we have a quadratic equation like x2 + 4x – 9 = 0, where x > 0 then we have to divide both sides by -2 ==> x2 =>x 0. So this method is a combination of those two things. By subtracting 2 from both sides and dividing both sides by -2, we get an equivalent linear equation which we can solve using our knowledge about what happens when you divide by -2. Step 1: Solve for x and y using the Quadratrix formulae Step 2: Solve for z using the Quadratrix formulae
The difference quotient (DQ) is a metric that measures how much the value of one asset differs from another. It is calculated by dividing the price of the first asset by its price. If the difference is positive, then the asset is undervalued relative to the other asset. If it is negative, then the asset is overvalued relative to the other asset. It can be used to identify undervalued and overvalued assets, as well as situations where an investment may be too early or too late. DQ helps investors determine when to buy an undervalued asset and when to sell an overvalued asset. A higher DQ indicates that the current valuation of an asset is out of whack with reality, whereas a lower DQ indicates that the current valuation of an asset is in line with reality. One approach to solving DQ involves comparing two assets and calculating the ratio between their prices. If one has a higher value than another, then this suggests that it is undervalued and therefore should be bought. Conversely, if one has a lower value than another, then this suggests that it is overvalued and therefore should be sold. To calculate DQ, divide each number by the other number: price>/other-price>. For example, if one stock costs \$100 while another costs \$120, then its DQ would be 0.60 (= \$100
There are two things you need to keep in mind when solving quadratic equations. First, remember that solutions will always involve a positive number (a solution with a negative number would be impossible). Second, remember that solutions may not be perfect. In other words, a solution may not be an exact value. This means that solutions will never be “x” exactly, but rather “x + b” or “x + b – c” where “b” and “c” are positive numbers. The formula for solving a quadratic equation is: math>left( frac{a}{x} - frac{b}{2} ight)^{2} = left( frac{a}{x} + frac{b}{2} ight)^{2}/math> where math>a/math> and math>b/math> are both positive numbers. To solve a quadratic equation step by step, you follow these three steps: Step 1 – Identify if your quadratic equation
The sine function is used to solve problems where you want to know the angle between two vectors. The formula for the sine function is : Where: Also, the sine of a number between 0 and π (ex: -1) is equal to 1. To calculate the sine of a number you can use the following formula: For example: If you wanted to calculate the sine of an angle of 15 degrees, you would use this formula: . You can also replace the angle with any other value by simply plugging in the numbers. For example, if you wanted to calculate the sine of a 30-degree angle, you would use this formula: . See below for an example of how to solve for a specific number.
This is extremely useful for my math class! Ever since I started, my grades immediately skyrocketed!! I especially love that it shows the steps so it actually helps me! Also, the fact that it lets you type what you need if it scans wrong is SOOO perfect!!!! My friend recommended it to me, and I would recommend it to someone else! I'd even rate higher if I could! It's so hard to believe this miracle is free, but it is! No fees to use, it’s PERFECT!!!!!!!
Valentina Bailey
Some people are angry that it can't solve ALL math problems at the moment, but come on, if you need some hardcore math solutions at unit, then how about you learn to solve it yourself. I for one am amazed that it solves in half a second from my handwriting, when a few years ago you couldn't get a program to scan and recognize text accurately from printed text. And it is free, shows all the steps, that is almost perfect.
Priscila Lopez
Solution Basic Input Solve for missing variable calculator Math helper online for free Solving functions Help with math problems online Area solver
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Q:
# An intelligence test that has a maximum completion time of 45 minutes was recently administered to a group of 9 people. Their respective completion times (in minutes) were as follows: 45, 40, 42, 39, 44, 40, 45, 33, 31 / 25 to Excel (a) What is the mean of this data set? If your answer is not an integer, round your answer to one decimal place. (b) What is the median of this data set? If your answer is not an integer, round your answer to one decimal place. (c) How many modes does the data set have, and what are their values? Indicate the number of modes by clicking in the appropriate circle, and then indicate the value(s) of the mode(s), if applicable. zero modes one mode two modes:and Clear Undo Help Next Question >>
Accepted Solution
A:
Answer:a) $$\bar X=39.9$$b) Median =40c) Bimodal distribution 40,45 with a frequency of 2 for each oneStep-by-step explanation:Part aThe statistical mean refers "to the mean or average that is used to derive the central tendency of the data in question. It is determined by adding all the data points in a population and then dividing the total by the number of points". And is defined:$$\bar X =\frac{\sum_{i=1}^n x_i}{n}$$And for this case if we apply this formula we got:$$\bar X =\frac{45+40+42+39+44+40+45+33+21}{9}=39.9$$Part bThe median is a "measure of central tendency. To find the median, we arrange the observations in order from smallest to largest value. And we have two possible cases:1) If there is an odd number of observations, the median is the middle value. 2) If there is an even number of observations, the median is the average of the two middle values."So if we order the dataset we got:31,33,39,40,40,42,44,45,45Since we have an odd number of observation n=9, the median would be just the middle value on position 5 and for this case Median =40Part cThe mode of a set of data values is the value that appears most often. As a set of data can have more than one mode, the mode does not necessarily indicate the centre of a data set.For this case we have a bimodal distribution and the corresponding values are 40 and 45 with a frequency of 2 for each one.
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## Want to keep learning?
This content is taken from the UNSW Sydney's online course, Maths for Humans: Linear, Quadratic & Inverse Relations. Join the course to learn more.
2.21
## UNSW Sydney
How high the ball?
# Vertices, axes and maximum values
How high does a basketball go? Because of the ubiquity of quadratic curves in motion and other applications, it is useful to be able to determine its maximum or minimum value. Al Khwarizmi’s formula is the algebraic key that unlocks the symmetry and position of a general parabola through understanding its vertex and axis.
In this step we will
• see how Khwarizmi’s formula determines the vertex and axis of a parabola
• show how to accurately predict the maximum height of a quadratic function.
## The vertex and axis of a parabola
Recall that by translating a standard parabola $$\normalsize{y = ax^2}$$ to the form $$\normalsize{y = a(x-h)^2+k}$$ we determined that its vertex was the point $$\normalsize{[h,k]}$$.
We are now going to see how to find the vertex of a general parabola $${\normalsize p(x) =ax^2+bx+c }$$. Al Khwarizmi’s formula
$\Large{ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}}. \label{1} \tag{1}$
is a rewrite of the general quadratic exactly in the form $$\normalsize{y = a(x-h)^2+k}$$, which immediately allows us to see that the vertex of the associated parabola is the point
$\Large{[h,k] = \Bigg[-\frac{b}{2a},\frac{4ac-b^2}{4a}\Bigg]} .$
We also deduce that the axis of symmetry of the parabola, which is the vertical line through the vertex, is
$\Large {x=-\frac{b}{2a}}.$
Since the square term $${\normalsize \left(x+\frac{b}{2a}\right)^2 \geq 0 }$$, from $$\normalsize{(\ref{1})}$$ we deduce that if $$a>0$$ then the minimum value of $$p(x)$$ is
$\Large {\frac{4ac-b^2}{4a}}.$
Similarly if $$a< 0$$ then the maximum possible value of $$p(x)$$ is the same quantity $$\normalsize{\frac{4ac-b^2}{4a}}.$$
Q1 (E): Where is the vertex and axis of the parabola $$\normalsize{y=-4(x-3)^2+5}$$ ?
Q2 (E): Find the vertex and axis of $$\normalsize{y=3x^2+4x+7 }$$.
Q3 (M): What is the maximum value of the polynomial function $$\normalsize{p(x)=-2x^2+3x-5 }$$. What is the minimum value?
Q4 (C): Suppose that $$\normalsize a<0$$. Show that if $$\normalsize d > c - \frac{b^2}{4a}$$ then there are no solutions to the equation $${\normalsize ax^2 + bx + c = d }.$$
A1. The parabola $$\normalsize{y=-4(x-3)^2+5}$$ has vertex $$\normalsize{[3,5]}$$ and axis the line $$\normalsize{x=3}$$.
A2. Let’s find al-Khwarizmi’s identity for $$\normalsize p(x)=3x^2+4x+7$$. We complete the square as follows: (please study this carefully!)
\Large{\begin{align} p(x)&=3\left(x^2+\frac{4}{3}x+\frac{7}{3}\right)\\ &= 3\left(x^2+\frac{4}{3}x+\Bigg(\frac{2}{3}\Bigg)^2+\frac{7}{3}-\Bigg(\frac{2}{3}\Bigg)^2\right)\\ &=3\bigg(\bigg(x+\frac{2}{3}\bigg)^2+\frac{17}{9}\bigg) \\ &=3\Bigg(x+\frac{2}{3}\Bigg)^2+\frac{17}{3}. \end{align}}
The vertex of the parabola is then the point $$\normalsize \big[-\frac{2}{3},\frac{17}{3}\big]$$. The axis is the line
$$\Large x=-\frac{2}{3}$$.
A3. For the polynomial $$\normalsize{p(x)=-2x^2+3x-5 }$$, the coefficient of $${\normalsize x^2}$$ is negative. So it has a maximum value which is
$\Large {\frac{4ac-b^2}{4a}=\frac{4(-2)(-5)-3^2}{4(-2)}=-\frac{31}{8}.}$
As for a minimum value, there is none, as the branches of the parabola become increasingly more negative for large values of $$\normalsize x$$.
A4. Suppose $$\normalsize ax^2 + bx + c = d > c - \frac{b^2}{4a}$$ where $$\normalsize a<0$$. This can be rearranged as
${\Large x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 < 0}.$
But the left hand side is a perfect square
${\Large \left(x + \frac{b}{2a}\right)^2 < 0 }$
and square numbers cannot be negative. Hence there can be no solutions to $$\normalsize ax^2 + bx + c = d$$ when $$\normalsize d > c - \frac{b^2}{4a}$$.
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2016 AMC 10A Problems/Problem 1
Problem
What is the value of $\dfrac{11!-10!}{9!}$?
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$
Solution 1
We can use subtraction of fractions to get $$\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.$$
Solution 2
Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}$.
Solution 3
$\dfrac{11!-10!}{9!}$ consider 10 as n $\dfrac{(n+1)!-n!}{(n-1)!}$ simpify $\dfrac{(n+1)n!-(-1)n!}{(n-1)!}$ = $\dfrac{n(n!)}{(n-1)!}$ = $\dfrac{n(n(n-1)!)}{(n-1)!}$ = $\dfrac{n(n)(1)}{(1}$ = $\dfrac{n^2}{1}$ subsitute n as 10 again $\dfrac{10^2}{1}$
answer is $10^2$ which is 100
Solution 4
We are given the equation $\frac{11!-10!}{9!}$
This is equivalent to $\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\frac{(11-1)(10!)}{9!}$, which equals $10 \cdot 10$.
Therefore, the answer is $10^2$ = $\boxed{\textbf{(B)}~100}$.
~TheGoldenRetriever
Video Solution (HOW TO THINK CREATIVELY!!!)
https://youtu.be/r5G98oPPyNM
~Education, the Study of Everything
~savannahsolver
Video Solution (FASTEST METHOD!)
~Veer Mahajan
2016 AMC 10A (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
2016 AMC 12A (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
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# How do you find (dy)/(dx) given x^2+y^2=1?
Oct 11, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$
#### Explanation:
${x}^{2} + {y}^{2} = 1$
Differentiate wrt $x$:
$\frac{d}{\mathrm{dx}} {x}^{2} + \frac{d}{\mathrm{dx}} {y}^{2} = \frac{d}{\mathrm{dx}} 1$
We already know how to deal with the first and third terms, so lets get them out the way:
$\frac{d}{\mathrm{dx}} {x}^{2} + \frac{d}{\mathrm{dx}} {y}^{2} = \frac{d}{\mathrm{dx}} 1$
$\therefore 2 x + \frac{d}{\mathrm{dx}} {y}^{2} = 0$
For the remaining term we use the chain rule, we don't know how to differentiate ${y}^{2}$ wrt $x$ but we do know how to differentiate ${y}^{2}$ wrt $y$ (it the same as differentiating ${x}^{2}$ wrt $x$!). The chain rule tells us that:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$, so we can rewrite:
$2 x + \frac{d}{\mathrm{dx}} {y}^{2} = 0$ as $2 x + \frac{d}{\mathrm{dy}} \left({y}^{2}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
We can know perform that final differentiation, as we are now differentiating a function of $y$ wrt $y$ so
$2 x + \frac{d}{\mathrm{dy}} \left({y}^{2}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 2 x + 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
We can then rearrange to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ as follows:
$2 x + 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x}{2 y}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$
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# How to Convert Decimals to Fractions for Math Problems
By Sandy Fleming
Jupiterimages/Photos.com/Getty Images
Decimals and fractions both represent portions of the whole. In fact, every decimal has a matching fraction name and every fraction has a matching decimal name. When decimal numbers are read correctly, the fraction name becomes very clear. Read ".5" as "five tenths" and ".02" as "two hundredths," for example. Resist the urge to read those numbers as "point five" and "point zero two" because those names are not only incorrect, they obscure the connection between decimals and fractions.
Read the decimal correctly, using the place value name. To do this, read the number to the right of the decimal point as if it were a whole number. Add the name of the final place after you have said the number. The names of the places to the right of the decimal point mirror the names of the places of whole numbers. They are tenths, hundredths, thousandths, ten thousandths, and so on. For example, read ".15" as "fifteen hundredths" because the digits "1" and "5" together are read as "fifteen" and the number takes up two places to the right of the decimal point: the tenths and the hundredths. ".157" would be read as "one hundred fifty-seven thousandths" by the same logic.
Consider the place value name that applies to the decimal you have read. That is the bottom number, or denominator, of your fraction. The top number, or numerator, is the actual number to the right of the decimal place. For the decimal ".25," place the "25" on the top of the fraction bar. Put a "100" on the bottom of the fraction since the decimal would be read as "twenty-five hundredths."
Simplify the fraction you have created if necessary. To put a fraction in simplest form, or lowest terms, divide numerator and denominator by the largest number possible, otherwise known as the greatest common factor. Both top and bottom numbers must be divided by the same number to simplify the fraction. In our example of "twenty-five hundredths," both 25 and 100 can be divided by 25. When that is done, the fraction in lowest terms is seen to be one-fourth.
Fractions can also be simplified by using multiple divisions instead of a single division by the greatest common factor. For the example fraction of 25/100, both numbers are divisible by five. Performing this operation would yield 5/20. Both of those numbers are also divisible by five. The answer to that calculation is 1/4. It is important to continue divisions in this manner until the numerator and denominator can no longer be divided by the same number.
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# Absolute Value of an Integer
Absolute value of an integer is its numerical value without taking the sign into consideration.
The absolute values of -9 = 9; the absolute value of 5 = 5 and so on.
The symbol used to denote the absolute value is, two vertical lines (| |), one on either side of an integer.
Therefore, if 'a' represents an integer, its absolute value is represented by |a| and is always non-negative.
Note:
(i) |a| = a; when 'a' is positive or zero.
(ii) |a| = -a; when 'a' is negative.
## Definition of Absolute Value of an Integer:
The numerical value of an integer regardless of its sign is known as its absolute value.
The two vertical bars | | represent the absolute value.
If x represents an integer, then
| x | = x if x is +ve or zero
| -x | = x if x is -ve.
The absolute value of 5, written as |5|, is 5 and the absolute value of -5, written as| -5|, is 5.
The absolute value of 15, written as | 15 |, is 15 and the absolute value of -15, written as | -15 |, is 15.
The absolute value of 0, written as | 0 |, is 0.
Find the absolute value of the following:
(i) -76
(ii) +50
(iii) -100
Solution:
(i) -76 = |76|
(ii) +50 = |50|
(iii) -100 = |100|
Examples on absolute value of an integer:
(i) Absolute value of - 7 is written as |- 7| = 7 [here mod of - 7 = 7]
(ii) Absolute value of + 2 is written as |+ 2| = 2 [here mod of + 2 = 2]
(iii) Absolute value of - 15 is written as |- 15| = 15 [here mod of - 15 = 15]
(iv) Absolute value of + 17 is written as |+ 17| = 17 [here mod of + 17 = 17]
On a number line the number indicates the distance from 0 and the sign before the number tells us whether the distance is to the right or left of 0. For example +5 is 5 units away to the right of 0 where as -5 is 5 units away to the left of 0 on the number line. The numerical value of the unit regardless of the sign is called absolute value of an integer. The absolute value of an integer is always positive. Thus, the absolute value of 5 and -5 is 5. It is written as |5|
So, |5| = 5 and |-5| = 5
Find the mod of:
(i) |14 - 6| = |8| = 8
(ii) - |- 10| = - 10
(iii) 15 - |- 6| = 15 - 6 = 9
(iv) 7 + |- 7| = 7 + 7 = 14
Note: (i) A positive number with a sign in front of its numerical value means increase or gain.
(ii) A negative number with a sign in front of its numerical value means decrease or loss.
Different Types of Solved Examples on Absolute Value of an Integer:
1. Write the opposites of the following statements:
(i) 28 m to the right
(ii) running 75 km towards East
(iii) loss of $250 (iv) 780 m above sea level (v) increase in population Solution: (i) 28 m to the left (ii) running 75 km towards West (iii) gain of$ 250
(iv) 780 m below sea level
(v) decrease in population
2. Represent the following numbers as integers with appropriate signs.
(i) 7°C above normal temperature
(ii) A deposit of $5690 (iii) 23°C below 0°C Solution: (i) +7°C (ii) +$5690
(iii) -23°C
3. Compare -2 and -6 using number line
Solution:
Since -2 is to the right of -6, therefore -2 > -6 or -6 < -2.
4. Find the absolute value of each of the following:
(i) -12
(ii) 30
(iii) 0
Solution:
(i) The absolute value of -12 = | -12 | = 12
(ii) The absolute value of 30 = | 30 | = 30
(ii) The absolute value of 0 = | 0 | = 0
[Since integer 0 is neither positive nor negative, the absolute value of zero is zero i.e., | 0 | = 0.
5. Find the value of 24 + | -14 |.
Solution:
24 + | -14 |.
= 24 + 14, since | -14 | = 14.
6. Write all the integers between
(i) -1 and 3
(ii) -3 and 4
Solution:
(i) The integers between -1 and 3 are 0, 1, 2.
(ii) The integers between -3 and 4 are -2, -1, 0, 1, 2, 3.
7. Which of the following pairs of integers is greater?
(i) 6, -6
(ii) 0, -9
(iii) 0, 8
(iv) -8, -3
(v) 4, -9
Solution:
(i) 6 > -6; since, every positive integer is greater than every negative integer.
(ii) 0 > -9; since, 0 is greater than every negative integer.
(iii) 0 < 8; Since, 0 is less than every positive integer.
(iv) -8 < -3; Since, If x = 8 and b = 3 then x > y, therefore, -x < -y.
(v) 4 > -9; Since, every positive integer is greater than every negative integer.
## You might like these
• ### Ordering Integers | Integers from Greater to Lesser, Lesser to Greater
In ordering integers we will learn how to order the integers on a number line. An integer on a number line is always greater than every integer on its left. Thus, 3 is greater than
• ### Representation of Integers on a Number Line | Integers on Number Line
Representation of integers on a number line is explained here step by step. In the number line the positive numbers are to the right side and the negative numbers are to the left side.
Numbers - Integers
Integers
Multiplication of Integers
Properties of Multiplication of Integers
Examples on Multiplication of Integers
Division of Integers
Absolute Value of an Integer
Comparison of Integers
Properties of Division of Integers
Examples on Division of Integers
Fundamental Operation
Examples on Fundamental Operations
Uses of Brackets
Removal of Brackets
Examples on Simplification
Numbers - Worksheets
Worksheet on Multiplication of Integers
Worksheet on Division of Integers
Worksheet on Fundamental Operation
Worksheet on Simplification
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# Descriptive and Inferential Statistics – A Comprehensive Guide For 2021
## Introduction
A branch of mathematics, statistics, deals with the collection of data, its analysis and interpretation, and the ways of presenting numerical data. Statistics basically is a collection of data that is quantitative in nature. The subject of Statistics is broadly categorized as theoretical Statistics and Applied Statistics. Applied Statistics is further divided into descriptive statistics and inferential statistics. Let us understand what descriptive and inferential Statistics are and then delve down deeper to understand the differences between descriptive and inferential statistics.
## 1) What is descriptive statistics?
Descriptive Statistics is also referred to as samples and it determines multiple observations that you take through the research. It is defined as a finding group that fits the research parameters and the groups that are being tested and the use of graphs and stats to summarise the groups’ findings. In simple words, descriptive Statistics pairs the findings from a group and reduces them to simple and small major points. In the case of descriptive Statistics, you only test for the results that you get from the relevant individuals. It needs you to test continuously if the result affects a larger population portion.
Conclusions that can be measured with descriptive Statistics are through:
• The central tendency is the method of making use of the mean and the median to find the data location on the graph
• Dispersion is a way to find the division of the data points from the graph centre. If the number is small then this tells that the dispersion is close to the centre. A large number means that there is a larger space from the graphs’ epicentre.
• Skewness basically highlights the data point separation that you have measured. It lets you conclude if the data points are skewed or symmetrical from the measurements.
## 2) What is inferential statistics?
Inferential Statistics is where you take the data got from a sample and make predictions if that influences the findings on a larger population. You can make use of random sampling in order to evaluate the ways in which the different variables lead to making generalizations to do experiments. To make an accurate analysis you will have to find out the population that is being measured and creates a population sample. You then need to incorporate analysis in order to find out the sampling error.
Some ways in which inferential Statistics can be measured is through:
• Hypothesis test that determines if the population that is being measured has a high value as compared to the other data points in the analysis. It can also help to conclude if the population varies which is based on the results that you have earned from various experiments.
• The confidence interval discovers the error margin in the research and finds out if it affects the testing. You will have to estimate the population range and if it falls under the median or means calculation.
• Regression analysis is basically an association between the dependent and the independent variables of the experiment. You need to know the hypothesis test results in order to perform a regression analysis. This lets you know the relationship that is there between the subject matter. A few things that the regression analysis helps to test are the comparison between two sets of populations or maybe the comparison between height and weight of the different genders.
## 3) Difference between them
Here we list down the differences between descriptive and inferential statistics.
• The descriptive analysis gives information about the raw data that describes the data in a particular manner. The inferential analysis makes the inference about a population which is done using the data that is drawn from a population.
• Descriptive Statistics helps to organize, analyze, and present the data in a meaningful way. Inferential statistics allows comparing data and making predictions and hypotheses with it.
• Descriptive Statistics is used in order to describe a situation whereas inferential Statistics is used to explain the chances of the occurrence of an event
• Descriptive Statistics explains the data that is already known and is limited to a population or a sample of a small size. Inferential Statistics tries to reach out to a conclusion about the population.
• Descriptive Statistics can be done using graphs, charts, and tables. Inferential Statistics is achieved through probability.
• Descriptive Statistics has a tabular or diagrammatic representation of the final result whereas inferential Statistics represents the result in the probability form.
• Descriptive Statistics describes the situation where inferential Statistics explains the likelihood if the event will occur.
• Descriptive Statistics measures only the group that is assigned for the experiment which means that when you do the descriptive analysis you decide to not consider in the variables. In the case of inferential Statistics, you account for the sampling errors which may make you conduct additional tests that need to be on a large population depending on the amount of data that is required. In other words, you are likely to get a definite calculation when you use descriptive statistics.
• Since you are testing the variables using inferential Statistics it is easy to arrive at conclusions when you use descriptive statistics.
## Conclusion
Statistics has a very important role to play in the research field. It helps to collect, analyze, and present the data in a form that is measurable. It is difficult to understand if the research lies in descriptive or inferential statistics. This is because people may not really be aware of these two Statistics branches. Descriptive statistics as the name suggests describes the population. On the contrary inferential Statistics is used in order to make a generalization of the population based on the sample. This shows that there is a lot of difference between descriptive and inferential Statistics which basically lies in what you do with the data.
Descriptive statistics is about how you illustrate the current data set whereas inferential stats focus on making an assumption about the extra population which is more than the state of data that is under study. Descriptive Statistics provides a summary of the data that the researcher has studied. Inferential Statistics, however, makes a generalization which is on the data that you have not studied actually. These are the differences between descriptive and inferential statistics.
If you are interested in making it big in the world of data and evolve as a Future Leader, you may consider our Integrated Program in Business Analytics, a 10-month online program, in collaboration with IIM Indore!
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