text stringlengths 22 1.01M |
|---|
## Related Articles
• CBSE Class 10 Syllabus (All Subjects)
• CBSE Class 10 Maths Notes
• CBSE Class 10 Science Revision Notes
• CBSE Class 10 Social Science Revision Notes
# Mean, Median and Mode of Grouped Data
• Difficulty Level : Medium
• Last Updated : 02 Nov, 2022
Suppose we want to compare the age of students in two schools and determine which school has more aged students. If we compare on the basis of individual students, we cannot conclude anything. However, if for the given data, we get a representative value that signifies the characteristics of the data, the comparison becomes easy.
A certain value representative of the whole data and signifying its characteristics is called an average of the data. Three types of averages are useful for analyzing data. They are:
• Mean
• Median
• Mode
In this article, we will study three types of averages for the analysis of the data.
## Mean
The mean (or average) of observations is the sum of the values of all the observations divided by the total number of observations. The mean of the data is given by x = f1x1 + f2x2 + …. + fnxn/f1 + f2+… + fn. The mean Formula is given by,
Mean = ∑(fi.xi)/∑fi
## Methods for Calculating Mean
### Method 1: Direct Method for Calculating Mean
Step 1: For each class, find the class mark xi, as
x=1/2(lower limit + upper limit)
Step 2: Calculate fi.xi for each i.
Step 3: Use the formula Mean = ∑(fi.xi)/∑fi.
Example: Find the mean of the following data.
Solution:
We may prepare the table given below:
Mean = ∑(fi.xi)/∑fi = 1100/50 = 22
### Method 2: Assumed – Mean Method For Calculating Mean
For calculating the mean in such cases we proceed as under.
Step 1: For each class interval, calculate the class mark x by using the formula: xi = 1/2 (lower limit + upper limit).
Step 2: Choose a suitable value of mean and denote it by A. x in the middle as the assumed mean and denote it by A.
Step 3: Calculate the deviations di = (x, -A) for each i.
Step 4: Calculate the product (fi x di) for each i.
Step 5: Find n = ∑fi
Step 6: Calculate the mean, x, by using the formula: X = A + ∑fidi/n.
Example: Using the assumed-mean method, find the mean of the following data:
Solution:
Let A=25 be the assumed mean. Then we have,
Mean = X = A + ∑fidi/n = (25 + 110/50) = 27.2
### Method 3: Step-Deviation method for Calculating Mean
When the values of x, and f are large, the calculation of the mean by the above methods becomes tedious. In such cases, we use the step-deviation method, given below.
Step 1: For each class interval, calculate the class mark x, where X = 1/2 (lower limit + upper limit).
Step 2: Choose a suitable value of x, in the middle of the x, column as the assumed mean and denote it by A.
Step 3: Calculate h = [(upper limit) – (lower limit)], which is the same for all the classes.
Step 4: Calculate ui = (xi -A) /h for each class.
Step 5: Calculate fu for each class and hence find ∑(fi × ui).
Step 6: Calculate the mean by using the formula: x = A + {h × ∑(fi × ui)/ ∑fi}
Example: Find the mean of the following frequency distribution:
Solution:
We prepare the given table below,
A = 100, h = 20, ∑fi = 100 and ∑(fi x ui) = 61
x = A + {h × ∑(fi x ui) / ∑fi}
=100 + {20 × 61/100} = (100 + 12.2) =112.2
## Median
We first arrange the given data values of the observations in ascending order. Then, if n is odd, the median is the (n+1/2). And if n is even, then the median will be the average of the n/2th and the (n/2 +1)th observation. The formula for Calculating Median:
Median, Me = l + {h x (N/2 – cf )/f}
Where,
• l = lower limit of median class.
• h = width of median class.
• f = frequency of median class,
• cf = cumulative frequency of the class preceding the median class.
• N = ∑fi
Example: Calculate the median for the following frequency distribution.
Solution:
We may prepare cumulative frequency table as given below,
Now, N = 80 = (N/2) = 40.
The cumulative frequency just greater than 40 is 58 and the corresponding class is 24-32.
Thus, the median class is 24-32.
l = 24, h = 8, f = 24, cf = c.f. of preceding class = 34, and (N/2) = 40.
Median, Me = l+ h{(N/2-cf)/f}
= 24 + 8 {(40 – 34)/ 24}
= 26
Hence, median = 26.
## Mode
It is that value of a variety that occurs most often. More precisely, the mode is the value of the variable at which the concentration of the data is maximum.
Modal Class: In a frequency distribution, the class having the maximum frequency is called the modal class. The formula for Calculating Mode:
Mo = xk + h{(fk – fk-1)/(2fk – fk-1 – fk+1)}
Where,
• xk = lower limit of the modal class interval.
• fk = frequency of the modal class.
• fk-1= frequency of the class preceding the modal class.
• fk+1 = frequency of the class succeeding the modal class.
• h = width of the class interval.
Example 1: Calculate the mode for the following frequency distribution.
Solution:
Class 40-50 has the maximum frequency, so it is called the modal class.
xk = 40, h = 10, fk = 28, fk-1 = 12, fk+1 = 20
Mode, Mo= xk + h{(fk – fk-1)/(2fk – fk-1 – fk+1)}
= 40 + 10{(28 – 12)/(2 × 28 – 12 – 20)}
= 46.67
Hence, mode = 46.67
Important Result
Relationship among mean, median and mode,
Mode = 3(Median) – 2(Mean)
Example 2: Find the mean, mode, and median for the following data,
Solution:
We have,
Mean = ∑(fi.xi)/∑f
= 2640/100
= 26.4
Here, N = 100 ⇒ N / 2 = 50.
Cumulative frequency just greater than 50 is 60 and corresponding class is 20-30.
Thus, the median class is 20-30.
Hence, l = 20, h = 10, f = 36, c = c. f. of preceding class = 24 and N/2=50
Median, Me = l + h{(N/2 – cf)/f}
= 20+10{(50-24)/36}
Median = 27.2.
Mode = 3(median) – 2(mean) = (3 × 27.2 – 2 × 26.4) = 28.8.
## Ogives
Let a grouped frequency distribution be given to us. On graph paper, we mark the upper-class limits along the x-axis and the corresponding cumulative frequencies along the y-axis. By joining these points successively by line segments, we get a polygon, called a cumulative frequency polygon. On joining these points successively by smooth curves, we get a curve, known as ogive.
The step-by-step process for the construction of an Ogive graph:
1. First, we mark the upper limits of class intervals on the horizontal x-axis and their corresponding cumulative frequencies on the y-axis.
2. Now plot all the corresponding points of the ordered pair given by (upper limit, cumulative frequency).
3. Join all the points with a free hand.
4. The curve we get is known as ogive.
### Types of Ogives
• Less Than Ogive
On graph paper, we mark the upper-class limits along the x-axis and the corresponding cumulative frequencies along the y-axis.
1. On joining these points successively by line segments, we get a polygon, called a cumulative frequency polygon.
2. On joining these points successively by smooth curves, we get a curve, known as less than ogive.
• More Than Ogive
On graph paper, we mark the lower class limits along the x-axis and the corresponding cumulative frequencies along the y-axis.
1. On joining these points successively by line segments, we get a polygon, called the cumulative frequency polygon.
2. On joining these points successively by smooth curves, we get a curve, known as more than ogive.
My Personal Notes arrow_drop_up |
# 1982 AHSME Problems/Problem 14
## Problem 14
In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB, BC$, and $CD$ are diameters of circle $O, N$, and $P$, respectively. Circles $O, N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has length
$[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("A", A, W); label("B", B, SE); label("C", C, NE); label("D", D, dir(0)); label("P", P, S); label("N", N, S); label("O", O, S); label("E", E, dir(120)); label("F", F, NE); label("G", G, dir(100));[/asy]$
## Solution
Drop a perpendicular line from $N$ to $AG$ at point $H$. $AN=45$, and since $\triangle{AGP}$ is similar to $\triangle{AHN}$. $NH=9$. $NE=NF=15$ so by the Pythagorean Theorem, $EH=HF=12$. Thus $EF=\boxed{24.}$ Answer is then $\boxed{C}$. |
Unveiling the Mystery: Understanding Logarithms as Inverse Functions” – A Comprehensive Guide
4.3
algorithms
March 25, 2023 1:02 pm
Logarithmic functions are one of the most important concepts in mathematics. A logarithm is an exponent that determines the power to which a given number, called the base, must be raised to produce a given value. In other words, the logarithm of a number is the power to which the base must be raised to get that number.
The concept of the logarithm is intimately related to that of the exponent. In fact, the logarithm is an inverse function of the exponential function. This means that if we apply the logarithm to a value that has been obtained by the exponentiation of a base, we obtain the exponent.
For example, consider the function f(x) = 2^x. This function maps any value of x to a value that is obtained by raising 2 to the power of x. If we want to know the value of x that corresponds to a given value of f(x), we need to apply the logarithm to that value. In this case, we get:
log2(f(x)) = x
This equation tells us that the logarithm of f(x) to the base 2 is equal to x. In other words, the logarithm is the inverse function of the exponential function. This relationship between the logarithm and the exponent is what makes logarithmic functions so powerful and useful.
Logarithmic functions have applications in many areas of mathematics, science, and engineering. For example, they are used to model exponential growth and decay in populations, to calculate interest rates in finance, and to measure the intensity of earthquakes and other natural phenomena.
There are several properties of logarithmic functions that are important to understand. One of the most important properties is the logarithmic identity:
logb(xy) = logb(x) + logb(y)
This property tells us that the logarithm of the product of two numbers is equal to the sum of the logarithms of those numbers. Another important property is the change of base formula:
logb(x) = loga(x) / loga(b)
This formula enables us to switch between different bases for logarithms.
Logarithmic functions also come in different bases. The most common base used in mathematics is the base 10. This is because we use the decimal system for numbers, which is based on powers of 10. Another commonly used base is the natural logarithm, which is denoted by ln. The natural logarithm is important in calculus and many other areas of mathematics.
In conclusion, logarithmic functions are an important concept in mathematics. They are the inverse function of the exponential function and are used to model many different phenomena in science and engineering. Understanding the properties of logarithmic functions is essential for anyone who wants to work with them effectively.
March 25, 2023 1:02 pm
Scroll to Top |
ACT Math : Basic Squaring / Square Roots
Example Questions
← Previous 1 3 4 5 6 7
Example Question #1 : How To Find The Common Factor Of Square Roots
Explanation:
To solve the equation , we can first factor the numbers under the square roots.
When a factor appears twice, we can take it out of the square root.
Now the numbers can be added directly because the expressions under the square roots match.
Example Question #2 : How To Find The Common Factor Of Square Roots
Solve for .
Explanation:
First, we can simplify the radicals by factoring.
Now, we can factor out the .
Now divide and simplify.
Example Question #21 : Arithmetic
Which of the following is equivalent to:
?
Explanation:
To begin with, factor out the contents of the radicals. This will make answering much easier:
They both have a common factor . This means that you could rewrite your equation like this:
This is the same as:
These have a common . Therefore, factor that out:
Example Question #1 : Factoring Common Factors Of Squares And Square Roots
Simplify:
Explanation:
These three roots all have a in common; therefore, you can rewrite them:
Now, this could be rewritten:
Now, note that
Therefore, you can simplify again:
Now, that looks messy! Still, if you look carefully, you see that all of your factors have ; therefore, factor that out:
This is the same as:
Example Question #1 : Factoring Common Factors Of Squares And Square Roots
Simplify:
Explanation:
Begin by factoring out the relevant squared data:
is the same as
This can be simplified to:
Since your various factors contain square roots of , you can simplify:
Technically, you can factor out a :
Example Question #4 : How To Find The Common Factor Of Square Roots
Solve for :
Explanation:
Begin by breaking apart the square roots on the left side of the equation:
This can be rewritten:
You can combine like terms on the left side:
Solve by dividing both sides by :
This simplifies to:
Example Question #1 : Basic Squaring / Square Roots
Solve for
Explanation:
To begin solving this problem, find the greatest common perfect square for all quantities under a radical.
--->
Pull out of each term on the left:
--->
Next, factor out from the left-hand side:
--->
Lastly, isolate :
--->
Example Question #1 : How To Find The Common Factor Of Square Roots
Solve for
Explanation:
Solving this one is tricky. At first glance, we have no common perfect square to work with. But since each term can produce the quantity , let's start there:
--->
Simplify the first term:
--->
Divide all terms by to simplify,
--->
Next, factor out from the left-hand side:
--->
Isolate by dividing by and simplifying:
--->
Last, simplify the denominator:
---->
Example Question #1 : Basic Squaring / Square Roots
Solve for :
Explanation:
Right away, we notice that is a prime radical, so no simplification is possible. Note, however, that both other radicals are divisible by .
Our first step then becomes simplifying the equation by dividing everything by :
--->
Next, factor out from the left-hand side:
--->
Lastly, isolate :
--->
Example Question #1 : How To Find The Common Factor Of Square Roots
Solve for
Explanation:
Once again, there are no common perfect squares under the radical, but with some simplification, the equation can still be solved for :
--->
Simplify:
--->
Factor out from the left-hand side:
--->
Lastly, isolate :
--->
← Previous 1 3 4 5 6 7 |
### Split Infinities
Today is the 165th birthday of Georg Ferdinand Ludwig Philipp Cantor, the mathematician who indirectly inspired me to major in math. In my first few semesters of college, I was at best an indifferent student, finding little inspiration in the humanities majors I was bouncing around among, playing a prodigious amount of pinball, and attaining (according to rumor) history’s first-ever grade of C in Peter Regenstrief‘s Poltical Science 101. Then one day, my friend Bob Hyman happened to mention that some infinities are larger than others, and set my life on track. This—the vision of Georg Cantor—was something I had to know more about. Before long I was immersed in math.
What does it mean for some infinities to be larger than others? Well, for starters, some infinite sets can be listed, while others are too big to list. The natural numbers, for example, are already packaged as a list:
The integers, by contrast (that is, the natural numbers plus their negatives) aren’t automatically listed because a list, by definition, has a starting point, whereas the integers stretch infinitely far in both directions. But we can fix that by rearranging them:
So the integers can also be listed.
The positive rational numbers (that is, numbers expressible as fractions) appear even harder to list than the integers, because they have no immediate successors. What is the next rational number after 1/2, for example? Answer: there is no next number. Between any two rationals lie infinitely many more.
We can still list them, though—after a suitable rearrangement of course. There are many ways to do this; here’s probably the simplest: First list all the fractions whose numerator and denominator add to 2, then all the fractions whose numerator and denominator add to 3, then all the fractions whose numerator and denominator add to 4, then 5, and so on, like so:
and finally string them all together:
There’s some repetition here (for example, 2/2 is the same number as 1/1), but just cross out the repeats and you’ve got your list.
What if you wanted to list all the rational numbers, both positive and negative? Easy! Just combine the two tricks we’ve already used. Start with the list just above, and stick in the negatives:
Now the only missing rational is zero, which you can throw in anyplace you like—say at the very beginning.
At this point you should begin to suspect that any infinite set can be listed, given enough cleverness. Not so, though. Let’s try to list all the real numbers—that is, all numbers expressible as (possibly infinite) decimals—between 0 and 1.
Now offhand, I can’t think of any way to do this, but that doesn’t prove anything about the real numbers; it might just prove I’m not as clever as I ought to be. But Cantor, with one incredibly simple argument, demonstrated that no attempt to list those real numbers can be successful.
Here’s the argument. Suppose you believe you have managed to list all those real numbers. Maybe your list looks something like this:
(For convenience, I’ve displayed this list vertically instead of horizontally, so the first item on the list is .8410729…., the second is .1415926…, and so on.) Now I’m going to prove you wrong—that is, I’m going to prove your list is incomplete—by writing down a number that’s definitely missing. First I write down a decimal point. Then I write down any first digit other than 8 (say 6). This insures that the number I’m writing down is not the same as .8410729….. Then I write down any second digit other than 4 (say 5). This insures that the number I’m writing down is not the same as .1415926….. Then I’ll write down any third digit other than 3 (say 7). This insures that the the number I’m writing down is not the same as .3333333….. Continuing in this way, I get a number
that is definitely nowhere on your list.
(If it bothers you to imagine that I could make infinitely many choices, just imagine that I make all the choices at once by applying some fixed rule. For example: Whenever I need a digit other than 1, I’ll pick 7; whenever I need a digit other than 2, I’ll pick 5…)
If you say “oops, I forgot that number; I’ll stick it in my list somewhere”, I’ll just pull the same trick again and find another number that’s not on your list. So no matter how clever you are, you can never list all the real numbers—or even just the real numbers between 0 and 1.
But we saw that there is a way to list all the rational numbers. So in what turns out to be a profound and fundamental sense, the infinity of real numbers is bigger than the infinity of rational numbers.
With that discovery, Cantor taught the world how think about infinity, rocked the foundations of mathematics, and, with a lag of a hundred and some years, changed my life.
#### 24 Responses to “Split Infinities”
1. 1 1 Ross Parker
I fear you are confusing or conflating ‘size’ and ‘listableness’ (susceptibility to being listed).
What does listing things have to do with their size?
I can list the planets in our solar system, but I can’t list the possible combinations of snowflakes. Still, I’m pretty sure most planets are bigger than most snowflakes.
What does it mean for an infinity to be ‘small’?
2. 2 2 Steve Landsburg
Ross Parker:
Two finite sets have the same size if and only if they can be put in one-to-one correspondence with each other. Bertrand Russell illustrated this point by observing that at a large table that has been set for dinner, you can know that the number of forks is equal to the number of knives (that is, the set of forks and the set of knives have the same size) without counting them, as long as there is one fork and one knife at each place setting; this establishes the needed one-to-one correspondence.
Cantor’s idea was to carry this criterion over to infinite sets. To list a set is the same thing as to establish a one-to-one correspondence between that set and the natural numbers. (The first item on the list corresponds to the number 1, the second to the number 2, etc.). By this criterion, sets that can be listed have the same size as the set of natural numbers; sets that can’t be listed are larger.
3. 3 3 Harold
I seem to remember something about pairing one element in an infinity with another in another infinity. You can pair an integer with a natural number and remove from the list. You can keep doing this for every member of the sets, so they are in some way “the same size”. This is somewhat against intuition, as the integers go backwards as well as forwards, so might be thought of a “larger”. You cannot pair a natural number with each member of the real number set, so the real number set is “larger” than the natural number set. In a sense, there will always be some real numbers “left over”.
4. 4 4 JLA
I’m still confused about why uncountable and countable sets are “different” sizes. We know that the rationals and the irrationals are dense in the reals. If there are more irrational numbers than rational numbers, then there must be at least one pair of irrational numbers that don’t have a rational number in between them, contradicting the density of the rationals in the reals.
5. 5 5 Steve Landsburg
JLA: The definition of “different sizes” is that they can’t be put into one-one correspondence. The rationals and irrationals can’t be put in one-one correspondence; that doesn’t contradict the fact that both are dense in the reals.
6. 6 6 Dan Grayson
JLA: Each rational number can be used many many times as the rational number between a pair of real numbers. If each pair were assigned exclusive use of a rational number in between, then, indeed, there would not be enough to go around.
7. 7 7 Al V.
So, if the number of natural numbers is n, for example, then the number of integers is 2n. The number of rationals is n*n, and the number of reals is infinitely more (n*n*infinity, since there are infinitely many reals between two rationals), and can’t be represented. Is that why we can say that the natural numbers, integers, and rational numbers are countable, while the reals are not?
8. 8 8 ScottN
I got the same shock when I read Infinity and the Mind, by Rudy Rucker.
Too late for a mathematics major, though.
http://www.amazon.com/Infinity-Mind-Philosophy-Infinite-Princeton/dp/0691121273/ref=sr_1_1?ie=UTF8&s=books&qid=1267627120&sr=8-1-spell
9. 9 9 Steve Landsburg
Al V.: If the number of natural numbers is n, then the number of integers is 2n. But we’ve shown that there are as many natural numbers as integers, so n=2n. The number of rationals is n*n, and we’ve shown there are as many rational numbers as natural numbers. Therefore n=n*n.
The number of real numbers turns out to be 10^n (because to specify a real number you have to choose one of 10 digits in each of n locations). (It’s also 2^n, because you could write your real numbers in binary notation instead of decimal). But you can’t jump to the conclusion that 10^n is greater than n; that requires proof, and the proof (due to Cantor) is in the blog post.
“Countable” means “the same size as the set of natural numbers” so yes, that’s why we say the naturals, integers and rationals are countable but the reals are not.
10. 10 10 Jon Shea
Imagine you run the front desk at a hotel with a countably infinite number of rooms. Imagine further that all of the rooms are occupied. A man shows up and asks for a room. Can you give him a room?
Assume that you can shuffle people around by assigning them to new rooms. But you must give everyone a key with a room number on it. You can’t give anyone the “∞” key, because there is no such key. And even if there were such a key, they’d have to walk forever to get to the room, so that doesn’t seem fair.
What if a bus with an infinite number of people in it shows up, and they all want rooms?
What if an infinite number of busses each with an infinite number of people show up, and they all want rooms?
11. 11 11 Jon Shea
(Sorry about the poor formatting above. Beta version of a web browser.)
Imagine that you run the front desk at a hotel with a countably infinite number of rooms. Imagine further that all of the rooms are occupied. A man shows up in the lobby and asks for a room. Can you give him a room?
Assume that you can shuffle people around by assigning them to new rooms. But you must give everyone a room key with a number on it. You can’t give anyone the “∞” key, because there is no such key. And even if there were such a key, they’d have to walk forever to get to the “∞” room, so that doesn’t seem fair.
What if a bus with an infinite number of people shows up, and they all want rooms?
What if an infinite number of buses each with an infinite number of people show up, and each and every one of them wants a room.
12. 12 12 neil wilson
When I was in high school and I learned about the different infinities in math class I was so confused that N+N=N. It just never made sense to me. I still think that when something is twice as big as something else then it is bigger. I still think that when one list contains all the numbers on a second list and contains a number that is not on the second list then the first list is bigger than the second list by at least one. N+1=N.
But then I realized that if you threw a dart at the wall of all real numbers then there is a 0% chance of it hitting a rational number. The real number has no chance of repeating (like 1/3) and no chance of ending in an infinite number of 0′s (like 1/2)
I just don’t remember why the idea of showing that your decimal list doesn’t contain a number because I will pick the number with a different digiit is any different than saying that -2 is not on a list of positive numbers.
Of course, now I spend all day playing with a non-linear optimizer to try and steal money from the government so I really don’t need to remember.
It seems
13. 13 13 Jerome
This is something i have always wondered. If you have a container with an infinite number of blue balls and an infinite number of red balls what is your chance of removing a red ball? What if the blue balls were equal to the number of reals and the red balls equal to the integers? What if the red balls were equal to the number of irrationals? What are your probabilities for picking a red or blue ball in each of these cases?
14. 14 14 Steve Landsburg
Neil Wilson:
I just don’t remember why the idea of showing that your decimal list doesn’t contain a number because I will pick the number with a different digiit is any different than saying that -2 is not on a list of positive numbers.
This is the crux of the entire matter. If you attempt to list the integers, and if you do it stupidly (say by listing all the positive integers first) then your list might omit -2. But if you do it intelligently, as in 0,1,-1,2,-2,…, then your list will be complete.
By contrast, if you attempt to list the real numbers, it doesn’t matter how smart your are. Any list you come up with will omit something or other (though of course different lists will omit different things.)
15. 15 15 Steve Landsburg
Jerome: Your question is not entirely well posed because you havent told me exactly how you’re going to choose one ball from an infinite collection. But on any reasonable interpretation, if there’s a blue ball for ever real and a red ball for every integer, then the probability of picking a blue ball is one.
One way to convince yourself that this is plausible is this: when you choose a ball, it’s got a number on it. What is the chance that the first digit to the right of the decimal point is zero? Answer: 1/10. What is the chance that the first *two* digits are both zeros? Answer: 1/100. The first three digits? 1/1000. And so on. Now what is the chance that ALL of the digits to the right of the decimal point are zeros? (That’s what it takes to have an integer). It’s got to be smaller than 1/10, smaller than 1/100, smaller than 1/1000, smaller than 1/10,000….. and the only number that’s smaller than ALL of these is zero. So that’s the chance of chossing a red ball.
16. 16 16 Bennett Haselton
You can prove that the infinite set of *subsets* of the natural numbers, is larger than the set of natural numbers. The method is basically the same as the diagonal method used to prove that the set of real numbers is larger than the naturals: You line up the natural numbers in the left-hand column, and the subsets in the right-hand column. Then you can construct a new subset by making it different from the nth subset at the nth counting number — in other words, if the number n IS in the nth subset of your list, then it’s NOT in the new subset you’re creating, and vice versa. Therefore it’s different from every subset of the list.
You can generalize this method to prove that the set of *subsets* of a set, must be larger than that set. So the infinite number of subsets of the real numbers, is larger than the real numbers. Therefore, for any set, you have a set which is larger.
Here’s the paradox: Consider now the set of *everything*. All objects, all atoms that make up those objects, all real numbers — and all *sets* of things, all subsets of those sets, and all subsets of the set of subsets, and so on. Every Thing. If it would make grammatical sense as a noun, then it’s in.
Nothing can be larger than the set of everything because it contains everything. But we just proved that for any set, the set of subsets is larger, right?
I have no idea what the resolution to this paradox is. I know that, technically, the “set of everything” is not considered a “set” — it’s too big, so it’s a “class”, not a set. But I’m not sure how that terminology resolves the paradox. Can it be put in one-to-one correspondence with its subsets, or not? On the one hand, it should be at least as big as its subsets, since it contains all of its own subsets (along with everything else). On the other hand, the diagonal method proves that you can’t put anything in one-to-one correspondence with its own subsets.
17. 17 17 Steve Landsburg
The bottom line of that blog post will be something like this:
When we set out to axiomatize the natural numbers, we write down some plausible axioms (e.g. the Peano axioms) and hope they will characterize the natural numbers. And we always fail, because there are always structures *other* than the natural numbers that also satisfy our axioms. Nevertheless, out of all those structures (which are called “models of arithmetic”) there is one standard model which is what we know we have in mind.
When we set out to axiomatize set theory, we write down some plausible axioms (e.g. the Zermelo Frankel axioms) and hope they will characterize the universe of sets. And we always fail, because there are always multiple structures that satisfy those axioms. We call those structures “models of set theory”. But here the analogy with arithmetic breaks down, because it’s not at all clear that we can pick out one “standard model” that corresponds to what we really had in mind all along. Instead, it seems likely that our thoughts were fuzzy enough that we never had any particular model in mind to begin with.
So when we talk about “the natural numbers” we have something very particular in mind, even though we can’t characterize it with axioms. By contrast, when we talk about “the universe of sets” we have only a very fuzzy picture in mind and are not really thinking about any one thing in particular.
18. 18 18 dave
“The integers, by contrast (that is, the natural numbers plus their negatives) aren’t automatically listed because a list, by definition, has a starting point, whereas the integers stretch infinitely far in both directions.”
i find it interesting that you gave these lists direction as if they were vectors.
theres another reason why numbers dont exist somewhere in this discussion. i can smell it.
19. 19 19 Ken Braithwaite
Steven
If you are interested you might be able to read Cohen’s book Set Theory and the Continuum Hypothesis, out in Dover. It deals with “forcing” a mehtod Cohen invented to show the CH is independent of ZF. (Godel proved consistency.)
20. 20 20 Steve Landsburg
Ken Braithwaite: I did work through a substantial part of Cohen’s book many many years ago. It also helped a lot to have the Lawvere/Tierney reworking of Cohen’s argument close at hand.
21. 21 21 Al
Prof. Landsburg,
Maybe I have missed something, but it appears to me that the only difference between attempting to list all of the real numbers and all of the natural numbers is that, with the former – if I am methodical about it – you will still be able to find numbers I have not listed between those I have.
It seems that it is still equal impossible to list all of the natural numbers however, owing to the fact that the list would be infinitely long. However many natural numbers you care to list I can still find one which you haven’t listed in much the same way as you can find real numbers which I have failed to list.
The only difference is that – again, you are methodical about your listing – the numbers I point out are not on your list are beyond the limit of the list rather than between numbers which are already included in the list.
22. 22 22 Steve Landsburg
Al: Of course it would take me forever to physically list all the natural numbers, but that’s not exactly what “list” means here. Instead, “list” means “give a rule for generating a list that will eventually get around to any given number.” In that sense, I can list the natural numbers with the simple rule “Name the numbers in their usual order”. Does this eventually get up to 17? Yes, after 17 steps. Does it eventually get up to 17,245,468? Yes, after 17,245,468 steps. And so on.
The integers and the rational numbers are listable by that criterion; I can specify “listing rules” that will eventually produce every given one of them. For the reals, that’s impossible.
1. 1 Weekend Roundup at Steven Landsburg | The Big Questions: Tackling the Problems of Philosophy with Ideas from Mathematics, Economics, and Physics
2. 2 countable infinity |
Courses
# Fractions - Objective Type Question Class 6 Notes | EduRev
## Class 6 : Fractions - Objective Type Question Class 6 Notes | EduRev
``` Page 1
Objective Type Questions page: 6.36
Mark the correct alternative in each of the following:
1. Which of the following is a proper fraction?
(a) 4/3
(b) 3/4
(c) 13/4
(d) 21/5
Solution:
The option (b) is correct answer.
We know that in a proper fraction, the numerator is less than the denominator.
2. Which of the following is an improper fraction?
(a) 1/2
(b) 3/7
(c) 7/3
(d) 3/15
Solution:
The option (c) is correct answer.
We know that in an improper fraction, the numerator is more than the denominator.
3. Which of the following is a fraction equivalent of 2/3?
(a) 4/5
(b) 8/6
(c) 10/25
(d) 10/15
Solution:
The option (d) is correct answer.
Consider
10/15=2/3
By cross multiplication
10 × 3 = 2 × 15
We get
30 = 30
4. A fraction equivalent to 3/5is
(a) 3+2/5+2
(b) 3-2/5-2
(c) 3×2/5×2
(d) None of these
Solution:
The option (c) is correct answer.
We know that by dividing the numerator and denominator by 2, we obtain 3/5.
5. If 5/12 is equivalent of x/3, then x =
Page 2
Objective Type Questions page: 6.36
Mark the correct alternative in each of the following:
1. Which of the following is a proper fraction?
(a) 4/3
(b) 3/4
(c) 13/4
(d) 21/5
Solution:
The option (b) is correct answer.
We know that in a proper fraction, the numerator is less than the denominator.
2. Which of the following is an improper fraction?
(a) 1/2
(b) 3/7
(c) 7/3
(d) 3/15
Solution:
The option (c) is correct answer.
We know that in an improper fraction, the numerator is more than the denominator.
3. Which of the following is a fraction equivalent of 2/3?
(a) 4/5
(b) 8/6
(c) 10/25
(d) 10/15
Solution:
The option (d) is correct answer.
Consider
10/15=2/3
By cross multiplication
10 × 3 = 2 × 15
We get
30 = 30
4. A fraction equivalent to 3/5is
(a) 3+2/5+2
(b) 3-2/5-2
(c) 3×2/5×2
(d) None of these
Solution:
The option (c) is correct answer.
We know that by dividing the numerator and denominator by 2, we obtain 3/5.
5. If 5/12 is equivalent of x/3, then x =
(a) 5/4
(b) 4/5
(c) 5/3
(d) 3/5
Solution:
The option (a) is correct answer.
Consider 5/12 = x/3
By cross multiplication
5 × 3 = 12 × x
So we get
x = (5 × 3)/12 = (5 × 3)/ (4 × 3) = 5/4
6. Which of the following are like fractions?
(a) 3/5, 3/7, 3/11, 3/16
(b) 5/11, 7/11, 15/11, 2/11
(c) 2/3, 3/4,4/5, 6/7
(d) None of these
Solution:
The option (b) is correct answer.
We know that like fractions are the fractions with the same denominator.
7. If 11/4 = 77/x, then x =
(a) 28
(b) 77/28
(c) 44
(d) 308
Solution:
The option (a) is correct answer.
11/4 = 77/x
By cross multiplication
11 × x = 77 × 4
x = (77 × 4)/ 11 = (7 × 11 × 4)/ 11
Dividing both the numerator & denominator by 11, we obtain 28.
8. 1/ (2 1/3) +1/ (1 3/4) is equal to
(a) 7/14
(b) 12/49
(c) 4 1/12
(d) None of these
Solution:
The option (d) is correct answer.
Page 3
Objective Type Questions page: 6.36
Mark the correct alternative in each of the following:
1. Which of the following is a proper fraction?
(a) 4/3
(b) 3/4
(c) 13/4
(d) 21/5
Solution:
The option (b) is correct answer.
We know that in a proper fraction, the numerator is less than the denominator.
2. Which of the following is an improper fraction?
(a) 1/2
(b) 3/7
(c) 7/3
(d) 3/15
Solution:
The option (c) is correct answer.
We know that in an improper fraction, the numerator is more than the denominator.
3. Which of the following is a fraction equivalent of 2/3?
(a) 4/5
(b) 8/6
(c) 10/25
(d) 10/15
Solution:
The option (d) is correct answer.
Consider
10/15=2/3
By cross multiplication
10 × 3 = 2 × 15
We get
30 = 30
4. A fraction equivalent to 3/5is
(a) 3+2/5+2
(b) 3-2/5-2
(c) 3×2/5×2
(d) None of these
Solution:
The option (c) is correct answer.
We know that by dividing the numerator and denominator by 2, we obtain 3/5.
5. If 5/12 is equivalent of x/3, then x =
(a) 5/4
(b) 4/5
(c) 5/3
(d) 3/5
Solution:
The option (a) is correct answer.
Consider 5/12 = x/3
By cross multiplication
5 × 3 = 12 × x
So we get
x = (5 × 3)/12 = (5 × 3)/ (4 × 3) = 5/4
6. Which of the following are like fractions?
(a) 3/5, 3/7, 3/11, 3/16
(b) 5/11, 7/11, 15/11, 2/11
(c) 2/3, 3/4,4/5, 6/7
(d) None of these
Solution:
The option (b) is correct answer.
We know that like fractions are the fractions with the same denominator.
7. If 11/4 = 77/x, then x =
(a) 28
(b) 77/28
(c) 44
(d) 308
Solution:
The option (a) is correct answer.
11/4 = 77/x
By cross multiplication
11 × x = 77 × 4
x = (77 × 4)/ 11 = (7 × 11 × 4)/ 11
Dividing both the numerator & denominator by 11, we obtain 28.
8. 1/ (2 1/3) +1/ (1 3/4) is equal to
(a) 7/14
(b) 12/49
(c) 4 1/12
(d) None of these
Solution:
The option (d) is correct answer.
9. If 1/3 + 1/2 + 1/x = 4, then x = ?
(a) 5/18
(b) 6/19
(c) 18/5
(d) 24/11
Solution:
The option (b) is correct answer.
It is given that
1/3 + ½ + 1/x = 4
On further calculation
1/x = 4 – 1/3 – 1/2
By taking LCM of 3 and 2 as 6
1/x = 24/6 – 2/6 – 3/6
So we get
1/x = (24 – 2 – 3)/ 6 = 19/6
Hence, x = 6/19
10. If 1/2 + 1/x = 2, then x =
(a) 2/5
(b) 5/2
(c) 3/2
(d) 2/3
Solution:
The option (d) is correct answer.
It is given that
½ + 1/x = 2
On further calculation
1/x = 2 – 1/2
By taking LCM as 2 we get
1/x = 4/2 – 1/2 = (4 – 1)/2 = 3/2
Hence, x = 2/3
11. Which of the following fractions is the smallest?
1/2,3/7,3/5,4/9
(a) 4/9
(b) 3/5
(c) 3/7
Page 4
Objective Type Questions page: 6.36
Mark the correct alternative in each of the following:
1. Which of the following is a proper fraction?
(a) 4/3
(b) 3/4
(c) 13/4
(d) 21/5
Solution:
The option (b) is correct answer.
We know that in a proper fraction, the numerator is less than the denominator.
2. Which of the following is an improper fraction?
(a) 1/2
(b) 3/7
(c) 7/3
(d) 3/15
Solution:
The option (c) is correct answer.
We know that in an improper fraction, the numerator is more than the denominator.
3. Which of the following is a fraction equivalent of 2/3?
(a) 4/5
(b) 8/6
(c) 10/25
(d) 10/15
Solution:
The option (d) is correct answer.
Consider
10/15=2/3
By cross multiplication
10 × 3 = 2 × 15
We get
30 = 30
4. A fraction equivalent to 3/5is
(a) 3+2/5+2
(b) 3-2/5-2
(c) 3×2/5×2
(d) None of these
Solution:
The option (c) is correct answer.
We know that by dividing the numerator and denominator by 2, we obtain 3/5.
5. If 5/12 is equivalent of x/3, then x =
(a) 5/4
(b) 4/5
(c) 5/3
(d) 3/5
Solution:
The option (a) is correct answer.
Consider 5/12 = x/3
By cross multiplication
5 × 3 = 12 × x
So we get
x = (5 × 3)/12 = (5 × 3)/ (4 × 3) = 5/4
6. Which of the following are like fractions?
(a) 3/5, 3/7, 3/11, 3/16
(b) 5/11, 7/11, 15/11, 2/11
(c) 2/3, 3/4,4/5, 6/7
(d) None of these
Solution:
The option (b) is correct answer.
We know that like fractions are the fractions with the same denominator.
7. If 11/4 = 77/x, then x =
(a) 28
(b) 77/28
(c) 44
(d) 308
Solution:
The option (a) is correct answer.
11/4 = 77/x
By cross multiplication
11 × x = 77 × 4
x = (77 × 4)/ 11 = (7 × 11 × 4)/ 11
Dividing both the numerator & denominator by 11, we obtain 28.
8. 1/ (2 1/3) +1/ (1 3/4) is equal to
(a) 7/14
(b) 12/49
(c) 4 1/12
(d) None of these
Solution:
The option (d) is correct answer.
9. If 1/3 + 1/2 + 1/x = 4, then x = ?
(a) 5/18
(b) 6/19
(c) 18/5
(d) 24/11
Solution:
The option (b) is correct answer.
It is given that
1/3 + ½ + 1/x = 4
On further calculation
1/x = 4 – 1/3 – 1/2
By taking LCM of 3 and 2 as 6
1/x = 24/6 – 2/6 – 3/6
So we get
1/x = (24 – 2 – 3)/ 6 = 19/6
Hence, x = 6/19
10. If 1/2 + 1/x = 2, then x =
(a) 2/5
(b) 5/2
(c) 3/2
(d) 2/3
Solution:
The option (d) is correct answer.
It is given that
½ + 1/x = 2
On further calculation
1/x = 2 – 1/2
By taking LCM as 2 we get
1/x = 4/2 – 1/2 = (4 – 1)/2 = 3/2
Hence, x = 2/3
11. Which of the following fractions is the smallest?
1/2,3/7,3/5,4/9
(a) 4/9
(b) 3/5
(c) 3/7
(d) 1/2
Solution:
The option (c) is correct answer.
We know that the LCM of numerator is 12
By converting each fraction to an equivalent fraction having 12 as numerator
1/2 = 1/2 × 12/12 = 12/24
3/7 = 3/7 × 4/4 = 12/28
3/5 = 3/5 × 4/4 = 12/20
4/9 = 4/9 × 3/3 = 12/27
We know that if the numerator is same the fraction having larger denominator is the smallest.
Hence, 3/7 is the smallest fraction.
12. Which of the following fractions is the greatest of all?
7/8, 6/7, 4/5, 5/6
(a) 6/7
(b) 4/5
(c) 5/6
(d) 7/8
Solution:
The option (d) is correct answer.
We know that the LCM of 8, 7, 6 and 5 is 840
By converting each fraction to an equivalent fraction having 840 as denominator
7/8 = 7/8 × 105/105 = 735/840
6/7 = 6/7 × 120/120 = 720/840
4/5 = 4/5 × 168/168 = 672/840
5/6 = 5/6 × 140/140 = 700/840
We know that if the denominator is same the fraction having larger numerator is the greatest.
Hence, 7/8 is the greatest fraction.
13. What is the value of a+b/ a-b, If a/b=4?
(a) 3/5
(b) 5/3
(c) 4/5
(d) 5/4
Solution:
The option (b) is correct answer.
It is given that a/b = 4
We can write it as a = 4b
By substituting the value of a in a+b/a-b
a+b/a-b = 4b+b/4b-b = 5b/3b
Dividing numerator and denominator by b, the value is 5/3.
14. If a/b = 4/3, then the value of 6a+4b/ 6a-5b is
(a) -1
(b) 3
(c) 4
(d) 5
Page 5
Objective Type Questions page: 6.36
Mark the correct alternative in each of the following:
1. Which of the following is a proper fraction?
(a) 4/3
(b) 3/4
(c) 13/4
(d) 21/5
Solution:
The option (b) is correct answer.
We know that in a proper fraction, the numerator is less than the denominator.
2. Which of the following is an improper fraction?
(a) 1/2
(b) 3/7
(c) 7/3
(d) 3/15
Solution:
The option (c) is correct answer.
We know that in an improper fraction, the numerator is more than the denominator.
3. Which of the following is a fraction equivalent of 2/3?
(a) 4/5
(b) 8/6
(c) 10/25
(d) 10/15
Solution:
The option (d) is correct answer.
Consider
10/15=2/3
By cross multiplication
10 × 3 = 2 × 15
We get
30 = 30
4. A fraction equivalent to 3/5is
(a) 3+2/5+2
(b) 3-2/5-2
(c) 3×2/5×2
(d) None of these
Solution:
The option (c) is correct answer.
We know that by dividing the numerator and denominator by 2, we obtain 3/5.
5. If 5/12 is equivalent of x/3, then x =
(a) 5/4
(b) 4/5
(c) 5/3
(d) 3/5
Solution:
The option (a) is correct answer.
Consider 5/12 = x/3
By cross multiplication
5 × 3 = 12 × x
So we get
x = (5 × 3)/12 = (5 × 3)/ (4 × 3) = 5/4
6. Which of the following are like fractions?
(a) 3/5, 3/7, 3/11, 3/16
(b) 5/11, 7/11, 15/11, 2/11
(c) 2/3, 3/4,4/5, 6/7
(d) None of these
Solution:
The option (b) is correct answer.
We know that like fractions are the fractions with the same denominator.
7. If 11/4 = 77/x, then x =
(a) 28
(b) 77/28
(c) 44
(d) 308
Solution:
The option (a) is correct answer.
11/4 = 77/x
By cross multiplication
11 × x = 77 × 4
x = (77 × 4)/ 11 = (7 × 11 × 4)/ 11
Dividing both the numerator & denominator by 11, we obtain 28.
8. 1/ (2 1/3) +1/ (1 3/4) is equal to
(a) 7/14
(b) 12/49
(c) 4 1/12
(d) None of these
Solution:
The option (d) is correct answer.
9. If 1/3 + 1/2 + 1/x = 4, then x = ?
(a) 5/18
(b) 6/19
(c) 18/5
(d) 24/11
Solution:
The option (b) is correct answer.
It is given that
1/3 + ½ + 1/x = 4
On further calculation
1/x = 4 – 1/3 – 1/2
By taking LCM of 3 and 2 as 6
1/x = 24/6 – 2/6 – 3/6
So we get
1/x = (24 – 2 – 3)/ 6 = 19/6
Hence, x = 6/19
10. If 1/2 + 1/x = 2, then x =
(a) 2/5
(b) 5/2
(c) 3/2
(d) 2/3
Solution:
The option (d) is correct answer.
It is given that
½ + 1/x = 2
On further calculation
1/x = 2 – 1/2
By taking LCM as 2 we get
1/x = 4/2 – 1/2 = (4 – 1)/2 = 3/2
Hence, x = 2/3
11. Which of the following fractions is the smallest?
1/2,3/7,3/5,4/9
(a) 4/9
(b) 3/5
(c) 3/7
(d) 1/2
Solution:
The option (c) is correct answer.
We know that the LCM of numerator is 12
By converting each fraction to an equivalent fraction having 12 as numerator
1/2 = 1/2 × 12/12 = 12/24
3/7 = 3/7 × 4/4 = 12/28
3/5 = 3/5 × 4/4 = 12/20
4/9 = 4/9 × 3/3 = 12/27
We know that if the numerator is same the fraction having larger denominator is the smallest.
Hence, 3/7 is the smallest fraction.
12. Which of the following fractions is the greatest of all?
7/8, 6/7, 4/5, 5/6
(a) 6/7
(b) 4/5
(c) 5/6
(d) 7/8
Solution:
The option (d) is correct answer.
We know that the LCM of 8, 7, 6 and 5 is 840
By converting each fraction to an equivalent fraction having 840 as denominator
7/8 = 7/8 × 105/105 = 735/840
6/7 = 6/7 × 120/120 = 720/840
4/5 = 4/5 × 168/168 = 672/840
5/6 = 5/6 × 140/140 = 700/840
We know that if the denominator is same the fraction having larger numerator is the greatest.
Hence, 7/8 is the greatest fraction.
13. What is the value of a+b/ a-b, If a/b=4?
(a) 3/5
(b) 5/3
(c) 4/5
(d) 5/4
Solution:
The option (b) is correct answer.
It is given that a/b = 4
We can write it as a = 4b
By substituting the value of a in a+b/a-b
a+b/a-b = 4b+b/4b-b = 5b/3b
Dividing numerator and denominator by b, the value is 5/3.
14. If a/b = 4/3, then the value of 6a+4b/ 6a-5b is
(a) -1
(b) 3
(c) 4
(d) 5
Solution:
The option (c) is correct answer.
It is given that a/b = 4/3
We can write it as a = 4b/3
By substituting the value of a in 6a+4b/6a-5b
15. If 1/5 - 1/6 = 4/x, then x =
(a) -120
(b) -100
(c) 100
(d) 120
Solution:
The option (d) is correct answer.
It is given that
1/5 – 1/6 = 4/x
LCM of 5 and 6 is 30
4/x = 6/30 – 5/30
On further calculation
4/x = 1/30
So we get
x = 4 (30) = 120
16. The fraction to be added to 6 7/15 to get 8 1/5 is equal to
(a) 11/15
(b) 1 1/15
(c) 44/3
(d) 3/44
Solution:
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
## Mathematics (Maths) Class 6
191 videos|224 docs|43 tests
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; |
# What Is the Least Common Multiple?
The Least Common Multiple is the smallest multiple that is common to two or more numbers.
# Understanding the Least Common Multiple
The multiples of a number are the numbers that result when that number is multiplied by an integer (a whole number).
The multiples of 2 are found by multiplying 2 by successive integers: 2 × 1 = 2 , 2 × 2 = 4...
The multiples of 3 are:
We see that 6 and 12 are both multiples of 2 and 3. They are common multiples.
6 is the smallest of the common multiples. It is the least common multiple of 2 and 3.
# How to Find the Least Common Multiple
The least common multiple of any numbers can be found.
If you wanted to find the least common multiple of 3 and 4, you would first list the multiples of each number...
Multiples of 3 = 3, 6, 9, 12, 15, 18
Multiples of 4 = 4, 8, 12, 16, 20, 24
...then find the smallest factor that appears in both lists:
Multiples of 3 = 3, 6, 9, 12, 15, 18
Multiples of 4 = 4, 8, 12, 16, 20, 24
12 is the least common multiple because it is the smallest multiple of both 3 and 4.
show
# What's in a Name?
The least common multiple (LCM) is so called because it is the smallest (least) multiple that is the same (common) to the numbers.
# What Is a Multiple?
A multiple is the result of multiplying a number by an integer (a whole number).
For example, the multiples of 3 are:
3, 6, 9, 12, 15
• 3 is the result of multiplying 3 by 1.
• 6 is the result of multiplying 3 by 2.
• 9 is the result of multiplying 3 by 3.
• 12 is the result of multiplying 3 by 4.
• 15 is the result of multiplying 3 by 5.
# Finding the Least Common Multiple of More than Two Numbers
In the examples here, the least common multiple has been been found for two numbers. But the same procedure extends to any number of numbers. |
Series and Parallel Resistors
We present examples of circuits with series and parallel resistors and the formulas to calculate the equivalent resistance of these groups of resistors.
Resistors in Series
The resistors R1, R2 ...., Rm in the circuit on the left side are said to be in series because the same current passes through them. They behave in the same way as the circuit on the right of resistance Req given by the sum of the resistances R1, R2 and R3.
Req = R1 + R2 ....+ Rm
The current I in the above circuit is given by
I = E / Req
Resistors in Parallel
The voltage across each of the resistors R1, R2 ...., Rm in the circuit on the left is the same and therefore these resistors are said to be in parallel. They behave in the same way as the circuit on the right of resistance Req that is given by the equation:
1 / Req = 1 / R1 + 1 / R2 + .... + 1 / Rm
Resistors in Series: Examples with Detailed Solutions
Example 1
Find the current I passing through and the voltage across each of the resistors in the circuit below.
The three resistor in series have a resistance Req given by the sum of the three resistances. Hence
Req = 100 + 400 + 200 = 700 Ω
The current I passing through R1, R3 and R3 is the same and is calculated as follows:
I = 7 v / 700 Ω = 0.01 A
The voltage across each resistance is calculated using
Ohm's law as follows:
The voltage across 100Ω: VR1 = 100 × I = 100 × 0.01 = 1 v
The voltage across 400Ω: VR2 = 400 × I = 400 × 0.01 = 4 v
The voltage across 200Ω: VR3 = 200 × I = 200 × 0.01 = 2 v
Resistors in Parallel: Examples with Detailed Solutions
Example 2
Find current I in the circuit below and the current passing through each of the resistors in the circuit.
Solution to Example 2
The three resistors are in parallel and behave like a resistor with resistance Req given by
1 / Req = 1 / 100 + 1 / 400 + 1 / 200
Multiply all terms by 400 and simplify to obtain
400 / Req = 4 + 1 + 2
Solve for Req to obtain
Req = 400 / 7 Ω
The main current I is given by
I = 7 / Req = 7 / (400 / 7) = 49 / 400 A
We now use Ohm's law to find the current passing through each resistor.
The current through the resistor of 100 Ω: I1 = 7 / 100 A
The current through the resistor of 400 Ω: I2 = 7 / 400 A
The current through the resistor of 200 Ω: I3 = 7 / 200 A
As an exercise; check that the sum of the three currents above is equal to the current I = 49 / 400 A.
Series and Parallel Resistors: Examples with Detailed Solutions
Example 3
Find current I in the circuit below.
Solution to Example 3
The two resistors that are in series are grouped as Req1 in the equivalent circuit below and their resistance is given by the sum
Req1 = 100 + 400 = 500 Ω
The two resistors that are in parallel are grouped as Req2 in the equivalent circuit below and their resistance is given by the equation
1 / Req2 = 1 / 100 + 1 / 200
Solve to obtain
Req2 = 200 / 3 Ω
Req1 and Req2 are in series and therefore are equivalent to R given by the sum
R = Req1 + Req2 = 500 + 200 / 3 = 1700 / 3 Ω
We now use Ohm's law to find current I.
I = 6 / R = 6 / (1700 / 3) = 18 / 1700 A
Example 4
What resistance x in parallel with resistances 100 Ω and 200 Ω gives an equivalent resistance of 50 Ω?
Solution to Example 4
Let x be the resistance to be found. The equivalent resistance of the all three resistor in parallel is known. We use the equation that gives the equivalent resistance of resistors in parallel as follows
1 / 50 = 1 / 100 + 1 / 200 + 1 / x
which gives
1 / x = 1 / 50 - 1 / 100 - 1 / 200
Set all fractions on the right to the common denominator 200 and rewrite the above equation as
1 / x = 4 / 200 - 2 / 200 - 1 / 200 = 1 / 200
solve for x to obtain
x = 200 Ω
Example 5
Show that if the resistors with resistances R1, R2 ,..., Rm are in parallel, then the equivalent resistance Req is always smaller than R1, R2, ..., Rm.
Solution to Example 5
The equivalent resistance Req is given by the equation
1 / Req = 1 / R1 + 1 / R2 + ... + 1 / Rm
Since R1, R2, ... Rm are positive quantities, we can write that
1 / Req > 1 / Ri , where Ri is any of the resistances.
Multiply all terms of the inequality above by (Req × Ri) and simplify to obtain
Ri > Req
or
Req < Ri , i = 1, 2, ... m. |
S k i l l
i n
A L G E B R A
4
# MULTIPLYING AND DIVIDING SIGNED NUMBERS
We can only do arithmetic in the usual way. To calculate 5(−2), we have to do 5· 2 = 10 -- and then decide on the sign. Is it +10 or −10? For the answer, we have the followng Rule of Signs.
1. What is the Rule of Signs for multiplying, dividing, and
fractions?
Do the problem yourself first!
Like signs produce a positive number;
unlike signs, a negative number.
Example 1. −5(−2) = 10. Like signs. 5(−2) = −10. Unlike signs. −12−4 = 3. Like signs. 12 −4 = −3. Unlike signs.
For an explaination of these rules, see below.
2. Write the formal Rule of Signs as it applies to fractions.
−a−b = ab −a b = − ab a −b = − ab
A formal rule is simply a rule we write with letters. We write it with letters because we want it to apply to any numbers. Algebra, after all, depends on how things look.
Example 2. −2−6 = 13 −26 = − 13 1 −3 = − 13
Problem 1. Calculate the following.
a) 7(−8) = −56 b) (−7)8 = −56 c) 8(−7) = −56 d) −8(−7) = 56 e) (−3)7 = −21 f) 5(−9) = −45 g) −6(−9) = 54 h) −8(−4) = 32
Problem 2. Evaluate the following. (Be careful to distinguish the operations.)
a) 4 − 6 = −2 b) 4(−6) = −24 c) (−4) − 6 = −10 d) (−4)(−6) = 24 e) 5 − 8· 2 = −11 f) (5 − 8)· 2 = −6 g) 5 − 8 + 2 = −1 h) 5 − (8 + 2) = −5 i) 2 − 3(−6) = 20 j) (2 − 3)(−6) = 6
Example 3. The form ab(−c). Consider a problem in this form:
3 − 5(−2).
We are to subtract 5 times −2:
3 − 5(−2) = 3 − (−10) = 3 + 10 = 13.
And so even though the problem means to subtract (5 times −2), we may interpret it to mean: −5 times −2 = +10. We may simply write
3 − 5(−2) = 3 + 10 = 13.
In other words, any problem that looks like this --
ab(−c)
-- we may evaluate like this:
a + bc.
Problem 3. Evaluate the following.
a) 8 − 2(−4) = 8 + 8 = 16 b) 9 − 5(−2) = 9 + 10 = 19 c) −20 − 3(−5) = −5 d) −70 − 9(−7) = −7 e) 3 + 4(−9) = −33 f) −6 + 5(−2) = −16 g) −10 − 2(4 −8) = −2 h) (−10 − 2)(4 −8) = (−12)(−4) = 48
Problem 4. Two variables. Let the value of y depend on the value
of x as follows:
y = 3x − 6.
Calculate the value of y that corresponds to each value of x:
When x = 0, y = 3· 0 − 6 = 0 − 6 = −6.
When x = 1, y = 3· 1 − 6 = 3 − 6 = −3 .
When x = −1, y = 3· −1 − 6 = −3 − 6 = −9.
When x = 2, y = 3· 2 − 6 = 6 − 6 = 0.
When x = −2, y = 3· −2 − 6 = −6 − 6 = −12.
When x = 3, y = 3· 3 − 6 = 9 − 6 = 3.
When x = −3, y = 3· −3 − 6 = −9 − 6 = −15.
Problem 5. Negative factors.
a) (−2)(−2) = 4 b) (−2)(−2)(−2) = −8 c) (−2)(−2)(−2)(−2) = 16 d) (−2)(−2)(−2)(−2)(−2) = −32
Problem 6. According to the previous problem:
An even number of negative factors produces a positive number. While an odd number of negative factors produces a negative number.
Example 4. Multiply −2(−9)7(−5).
Solution. Before even multiplying, we can see that because there are an odd number of negative factors, the sign will be negative.
−2(−9)7(−5) = −2· 9· 7· 5.
We have now only to multiply those numbers.
But the order of factors does not matter. (Lesson 1.) The multiplication therefore will be simpler if we first multiply 2· 5 --
−2· 9· 7· 5 = −2· 5· 9· 7 = −10· 63 = −630.
Multiplication is always simpler if factors will produce 10, or 100, or any power of 10.
Problem 7. Multiply.
a) 2(−3)4 = −24 b) (−2)3(−4) = 24 c) 2(−3 −4) = −14 d) (−3)(−4)(−5) = −60 e) (−1)(−2)(−3)(−4) = 24 f) (−2)8(−5)7 = 560 g) 25(−8)(−3)(−4) = −100· 24 = −2400 h) (−1)(−1)(−1) = −1 i) (−1)(−1)(−1)(−1) = 1
Problem 8. Evaluate each of the following as a positive or negative fraction in lowest terms, or as an integer.
a) −24 6 = − 4 b) 24−6 = − 4 c) −24 −6 = 4 d) 3 −12 = − 14 e) −8−20 = 25 f) −18 42 = − 37 g) −23 = − 23 h) 2 −3 = − 23 i) −2−3 = 23 j) −12 3 = − 4 k) −5 −20 = 14 l) 3 −4 = − 34
Example 5. Multiplying fractions.
To multiply fractions, multiply the numerators
and multiply the denominators, as in arithmetic.
Observe the Rule of Signs.
Problem 9. Multiply.
a) −3 5 · 78 = −21 40 = − 2140 b) 12 · −x 4 = −x 8 = − x8
c) −2 3 · x −8 = −2x−24 = x12 d) x −6 · 3 −5 = 3x30 = x10
An explanation of the Rule of Signs
To decide how negative numbers should behave, we are not able to copy arithmetic. Rather, we have to respect the either-or, yes-or-no nature of logic.
For example, the introduction of the word "not" into a statement changes its truth value. If the statement was true, "not" makes it false, and vice-versa. If this statement
Today is Monday
is true, then
Today is not Monday
is false. But if we write
Today is not not Monday,
then that changes its truth value again. That statement is true
Now in algebra we do not have true or false, but we do have the logical equivalent: positive or negative. Thus if the value of x is positive, then the value of −x must be negative, and vice-versa.
And so since we call the positive or negative value of a number its sign, then we can state the following principle:
A minus sign changes the sign of a number.
A minus sign is the logical equivalent of "not." Geometrically, a minus sign reflects a number symmetrically about 0.
We saw that with −(−3) = 3.
(As for 0, it is best to say that it has both signs: −0 = +0 = 0. See for example Lesson 11, Problem 11.)
If we now apply this principle to multiplication:
A negative factor changes the sign of a product.
Thus if ab is positive, then (−a)b cannot also be positive. It must be negative -- it must be the negative of ab.
(−a)b = −ab.
That is, "Unlike signs produce a negative number."
And upon introducing another negative factor, the sign changes back:
(−a)(−b) = ab.
"Like signs produce a positive number."
This same logical principle will apply to division and fractions. Hence we have the Rule of Signs.
To prove that (−a)b is the negative of ab in what some call a rigorous manner, we would have to apply the definition of the negative of a number. We would have to prove:
ab + (−a)b = 0.
That will be Problem 13 of the Lesson on Common Factor.
Next Lesson: Reciprocals and zero
Please make a donation to keep TheMathPage online.
Even \$1 will help. |
# Matrix Multiplication – Matlab
Matrix multiplication is likely to be a source of a headache when you fail to grasp conditions and motives behind them. Here, we will talk about two types of matrix multiplication and how you can handle them both manually and using Matlab.
One of the best ways to test your understanding of the following is to work them out manually and then using Matlab to check your result.
## Matrix multiplication theory reminder
Let’s consider the following matrices.
### Matrix product
Here is the formula for multiplying the above matrices, and I will highly recommend you check the properties of Matrix multiplication and most importantly the dimension agreement that is crucial between two matrices that need to be part of a multiplication.
Generally speaking, if A is an n × m matrix and B is an m × p matrix, their matrix product AB is an n × p matrix, in which the m elements across the rows of A are multiplied with the m elements down the columns of B
### Matrix multiplication element by element
In the other side, we have the element by element matrix multiplication, which is rather a straightforward operation, here is the formula used.
It is simply the product of matrices, element by element, this type works only when the dimensions of the matrices are equal. To be more specific, if A is an n × m matrix, B has to be an n × m matrix for this to work.
## Matlab Matrix Multiplication
The following code allows finding a matrix product in Matlab
`C=A*B`
and this one is the code to find the product of matrices, element by element
`C=A.*B`
## Matrix multiplication examples
### Example 1
If we keep the same logic as above while varying the value of A and B, but knowing that C is the matrix product and D is the element by element matrix multiplication.
Matlab code
```A=[1 2 2; 1 0 5; 3 1 2];
B=[3 2 5;3 0 0; 1 1 2];
C=A*B
D=A.*B```
Results
### Example 2
Matlab code
```A=[1 2;1 5;3 2];
B=[3 2;1 1];
C=A*B
```
Results
### Example 3
Matlab code
```A=[1 2;1 5;3 2];
B=[3 2;3 0; 1 1];
D=A.*B```
Results |
## Understanding Domain and Range Part 3
In the previous post, we have learned how to analyze equations of functions and determine their domain and range. We have observed that the range of the functions $y = x^2$ and $y = |x|$ are the set of real numbers greater than or equal to $0$ since squaring a number or getting its absolute value results to $0$ or a positive real number. We also learned that for a function to be defined, the number under the square root sign must be greater than or equal to 0. Lastly, we have learned that we cannot divide by zero because it will make the function undefined.
In this post, we are going to continue our discussion by examining functions with equations more complicated than those in the second part of this series.
Squares and Absolute Values
1. $f(x) = x^2 - 3$
Domain: The function is defined for any real number $x$, so the domain of $f$ is the set of real numbers.
Range: The minimum value of $x^2$ is $0$ for any real number $x$ and $f(0) - 3 = 0^2 - 3 = -3$. So, the minimum value of the function is $-3$. We can make the value of the function as large as possible by increasing the absolute value of $x$. So, the range of the function is the set of real numbers greater than or equal to $-3$ or $[-3, \infty)$ in interval notation. » Read more
## Understanding Domain and Range Part 1
The domain of a function is the set of x-coordinates of the points in the function. The range of the function f is the set of y-coordinates of the points in the function. So if we have a function f with points (-3, -2), (-1, 3), (2, 3), and (5,4), then the domain of the function f is the set {-3, -1, 2, 5} and the range of f is the set {-2, 3, 4). Graphically, we can say that the domain is the “projection” of the points to the x-axis (see red points in the following figure).
The range of f is the projection of the points to the y-axis (see green points in the following figure). » Read more
## What exactly is the vertical line test?
A function as we have discussed is a relationship between two sets, where each element in the first set has exactly one corresponding element in the second set. If we think of candies which cost 10 cents each, then we can say that 1 candy costs 10 cents, 2 candies cost 20 cents, 3 candies cost 30 cents, and so on. We can think of this relationship as a function since for each number of candies, there is only one possible price.
If we consider the relation y = 2x, then we can say that it is a function since for every value we substitute to x, there is one and only one corresponding value for y. For instance, if x = -3, then y = -6 and and if x = 9, then y = 18 (one y for each x). » Read more
1 2 |
# How many 3-digit passwords can be formed from the digits 0 through 9?
## How many three digit codes can be formed using digits 0 9?
There would be about 720 possible three – digit passwords that can be formed using digits 0 through 9 if digits are repeated. THis can be calculated by getting 10P3 in a calculator.
## How many 3 digit numbers can be formed by using the digits 0 to 9 if no digit is repeated?
How many 3 – digit numbers can be formed by using the digits 1 to 9 if no digit is repeated ? Therefore, total numbers =89×8×7504.
## How many codes can be formed using digits 0 9?
Now, there are 105 ways in which the digits 0 – 9 can be chosen for the five places of a five digit number. Out of these, 104 start with zero (once we start with 0 , there are only 4 slots to fill, where we have 10 choices each).
## How many 3 digit numbers can be formed from the digits?
So, required number of ways in which three digit numbers can be formed from the given digits is 5× 4 ×3=60.
## What are all the combinations for a 3 Number Lock?
2 Picking a Lock By comparison, this 3 -dial lock (three wheels, each with digits 0-9) has 10 10 10 = 1, 000 possible combinations .
## How many 3 digit numbers can be made using the digits 4 6 7 and 9?
There is 24 possible 3 digit numbers using 3 , 6 , 7 , 9 with no repeated digits .
## How many two digit numbers are there?
How many two-digit numbers are there? There are 90 two -digit numbers in all.
## How many 6 digit numbers are there in all?
∴ there are 900,000 6-digit numbers in all.
You might be interested: What is homegroup?
## How many 6 digit combinations are there using 0 9?
There are 900000 possible ways to get 6 digit numbers using 0 – 9 .
## What is every possible combination of 0 9?
There are 10,000 possible combinations that the digits 0-9 can be arranged into to form a four-digit code.
## How many 4 Number passwords can be chosen from the digits 0 9?
For each choice of the first two digits you have 10 choices for the third digit . Thus you have 10x10x10 = 1000 choices for the first three digits . Finally you have 10 choices for the fourth digit and thus there are 10x10x10x10 = 10 000 possible 4 digit combinations from 0 – 9 . |
# Is 172 a composite number?
## Is 172 a composite number?
In mathematics 172 is an even number, a composite number and a deficient number.
## What are the Factors of 173?
What are the Factors of 173? The factors of 173 are 1, 173 and its negative factors are -1, -173.
What is the prime factorization for 171?
So, the prime factorization of 171 can be written as 32 × 191 where 3, 19 are prime.
How do you find the factors of big numbers?
To calculate the factors of large numbers, divide the numbers with the least prime number, i.e. 2. If the number is not divisible by 2, move to the next prime numbers, i.e. 3 and so on until 1 is reached. Below is an example to find the factors of a large number.
### What are the natural numbers from 1 to 12?
First 12 natural numbers are: 1,2,3,4,5,6…. 12.
### Is 173 a prime number Why?
Yes, 173 is a prime number. The number 173 is divisible only by 1 and the number itself. For a number to be classified as a prime number, it should have exactly two factors. Since 173 has exactly two factors, i.e. 1 and 173, it is a prime number.
What is the prime factorization of 270?
Factors of 270: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, and 270. What Are the Factors of 270?…Factors of 270 in Pairs.
Factorization of 270 Factor Pairs
1 × 270 = 270 (1,270)
2 × 135 = 270 (2,135)
3 × 90 = 270 (3,90)
5 × 54 = 270 (5,54)
What are the factors of 177?
Factors of 177
• All Factors of 177: 1, 3, 59 and 177.
• Negative Factors of 177: -1, -3, -59 and -177.
• Prime Factors of 177: 3, 59.
• Prime Factorization of 177: 31 × 591
• Sum of Factors of 177: 240.
#### What are the factors of 170?
Factors of 170
• All Factors of 170: 1, 2, 5, 10, 17, 34, 85 and 170.
• Negative Factors of 170: -1, -2, -5, -10, -17, -34, -85 and -170.
• Prime Factors of 170: 2, 5, 17.
• Prime Factorization of 170: 21 × 51 × 171
• Sum of Factors of 170: 324.
#### What are the prime factors of 170?
The Prime Factors and Pair Factors of 170 are 2 × 5 × 17 and (1, 170), (2, 85), (5, 34), (10, 17) respectively. |
Finding the subtraction rules
Here are some very important things we have discovered so far:
Important understandings of ADDITION Adding a positive number goes to the right. Adding a negative number goes to the left.
Important understandings of SUBTRACTION Subtracting a positive number goes to the left. Subtracting a negative number goes to the right.
Putting these understandings together we find that
• Subtracting a negative number is the same as adding a positive number.
• Subtracting a positive number is the same as adding a negative number.
Here are two examples showing that each subtraction problem has a related addition problem that gives the exact same answer.
5 – (-8) = 13 5 + 8 = 13
5 – 2 = 3 5 + (-2) = 3
Let's practice a little:
Change the subtraction expression into an addition expression. Then simplify. (-4) – (+6) (-4) – (+6) will result in the same answer as (-4) + (-6). (-4) – (+6) = (-4) + (-6) = -10 (-7) – (-9) (-7) – (-9) will result in the same answer as (-7) + (+9) (-7) – (-9) = (-7) + (+9) = +2 (+3) – (+5) (+3) – (+5) will result in the same answer as (+3) + (-5) (+3) – (+5) = (+3) + (-5) = -2 (+8) – (-2) (+8) – (-2) will result in the same answer as (+8) + (+2) (+8) – (-2) = (+8) + (+2) = +10
By looking at the four examples above, we can see a pattern that occurs in each of the examples:
• The first integer stays the same;
• The subtraction sign turns into an addition sign;
• The second integer turns into its opposite.
In fact, the rule for subtracting positives and negatives is:
To subtract two integers: The first integer stays the same and add the opposite of the second integer
Self-Check
Question 1 Write the following expression using addition. Then simplify. (+8) – (-5) [show answer] (+8) – (-5) (+8) + (+5) = +13
Question 2 Write the following expression using addition. Then simplify. (-3) – (-4) [show answer] (-3) – (-4) (-3) + (+4) = +1
Question 3 Write the following expression using addition. Then simplify. (-6) – (+7) [show answer] (-6) – (+7) (-6) + (-7) = -13 |
Cynthia Lanius
# Finding the Formulas
Fitting a formula to data is an important skill, especially for making predictions. If you have a formula, you can use it to predict what would happen at the 99th stage or the 1000th stage, by just plugging that number into the formula. A word of caution: predictions are based upon the function maintaining the same rate of growth all the way to that stage. Let's examine each color individually and find the formulas for the Rectangle Pattern Challenge.
Blue The terms are 6, 10, and 14. Notice that the difference between each term is a constant - 4. That means the function is growing at a constant rate; its graph is a straight line; and its formula is in the form of y = mn + b where m is the slope and b is the y intercept (where the graph crosses the y axis).
How do we find m and b? Slope is the rate at which the function is changing. Well, that's 4, so that's m. The y intercept occurs where n = 0, so go back to n = 0 by subtracting 4 from the n = 1 stage, and we get 2. So the formula for blue is y = 4n + 2. Try all 3 stages and confirm that our formula works for all stages.
Red The terms are 2, 8, and 18. The difference between the terms are 6 and 10. It's not constant, so it's not linear. But what is it?
Let's try a difference of the differences. I need to know how many subtractions it takes to get a constant difference, but I don't really have enough information to determine that. I need another stage. Go back and look at Stage 3 and count the number of reds that we'd have on Stage 4 (add a row of reds on each side). Do you agree that it would be 32? Now our terms are 2, 8, 18, and 32. Our first differences are 6, 10, 14. And our second differences are 4 and 4. Aha! We now have our constant difference after 2 subtractions. That means we have a quadratic equation. (Yes, you are right - the rule is that the number of subtractions that it takes to get a constant difference is the largest power of the polynomial formula).
If the formula's quadratic, it's in the form of y = an2 + bn + c, and we must find the values of a, b, and c. Look at the red terms again in the table below. Notice that I added the 0 row, since there are 0 reds in Stage 0.
Stage No. (n) 0 1 2 3 4
No. Red Squares (y) 0 2 8 18 32
Since y = 0, when n = 0, substitute those values into y = an2 + bn + c and get
c = 0
When n = 1, y = 2, so let's substitute those values (and remember, c = 0) and get
2 = a + b
Substituting the next pair of n and y gives
8 = 4a +2b
This gives a system of two equations with two variables that we can solve.
2 = a + b
8 = 4a +2b
So, a = 2, b = 0, and c = 0, and the formula is y = 2n2. |
Chapter 9
# Chapter 9 - Chapter 9 Static Equilibrium Statics and...
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: Chapter 9 Static Equilibrium Statics and Equilibrium Equilibrium when the net force and the net torque acting on an object is zero. Object is either at rest or in motion at a constant velocity. Statics study of the forces acting on and within objects in equilibrium. Types of Equilibrium Stable Equilibrium once displaced the object returns to its original position. Unstable Equilibrium once displaced the object loses its equilibrium and accelerates away from its original position. Neutral Equilibrium once displace the object remains stationary in a new equilibrium position Types of Equilibrium Example: What type of equilibrium exists for the ball in each of the three locations below? 1st Condition of Equilibrium sum of all the forces acting on an object must be equal to zero. Conditions of Equilibrium Fy=0 Fz=0 2nd Condition of Equilibrium the sum of all the torques acting on an object must be equal to zero. =0 In order to calculate the torques an arbitrary axis of rotation must be chosen. ANY point may be taken to be the axis since the object will not rotate if in equilibrium. Chose a point that makes the problem easiest. Fx=0 Problem Solving with Statics Step 1: Draw a FBD (chose an xy plane, resolve forces in to x and y components) Step 2: If an extended object, chose an axis of rotation, usually one that eliminates a torque or one for which torque arms are easily measured. Step 3: Apply the equilibrium equations for forces and torque. Step 4: Solve for unknowns. You may need multiple equations for multiple unknowns. Problem Solving with Statics Example: Find the tension in the two ropes. Example: How far from the pivot point on a seesaw must an 80 kg man sit to exactly balance a 150 kg man sitting 2 meters from the pivot point on the opposite side? (Ignore the mass of the seesaw) Problem Solving with Statics Example: A 50.0 N weight is held in a person's hand with the forearm horizontal as shown. The biceps muscle is attached 0.03 m from the joint and the weight is 0.35 m from the joint. Calculate the upward force exerted by the biceps muscle and the downward force exerted by the humerus bone. Problem Solving with Statics Example: A uniform section of a bridge is 80 m long with a mass of 12000 kg and is supported at either end. A 1000 kg car sits 10 meters from one end of the section. What is the force in exerted by each support? Problem Solving with Statics A 700 N bear walks out along a uniform beam that is 6.00 m long and weighs 200 N toward a basket of goodies that weigh 80 N. The end of the beam is supported by a wire that makes and angle of 60.0 degrees. If the maximum tension the wire can withstand is 900 N, can the bear reach the goodies? ...
View Full Document
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern |
# How do you factor x^3-27v^3?
Apr 10, 2016
$\left(x - 3 v\right) \left({x}^{2} + 3 v x + 9 {v}^{2}\right)$
#### Explanation:
This problem is what is described as a Difference of Cubes
If we consider the equation:
${a}^{3} - {b}^{3}$ where $a$ and $b$ are perfect cubes, we can factorise the problem using the formula:
(a – b)(a^2 + ab + b^2)
In your case, if we consider the equation:
x^3−27v^3
We notice that not only are the $x$ and $y$ terms cubes, but 27 is also a cube $\left({3}^{3} = 27\right)$.
Therefore, if we substitute the values in to the Difference of Cubes formula, letting $x = a$ and $3 v = b$ we get:
$\left(x - 3 v\right) \left({x}^{2} + 3 v x + {\left[3 v\right]}^{2}\right)$
Simplifying this we get the factored equation:
$\left(x - 3 v\right) \left({x}^{2} + 3 v x + 9 {v}^{2}\right)$ |
# Multiply And Divide Fractions
## Multiplying Fractions:
Real life applications where common fractions are multiplied or divided include:
• image reduction
• adapting recipes for larger groups of people.
Steps to multiply fractions are:
• Step 1 - Change mixed numbers to improper fractions
• Step 2 - Multiply the numerators(top numbers) and multiply the denominators (bottom numbers)
• Step 3 - Simplify if needed.
The extra step to divide fractions is to multiply by the reciprocal of the second fraction (that is, turn the second fraction upside and then multiply).
## Example One - Image Reduction
An image is reduced to one half of its original size. If the original measurement is 34 cm, what is the new reduced measurement?
12 × 34
= 38 cm
## Example Two - Adapting Recipes for Large Groups
A risotto recipe for 5 people uses 2 12 cups of rice. How many cups of rice are required to feed 20 people?
Multiplier
= 20 ÷ 5
= 4
No. of cups of rice
= 2 12 × 4
= 10 cups
Q1. 13 × 25
Q2. 38 × 45
Q3. 2 14 × 3 23
A1. 215
A2. 310
A3. 8 14
## Example Three - Gym Training
In a gym the whole 2 12 hour training session is divided into 14 hour parts. How many parts is this?
2 12 ÷ 14
= 52 ÷ 14
= 52 × 41
= 202
= 10 parts
Q1. 67 ÷ 14
Q2. 38 ÷ 12
Q3. 2 45 ÷ 110 |
# The Mean Value Theorem and Rolle’s Theorem
The Mean Value Theorem and its’ special case Rolle’s Theorem are two of the fundamental theorems in differential calculus. These theorems are usually learned at schools right after the derivatives and sometimes after the extreme values. They both have very clear physical and geometrical interpretations. This is why you’ll often come across problems which are based on these theorems. Aside from ‘standard’ math problems, these theorems are often required to solve ‘theoretical’ problems which aren’t so rare in tests and those types of problems are usually the ones that students hate the most. They are widely considered difficult and boring which means it is essential to understand these theorems. But don’t be fooled – there are also some very interesting problems involving these theorems and we’ll cover those too. Here, we’ll start with Rolle’s Theorem. You might wonder why we would do something like that since it is a special case of The Mean Value Theorem, but we do it because it’s the simplest special case – and also because it helps understanding our main theorem. So without further a do, here it is:
Let’s see what this means in plain English. If you remember our lesson about extreme values (link placeholder) we said that they appear at points where the first derivative is zero. This also means that the tangent line of the function at that point is horizontal – parallel to x-axis. Knowing this, the theorem simply tells us that if the function has the same value on both ends of the interval then there must be a point inside interval at which the function takes turn – there must be some local extreme value inside. Let’s take a look at geometrical interpretation:
So the geometrical interpretation of Rolle’s theorem is this: if we observe two points of some contionous and differentiable (you can tell that it’s differentiable if it is smooth – no spikes) curve at which the function has the same value, then there must be some point inbetween at which the tangent line is horizontal – the first derivative is zero. Physicaly, the example of interpretation is a ball being tossed up in the air. Based on the theorem (and it’s very logical too), there will be some point in time at which the ball’s speed will be zero. This of course will happen at the ball’s highest point – the extreme value. This example’s intepretation is understandable from the graph if you look at x-axis as representation of time, and the curve is behaviour of the ball at different points in time. At point A, the ball is tossed and at point B the ball comes back down on the ground. In between, at some point it reaches maximum height – that’s when its’ speed is zero.
So after some simple factoring we got two solutions. However, we need our point c to be inside given interval, which means that the only solution will be x=2. That’s the point at which this functions’ tangent is horizontal.
Lagrange’s Mean Value Theorem
Or often just called The Mean Value Theorem can be viewed as a more general version of Rolle’s Theorem, since this one doesn’t require the function to have equal values at endpoints of interval. Here’s what it has to say:
Let’s slowly go through the theorem in order to understand it. So we have two endpoints, a and b. What would be the slope of a straight line passing through these two points? The slope of the line is a change in y divided by change in x, and in our case change in y is the difference between values of a function at endpoints f(b)-f(a) and change in x is difference between those endpoints, b – a. So the slope is . Since point c is between a and b, then the theorem tells us that for every continuos and differentiable function there is that point through which the tangent line of a function has the same slope as the slope of a secant line through endpoints. Here’s how it looks on a graph:
So geometricaly, the mean value theorem says that for a continous and differentiable (smooth) curve we can always find some point c, between two points a and b, such that the slope of a tangent line at that point is the same as the slope of a secant line through the endpoints. Physical interpretation would be something like this. Imagine we were driving from City A to City B which are 145 miles appart, and we got from one city to another in one hour. The mean value theorem says that if that is the case, then at some point in time during our journey we were driving at exactly 145 miles per hour (and greatly exceeding speed limit by doing so). In the graph, the distance between our two cities is the distance between points A and B, and the blue line would represent our speed at different points in time if we were driving at a constant speed of 145 miles per hour. Since we weren’t driving at constant speed, then our speed is represented by the curve, and at point c we were indeed driving at exactly 145 miles per hour.
To better understand the theorem it is important to know why our assumtions are that a function is continous and differentiable at given interval. Here are two examples at what could happen if our function weren’t continous or differentiable:
On the left side we have a function that is not contionous at (a,b), and as you can see there is no point inbetween at which the slope of the tangent line would be the same as the slope of the line passing through A and B. On the right side we have a function that is not differentiable on (a,b) (it has a spike, it’s not smooth), and therefore we also can’t find such point c. |
Top
# Bisector
Bisector is a line that divides a line or an angle in to two equivalent parts. There are two types of Bisectors based on what geometrical shape it bisects.
• Bisector of a Line
• Angle Bisector
Bisector of a line divides a line in to two equal parts. Angle bisector divides the angle in to two angles of equal measures.
## Bisector Definition
A bisector is a straight line that divides any figure into two equal parts. An angle or a line is bisected when they are cut into 2 equal parts.
## Segment Bisector
A bisector divides a figure into two congruent parts. The bisector of a segment always contains the midpoint of the segment.
### Line Segment Bisector
A line bisector divides the line segment into 2 equal parts. It passes through the midpoint of the line segment. In the below figure line PQ is the bisector of AB.
## Angle Bisector
An angle bisector divides an angle into equal angles. If the angle is x$^o$, each of the two angles formed will now be $(\frac{x}{2})^o$. The angle bisector passes through the vertex of the angle.
The figure below shows an angle bisector of $\angle$ ABC.
### Angle Bisector Theorem
The angle bisector of an angle in a triangle divides the opposite side in the same ratio as the sides adjacent to the angle.
### Angle Bisector Conjecture
If a point is on the bisector of an angle, then it is equidistant from the sides of the angle.
### Angle Bisector Problems
Find the value of x, if $\bar{BD}$ is the angle bisector of $\angle$ABC.
Since $\bar{BD}$ is the angle bisector of $\angle$ABC
=> 2x = 80$^o$
x = 40$^o$ (Divide each side by 2)
Thus, the value of x is 40$^o$.
## Triangle Bisector
An angle bisector of a triangle is a bisector of an angle of the triangle. An angular bisector of a triangle is the line segment which bisect an angle of the triangle and has its other endpoint on the opposite side of that angle.
### Triangle Bisector Theorem
An angle bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides.
In this figure, $\frac{AB}{AC} = \frac{BD}{DC}$
## Perpendicular Bisector
A perpendicular bisector is a segment, ray, or line that intersects a given segment at a 90$^o$, and passes through its mid point.
### Construct Perpendicular Bisector
To construct the perpendicular bisector of a segment follow the following steps:
Step 1: Draw a line segment $\bar{PQ}$.
Step 2: Set compass to more than half the distance between P and Q. Using P as center, swing an arc one side of the segment.
Step 3
: With same setting of the compass, swing another arc intersecting the first, using Q as a center. Label the intersecting point as R.
Step 4:
The point where the two arcs intersect is equidistant from the endpoints of PQ. Join R and the mid point of PQ, named as M.
### Perpendicular Bisector Theorem
A point lies on the perpendicular bisector of a segment if and only if it is equidistant from the endpoints of the segment.
In this figure: AH = BH, AG = BG, AF = BF and AE = BE.
### Perpendicular Bisector Equation
The equation of the perpendicular bisector of the segment joining the points ($x_1, y_1$) and ($x_2, y_2$).
Equation of $\bar{RM}$ is:
y - $\frac{y_1 + y _2}{2}$ = - $\frac{x_2 - x_1}{y_2 - y _1}$ (x - $\frac{x_1 + x _2}{2}$)
The slope of perpendicular bisector = Negative reciprocal of line segment's slope.
### Perpendicular Bisector Triangle
The perpendicular bisectors of the sides of a triangle intersect at a point. The point of concurrency of the perpendicular bisectors of a triangle is called the circumcenter of the triangle, which is equidistant from the vertices of the triangle.
### Perpendicular bisector conjecture
If a point is on the bisector of an line, then it is equidistant from the endpoints of the segment.
### Find Perpendicular Bisector
Find the equation of the perpendicular bisector of the line segment joining the points (1, 2) and (3, 6).
Step 1:
Find the slope and mid point of the line segment.
Slope of the line segment joining the points (1, 2) and (3, 6)
Slope (m)
= $\frac{6-2}{3-1}$ = $\frac{4}{2}$ = 2
Mid point: ($\frac{1+3}{2}$, $\frac{2+6}{2}$) = (2, 4)
Step 2:
Slope of perpendicular bisector, $m_1$ = $\frac{-1}{m}$ = $\frac{-1}{2}$
Step 3:
Equation of the bisector passing through mid point of line segment is
y - 4 = $\frac{-1}{2}$(x - 2)
2y - 8 = -x + 2
2y = -x + 10
or y = -0.5x + 5, which is required equation.
### Perpendicular Bisector Problems
Given below are some of the practice problems on perpendicular bisector.
### Practice Problems
Question 1: Construct a perpendicular bisector for $\bar{AB}$. If a line segment of length 12 cm.
Question 2: Find the equation of the perpendicular bisector of the line segment joining the points (4, 9) and (10, 3).
## Bisector Examples
Given below are the examples on bisector.
### Solved Examples
Question 1: Find the missing side length
$\bar{CA} = 46, \bar{CD} = 49, \bar{BA} = 20, \bar{DB} = ?$
Solution:
The given sides are $\bar{CA} = 46, \bar{CD} = 49$ and $\bar{BA} = 20$
So, we have to find out $\bar{DB}$
Using the angle bisector theorem, $\frac{\bar{CA}}{\bar{CD}}$ = $\frac{\bar{BA}}{\bar{DB}}$
$\frac{46}{49}$ = $\frac{20}{\bar{DB}}$
$46 \bar{DB} = 980$
$\bar{DB} = 21.30$
Question 2:
Find the missing side length
$\bar{CA} = 32, \bar{CD} = 57, \bar{BA} = 30$ and $\bar{DB} = ?$
Solution:
The given sides are $\bar{CA} = 32, \bar{CD} = 57$ and $\bar{BA} = 30$
So, we have to find out $\bar{DB}$
Using the angle bisector theorem, $\frac{\bar{CA}}{\bar{CD}}$ = $\frac{\bar{BA}}{\bar{DB}}$
$\frac{32}{57}$ = $\frac{30}{\bar{DB}}$
$32 \bar{DB} = 1710$
$\bar{DB} = 53.43$
More topics in Bisector Angle Bisector Bisector of a Line
NCERT Solutions NCERT Solutions NCERT Solutions CLASS 6 NCERT Solutions CLASS 7 NCERT Solutions CLASS 8 NCERT Solutions CLASS 9 NCERT Solutions CLASS 10 NCERT Solutions CLASS 11 NCERT Solutions CLASS 12
Related Topics Math Help Online Online Math Tutor
*AP and SAT are registered trademarks of the College Board. |
# Isosceles Triangle
How to solve an isosceles triangle by using its definition: definition, 3 examples, and their solutions.
## Definition
### Definition
An isosceles triangle is a triangle
that has two congruent sides (= legs).
The non-congruent side is the base.
The two angles
that are adjacent to the base
are congruent.
## Example 1
### Solution
The given triangle is an isosceles triangle.
So these two angles are congruent.
So [3x + 25] = [55].
Move +25 to the right side.
Then 3x = 30.
Divide both sides by 3.
Then x = 10.
So x = 10.
## Example 2
### Solution
The given triangle is an isosceles triangle.
Then the bottom two angles are congruent.
So the bottom right angle is [4x + 3]º.
These three angles
are the interior angles of the given triangle.
So 2[4x + 3] + [3x - 2] = 180.
2(4x + 3) = 8x + 6
8x + 3x = 11x
+6 - 2 = +4
Move +4 to the right side.
Then 11x = 176.
Divide both sides by 11.
Then x = 16.
So x = 16.
## Example 3
### Solution
See the left triangle.
This is an isosceles triangle.
So the upper angle
is congruent to the bottom angle:
xº.
So the upper angle is
xº.
Next, see the right triangle.
This is also an isosceles triangle.
So the bottom left angle
is congruent to the bottom right angle:
70º.
So the bottom left angle is
70º.
70º is the exterior angle.
And the two xº angles |
# How do you find the coordinates of the center, foci, the length of the major and minor axis given x^2+5y^2+4x-70y+209=0?
Dec 4, 2017
Complete the squares so that the equation fits one these two forms:
(x-h)^2/a^2+(y-k)^2/b^2= 1; a > b" [1]"
(x-h)^2/b^2+(y-k)^2/a^2= 1; a > b" [2]"
Then the desired information can be obtained.
#### Explanation:
Given:
${x}^{2} + 5 {y}^{2} + 4 x - 70 y + 209 = 0$
Group, the x terms together, the y terms together, and move the constant term to the right:
${x}^{2} + 4 x + 5 {y}^{2} - 70 y = - 209$
${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$.
To make the x terms look like the pattern, we must insert $+ {h}^{2}$ on the left but, to maintain equality, we must, also add ${h}^{2}$ to the right:
${x}^{2} + 4 x + {h}^{2} + 5 {y}^{2} - 70 y = - 209 + {h}^{2}$
We can solve for the value of h, if we set the middle term in the pattern equal to the middle term in the equation:
$- 2 h x = 4 x$
$h = - 2$
This allows us to substitute ${\left(x - \left(- 2\right)\right)}^{2}$ for the x terms on the left and 4 for ${h}^{2}$ on the right:
${\left(x - \left(- 2\right)\right)}^{2} + 5 {y}^{2} - 70 y = - 209 + 4$
${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$
We must multiply both sides by 5 so that the pattern matches the equation:
$5 {\left(y - k\right)}^{2} = 5 {y}^{2} - 10 k y + 5 {k}^{2}$
This means that we must add $5 {k}^{2}$ to both sides of the equation:
${\left(x - \left(- 2\right)\right)}^{2} + 5 {y}^{2} - 70 y + 5 {k}^{2} = - 209 + 4 + 5 {k}^{2}$
As we did with h, we can use the middle terms to solve for k:
$- 10 k y = - 70 y$
$k = 7$
This means that we can substitute 5(y-7)^2 for 5y^2-70y+5k^2 and 245 for $5 {k}^{2}$ on the right:
${\left(x - \left(- 2\right)\right)}^{2} + 5 {\left(y - 7\right)}^{2} = - 209 + 4 + 245$
Simplify the right:
${\left(x - \left(- 2\right)\right)}^{2} + 5 {\left(y - 7\right)}^{2} = 40$
Divide both sides by 40:
${\left(x - \left(- 2\right)\right)}^{2} / 40 + {\left(y - 7\right)}^{2} / 8 = 1$
Write the denominators as squares:
${\left(x - \left(- 2\right)\right)}^{2} / {\left(2 \sqrt{10}\right)}^{2} + {\left(y - 7\right)}^{2} / {\left(2 \sqrt{2}\right)}^{2} = 1$
Here is the corresponding form:
(x-h)^2/a^2+(y-k)^2/b^2= 1; a > b" [1]"
This allows us to obtain the desired information by observation.
The center is:
$\left(h , k\right) = \left(- 2 , 5\right)$
The left focus is:
$\left(h - \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(- 2 - \sqrt{40 + 8} , 7\right)$
$\left(h - \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(- 2 - 4 \sqrt{3} , 7\right)$
The right focus is:
$\left(h + \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(- 2 + 4 \sqrt{3} , 7\right)$
The length of the major axis is:
$2 a = 4 \sqrt{10}$
The length of the minor axis is:
$2 b = 4 \sqrt{2}$ |
# What is the degree of a polynomial in factored form?
## What is the degree of a polynomial in factored form?
The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading term is the term containing the highest power of the variable, or the term with the highest degree.
What is polynomial degree of a quadratic equation?
A quadratic polynomial is a type of polynomial which has a degree of 2. So, a quadratic polynomial has a degree of 2.
What is factored form example?
A fully factored form means the given number or polynomial is expressed as a product of the simplest possible form. For example, if we write 12y2−27=3(4y2−9) 12 y 2 − 27 = 3 ( 4 y 2 − 9 ) , then it is not considered as fully factored form as (4y2−9) ( 4 y 2 − 9 ) can be factored further.
### What is the formula for factored form?
When the given equation can be expressed in the form a2 – b2, it can be factored as (a+b)(a−b) ( a + b ) ( a − b ) . Example: Consider y2−100 y 2 − 100 . Each of the terms here can be expressed in the form of square. Here the factors are (y+10 ) and (y−10) .
How do you write an expression in a quadratic equation?
The standard form is ax² + bx + c = 0 with a, b and c being constants, or numerical coefficients, and x being an unknown variable. Keep reading for examples of quadratic equations in standard and non-standard forms, as well as a list of quadratic equation terms.
Which equation is quadratic in form?
Any equation in the form ax 2 + bx + c = 0 is said to be in quadratic form. This equation then can be solved by using the quadratic formula, by completing the square, or by factoring if it is factorable.
## What is the degree of polynomial 7?
zero
As given polynomial does not contain variable its degree is zero.
What is the degree of 0 polynomial?
Degree of the zero polynomial Like any constant value, the value 0 can be considered as a (constant) polynomial, called the zero polynomial. It has no nonzero terms, and so, strictly speaking, it has no degree either. As such, its degree is usually undefined. |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# What is the probability of getting a sum of $22$ or more when four dice are thrown?(A) $\dfrac{1}{{456}}$ (B) $\dfrac{5}{{432}}$ (C) $\dfrac{9}{{756}}$ (D) $\dfrac{1}{{756}}$
Last updated date: 22nd Jun 2024
Total views: 394.8k
Views today: 3.94k
Verified
394.8k+ views
Hint: Probability is the ratios of favourable outcomes and total outcomes of this event. Favorable outcomes will be arrangements where the sum is greater than or equal to $22$ . And total outcomes can be calculated by multiplying the number of choices of each place together. The maximum sum that can be obtained is $24$ . So find the number of arrangements to obtain the sum of $24,23{\text{ and 22}}$ for favorable outcomes.
Here in this problem, we are given that four dice are rolled and we need to find the probability of getting a sum of all four numbers on top of $22$ or more.
Before starting with the solution we need to understand the concept of probability first.
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it. The value is expressed from zero to one. The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory, which is also used in the probability distribution, where you will learn the possibility of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
$\Rightarrow$ Probability $= \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}$
When four dice are rolled together, with each having possibilities of $1,2,3,4,5{\text{ and }}6$.
So, when all four dices are rolled together then the total number of outcomes possible can be calculated by multiplying the number of choices on each of the places, i.e.
$\Rightarrow$ Total possible outcomes $= 6 \times 6 \times 6 \times 6 = {6^4} = 1296$
Now let’s try to figure out the arrangements where the sum is greater than or equal to $22$ .
$\Rightarrow 6 + 6 + 6 + 6 = 24$ , this is the maximum possible sum that can be obtained here
$\Rightarrow 6 + 5 + 6 + 6 = 23$ , this is the sum that can be obtained by $4$ ways, with different positions of the number $5$ to appear.
We should now write all the possible arrangements for sum $22$ using all four positions:
$\Rightarrow 6 + 6 + 6 + 4 = 22$ , this arrangement is possible in $4$ ways with different positions of numbers $4$ to appear.
$\Rightarrow 6 + 5 + 5 + 6 = 22$ , this arrangement is only possible in $6$ ways
Therefore, the total number of favourable outcomes is $= 1 + 4 + 4 + 6 = 15$
Thus, by using the formula for probability, we get:
$\Rightarrow$ Probability $= \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{15}}{{1296}} = \dfrac{5}{{432}}$
Hence, the option (B) is the correct answer.
Note: In probability, questions like this can be solved with careful arrangement of favourable outcomes. Be careful while finding them. For finding a number of arrangements of $6 + 5 + 5 + 6 = 22$ , we can use the arrangement of four things at four places but with two similar things. And this can be represented by: $\dfrac{{4!}}{{2! \times 2!}} = \dfrac{{1 \times 2 \times 3 \times 4}}{{1 \times 2 \times 1 \times 2}} = 6$ . |
# Basic Geometrical Ideas (Exercise – 4.1)
## Ex 4.1 Question 1.
1. Use the figure to name:
(a) Five points
(b) A line
(c) Four rays
(d) Five line segments
(a) Five points = B ,C ,D ,E and O
(b) A line
(c) Four rays
,
,
and
.
(d) Five line segments
,
,
,
and
## Ex 4.1 Question 2.
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
Solutions:
The lines are
,
,
,
,
,
,
,
,
,
,
,
## Ex 4.1 Question 3.
Use the figure to name:
(a) Line containing point E.
(b) Line passing through A.
(c) Line on which O lies
(d) Two pairs of intersecting lines.
Solutions:
(a) Line containing point E
(b) Line passing through A
(c) Line on which O lies
(d) Two pairs of intersecting lines
,
and
,
## Ex 4.1 Question 4.
How many lines can pass through (a) one given point? (b) two given points?
Solutions:
(a) An infinite number of lines can pass through one given point.
(b) Only one line can pass through two given points.
## Ex 4.1 Question 5.
Draw a rough figure and label suitably in each of the following cases:
(a) Point P lies on .
(b) and intersect at M.
(c) Line l contains E and F but not D.
(d) and meet at O.
Solutions:
## Ex 4.1 Question 6.
6. Consider the following figure of line . Say whether following statements are true or false in context of the given figure.
(a) Q, M, O, N, P are points on the line .
(b) M, O, N are points on a line segment .
(c) M and N are end points of line segment.
(d) O and N are end points of line segment .
(e) M is one of the end points of line segment .
(f) M is point on ray .
(g) Ray is different from ray .
(h) Ray is same as ray .
(i) Ray is not opposite to ray .
(j) O is not an initial point of
(k) N is the initial point of and .
Solutions:
(a) True
(b) True
(c) True
(d) False
(e) False
(f) False
(g) True
(h) False
(i) False
(j) False
(k) True
error: Content is protected !! |
### Scientific Notation
```Scientific Notation
2-6-4
Writing Numbers
• Scientific notation is used to write very
large and very small numbers.
• The distance to the sun is 150,000,000km
or 1.5x108km.
• The bacteria Streptococcus pyogenes is
0.0000625 cm or 6.25x10-5cm long
Moving Places
• 4x103=4x10x10x10=4x1000=4000
Notice: the exponent on the 10 is 3 and the decimal
after the 4 moved to the right 3 spaces.
• 2.3x105 = 230000 moving the decimal right 5
spaces.
• 2.3x105 is scientific notation and 230000 is decimal
notation.
Writing in Scientific
Notation
Rule:
is scientific notation
a 10
if 1 a 10 and m is an integer.
m
2.3x108 = 23x107 =
0.23x109 = 230x106
But only one of these is in
scientific notation.
The Reason
• 2.3 is between 1 and 10 so the correct form of
the number in scientific notation is 2.3x108.
• Although the other numbers are the same as
2.3x108, they are not scientific notation.
• 3.21x102 is the scientific form of 321.
• To write 45600000 in scientific notation,
4.56x107 = 45600000
• move the decimal so only one digit is in
front and drop ending zeros. 4.56
• Multiply by ten to the power that matches
the number of spaces the decimal was
moved.
Small Numbers
1
4
Recall 10
4
so 6.8 10
10
6.8
4
1
10000
= 0.00068.
10000
Dividing by 10000 moves the decimal 4
spaces to the left.
Converting
To change a small number in scientific notation to
decimal notation,
move the decimal to the left
the number of spaces indicated
by the power of 10.
5.43x10-3 = 0.00543 (left)
and 6.43x103= 6430 (right)
Converting
• To write a small decimal notation number
to scientific notation, move the decimal
after the first nonzero digit.
– Count how many spaces the decimal was
moved. This is the power of 10 the number
would be divided by.
– Make it a negative exponent.
0.00003123 =
0.0000 3 123 3.123 10
5
Multiplication
and Scientific Notation
• To multiply 3.1 x 104 X 2.3x105,
– use the commutative property 3.1x2.3x104x105
– the associative property (grouping of multiplication
doesn’t matter) to get 7.13x104x105.
• Then use the exponent rules to get the
final result 7.13x109.
Examples
(3x108)(1.5x1023)=4.5x1031
(2.1x104) (4x10-12) =8.4x10-8
(4.6x108) (5.8x106) =26.68x1014
Notice: What is wrong with the last example?
Although the answer is correct, the number
is not in scientific notation.
To finish the problem, move the decimal one
space left and increase the exponent by
one.
26.68x1014 = 2.668x1015
For Your
Information
If the decimal moves right,
the exponent moves down.
0.00042x109=4.2x105
If the decimal moves left,
the exponent moves up.
7890x103=7.89x106
Division
and Scientific Notation
Division works in a similar way.
4.5 10
3 10
5
8
4.5
3
10
85
1.5 10
3
Division
and Scientific Notation
Division works in a similar way.
1.092 10
2.1 10
13
5
1.092
10
13 ( 5 )
2.1
8
0.52 10 5.2 10
9
Notice the decimal and exponent.
6.8 10 3.4 10
3
8
6.8 10
3
3.4 10
8
2 10
3 8
2 10
5
Division
and Scientific Notation
Division works in a similar way.
6.8 10 3.4 10
3
6.8 10
3
3.4 10
8
8
2 10
3 8
2 10
5
Notice the decimal and exponent.
Calculator notation
• Some calculators show scientific
notation the same way we have written
them here.
• Some use EE. They write 3.456 EE5 to
mean 3.456 x105.
• Some write the same thing with this in
the screen 3.4560505.
There are several ways electronic
equipment displays scientific notation.
Look in your user’s manual to see how
it is displayed on your calculator.
``` |
+0
# HELP! ME!
0
365
2
+3994
Find the sum of all the perfect squares (positive integers) of the form $$\frac{x}{4000-x}$$
where x is an integer
Aug 2, 2017
### 2+0 Answers
#1
+2340
+1
Let's think about this problem; I've been begged to attempt to solve this problem, so I guess I will give it a shot:
$$\frac{x}{4000-x}=a^2$$
This represents the problem exactly.
Let's solve for x:
$$\frac{x}{4000-x}=a^2$$ Multiply by 4000-x on both sides of the equation. $$x=a^2(4000-x)$$ Distribute the a^2 to both terms inside of the parentheses. $$x=4000a^2-a^2x$$ Add a^2*x to both sides of the equation. $$x+a^2x=4000a^2$$ Let's factor out an x from both of the terms on the left hand side of the equation. $$x(1+a^2)=4000a^2$$ Divide by 1+a^2 on both sides of the equation. $$x=\frac{4000a^2}{1+a^2}$$
Ok, now let's try and think about this problem logically. $$a^2$$ will never be divisible by $$1+a^2$$ because adding one to a number means that the numbers are co-prime. Because of this, we do not have to consider any of the cases. The only time there will be integer solutions for is when 4000 is divisible by $$1+a^2$$; in other words, we need to know the factors of 4000.
The factors of 4000, in order, are $${1,2,4,5,8,10,16,20,25,32,40,50,80,100,125,160,200,250,400,500,800,1000,2000,4000}$$. Now, we must set the denominator to all these values and solve:
First is 1. Here we go:
$$1+a^2=1$$ Subtract 1 on both sides of the equation $$a^2=0$$ Take the square root of both sides $$|a|=0$$ Split your answers into plus or minus. $$a=0$$
Here's a question, though. Is 0 a perfect square? Unfortunately, I am unsure about whether or not it is indeed a perfect square because definitions vary. Many people agree with the definition of if $$\sqrt{a}\in\mathbb{Z}$$where $${a}\in\mathbb{Z}$$ , then that number is a perfect square. With that definition, 0 is a perfect square. However, some define it as $$\sqrt{a}\in\mathbb{N}$$ where $${a}\in\mathbb{Z}$$. In this case, 0 would be excluded. It's an open question, really, without a concrete answer. If you know whether or not an answer exists to this question, I would appreciate feedback. Anyway, let's try the next factor, 2:
You know what? There is a method to make this go even faster, anyway! Let's take the following set and subtract 1 from every term.
Now,$${1,2,4,5,8,10,16,20,25,32,40,50,80,100,125,160,200,250,400,500,800,1000,2000,4000}$$
transforms into $${0,1,3,4,7,9,15,19,24,31,39,49,79,99,124,159,199,249,399,499,799,999,1999,3999}$$. Now, out of those, which ones are perfects squares? They are the following:
$$0,1,4,9,49$$
Now, take the square roots of the numbers in the set:
$$0,1,2,3,7$$
Therefore, the sum of the positive perfect square integers is $$0+1+2+3+7=13$$.
Aug 3, 2017
#2
+98130
0
Nice, X2.....!!!!
Aug 3, 2017 |
You are currently browsing the tag archive for the ‘trolls’ tag.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).
How to Avoid The Troll: A Puzzle
In these situations it’s best to look at the extreme case first so you get some idea of the problem. If you are feeling particularly pessimistic you could assume that the troll is always going to be there. Therefore you would head to the top of the barrier each time. This situation is represented below:
The Pessimistic Solution:
Another basic strategy would be the optimistic strategy. Basically head in a straight line hoping that the troll is not there. If it’s not, then the journey is only 2km. If it is then you have to make a lengthy detour. This situation is shown below:
The Optimistic Solution:
The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.
The question is now, is there a better strategy than either of these? An obvious possibility is heading for the point halfway along where the barrier might be. This would make a triangle of base 1 and height 1/2. This has a hypotenuse of root (5/4). In the best case scenario we would then have a total distance of 2 x root (5/4). In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2. We find the expected value by multiply both by 0.5 and adding. This gives 2.63 (2 dp). But can we do any better? Yes – by using some algebra and then optimising to find a minimum.
The Optimisation Solution:
To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum. Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us. Wolfram Alpha is incredibly powerful -and also very easy to use. Here is what I entered:
and here is the output:
So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.
So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint. Not a huge difference – but nevertheless we’ll save ourselves a few seconds!
This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across? Maybe the troll appears only 1/3 of the time? Could you even generalise the results for when the troll is y distance away or appears z percent of the time?
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
### Website Stats
• 9,162,705 views
### About
All content on this site has been written by Andrew Chambers (MSc. Mathematics, IB Mathematics Examiner).
### New website for International teachers
I’ve just launched a brand new maths site for international schools – over 2000 pdf pages of resources to support IB teachers. If you are an IB teacher this could save you 200+ hours of preparation time.
Explore here!
### Free HL Paper 3 Questions
P3 investigation questions and fully typed mark scheme. Packs for both Applications students and Analysis students.
Available to download here
### IB Maths Super Exploration Guide
A Super Exploration Guide with 168 pages of essential advice from a current IB examiner to ensure you get great marks on your coursework.
Available to download here. |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 1.1: Introduction to Problem Solving
[ "article:topic", "Problem Solving", "authorname:mmanes" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
The Common Core State Standards for Mathematics (http://www.corestandards.org/Math/Practice) identify eight “Mathematical Practices” — the kinds of expertise that all teachers should try to foster in their students, but they go far beyond any particular piece of mathematics content. They describe what mathematics is really about, and why it is so valuable for students to master. The very first Mathematical Practice is:
Make sense of problems and persevere in solving them. Mathematically proficient students start by explaining to themselves the meaning of a problem and looking for entry points to its solution. They analyze givens, constraints, relationships, and goals. They make conjectures about the form and meaning of the solution and plan a solution pathway rather than simply jumping into a solution attempt. They consider analogous problems, and try special cases and simpler forms of the original problem in order to gain insight into its solution. They monitor and evaluate their progress and change course if necessary.
Problem 1 (ABC)
Draw curves connecting A to A, B to B, and C to C. Your curves cannot cross or even touch each other,they cannot cross through any of the lettered boxes, and they cannot go outside the large box or even touch it’s sides.
Think / Pair / Share
After you have worked on the problem on your own for a while, talk through your ideas with a partner (even if you have not solved it).
• What did you try?
• What makes this problem difficult?
• Can you change the problem slightly so that it would be easier to solve?
##### Problem Solving Strategy 1 (Wishful Thinking).
Do you wish something in the problem was different? Would it then be easier to solve the problem?
For example, what if ABC problem had a picture like this:
Can you solve this case and use it to help you solve the original case? Think about moving the boxes around once the lines are already drawn.
Here is one possible solution. |
# Lesson 6
More Linear Relationships
## 6.1: Growing (5 minutes)
### Warm-up
This warm-up encourages students to look for regularity in how the number of tiles in the diagram are growing. This relates naturally to the work that they are doing with understanding the linear relationships as two of the three patterns students are likely to observe are linear and, in fact, proportional.
### Launch
Arrange students in groups of 2. Display the image for all to see and ask students to look for a pattern in the way the collection of red, blue, and yellow tiles are growing. Ask how many tiles of each color will be in the 4th, 5th, and 10th diagrams if the diagrams keep growing in the same way. Tell students to give a signal when they have an answer and strategy. Give students 1 minute of quiet think time, and then time to discuss their responses and reasoning with their partner.
### Student Facing
Look for a growing pattern. Describe the pattern you see.
1. If your pattern continues growing in the same way, how many tiles of each color will be in the 4th and 5th diagram? The 10th diagram?
2. How many tiles of each color will be in the $$n$$th diagram? Be prepared to explain how your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
Invite students to share their responses and reasoning. Record and display the different ways of thinking for all to see. If possible, record the relevant reasoning on or near the images themselves. After each explanation, ask the class if they agree or disagree and to explain alternative ways of thinking, referring back to what is happening in the images each time.
Time permitting, ask students which patterns represent a linear relationship? Which ones represent a proportional relationship? The patterns for the blue and red blocks are both proportional (hence linear) while the pattern for the yellow blocks is neither proportional nor linear.
## 6.2: Slopes, Vertical Intercepts, and Graphs (20 minutes)
### Activity
In the previous lesson, students analyzed the graph of a linear, non-proportional relationship (number of cups in a stack versus the height of the stack). This task focuses on interpreting the slope of a graph and where it crosses the $$y$$-axis in context. Students are given cards describing situations with a given rate of change and cards with graphs. Students match each graph with a situation it could represent, and then use the context to interpret the meaning of the slope. They find where the line crosses the vertical axis, i.e., the vertical intercept, and interpret its meaning in each situation. They also decide if the two quantities in each situation are in a proportional relationship.
Make sure that students draw the triangle they use to compute the slope. There are strategic choices that can be made to make the computation easier and more precise. Watch for students who use different triangles for the same slope computation, and ask them to share their reasoning during the whole-group discussion.
In the whole-group discussion at the end of the task, discuss and emphasize the meaning of the terms slope and vertical intercept or $$y$$-intercept (in situations where the name of the variable graphed on the vertical axis is $$y$$).
You will need the Slopes and Graphs blackline master for this lesson.
### Launch
Tell students that they will match a set of cards describing different relationships with a set of cards showing graphs of lines. The axes on the graphs are not labeled (since this could be used as an aid in the matching). Instruct students to add labels to the axes as they make their matches.
Arrange students in groups of 2. Distribute a set of 12 cards to each group. 10 minutes of group work and then whole-class discussion.
### Student Facing
Your teacher will give you 6 cards describing different situations and 6 cards with graphs.
1. Match each situation to a graph.
2. Pick one proportional relationship and one non-proportional relationship and answer the following questions about them.
1. How can you find the slope from the graph? Explain or show your reasoning.
2. Explain what the slope means in the situation.
3. Find the point where the line crosses the vertical axis. What does that point tell you about the situation?
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
The slopes of the 6 lines given on the situation card are all different, so the matching part of the task can be accomplished by examining the slopes of the different lines. Invite students who have made strategic choices of slope triangles for calculating the slopes (for example, Graph 6 contains the points $$(0,0)$$ and $$(5,20)$$, which give a value of $$\frac{20}{5}$$ for the slope), and ask them to share.
Next, focus the discussion on the interpretation of the point where the line crosses (or touches) the $$y$$-axis. For some situations, the contextual meaning of this point is abstract. For example, in Situation B, a square with side length 0 is just a point that has no perimeter and so it “makes sense” that $$(0,0)$$ is on the graph. Some students may argue that a point is not a square at all but if we consider it to be a square then it definitely has 0 perimeter. In other situations, the point where the line touches the $$y$$-axis has a very natural meaning. For example, in Situation A, it is the amount Lin’s dad spent on the tablet and 0 months of service: so this is the cost of the tablet.
Define the vertical intercept or $$y$$-intercept as the point where a line crosses the $$y$$-axis. Note that sometimes “$$y$$-intercept” refers to the numerical value of the $$y$$-coordinate where the line crosses the $$y$$-axis. Go over each of the situations and ask students for the meaning of the vertical intercept in the situation (A: cost of the device, B: perimeter of a square with 0 side length, C: amount of money in Diego’s piggy bank before he started adding \$5 each week, D: money Noah saved helping his neighbor, E: amount of money in Elena's piggy bank before she started adding money, F: amount Lin’s mom has paid for internet service before her service begins. Representation: Develop Language and Symbols. Create a display of important terms and vocabulary. Include the following terms and maintain the display for reference throughout the unit: vertical intercept, y-intercept. Invite students to suggest language or diagrams to include on the display that will support their understanding of these terms. Supports accessibility for: Memory; Language Representing, Conversing: MLR7 Compare and Connect. As students share their matches with the class, call students’ attention to the different ways the vertical intercept is represented graphically and within the context of each situation. Take a close look at Graphs 2 and 3 to distinguish what the 40 represents in each corresponding situation. Wherever possible, amplify student words and actions that describe the correspondence between specific features of one mathematical representation and a specific feature of another representation. Design Principle(s): Maximize meta-awareness; Support sense-making ## 6.3: Summer Reading (10 minutes) ### Activity Students have just learned the meaning of the $$y$$-intercept for a line and have been interpreting slope in context. In this activity, they investigate the $$y$$-intercept and slope together and investigate what happens when their values are switched. In contexts like this one, the $$y$$-intercept and the slope come with natural units and understanding this can help graph accurately. Specifically, the $$y$$-intercept is the number of pages Lin read before she starts gathering data for the graph. The slope, on the other hand, is a rate: it’s the number of pages Lin reads per day Watch for students who understand the source of Lin’s error (confusing the $$y$$-intercept with the slope) and invite them to share this observation during the discussion. ### Launch Work time followed by whole-class discussion. Representation: Internalize Comprehension. Use MLR6 Three Reads to supporting reading comprehension of the word problem. Use the first read to orient students to the situation. Ask students to describe what the situation is about without using numbers (Lin’s reading assignment). Use the second read to identify quantities and relationships. Ask students what can be counted or measured without focusing on the values. Listen for and amplify the important quantities that vary in relation to each other in this situation: first 30 pages and 40 pages each day. After the third read, ask students to brainstorm possible strategies to answer the question, “What does the vertical intercept mean in this situation?” Supports accessibility for: Language; Conceptual processing Speaking: MLR8 Discussion Supports. Use this to amplify mathematical uses of language to communicate about vertical intercepts, slope, and constant rate. Invite students to use these words when describing their ideas. Ask students to chorally repeat phrases that include these words in context. Design Principle(s): Support sense-making, Optimize output (for explanation) ### Student Facing Lin has a summer reading assignment. After reading the first 30 pages of the book, she plans to read 40 pages each day until she finishes. Lin makes the graph shown here to track how many total pages she'll read over the next few days. After day 1, Lin reaches page 70, which matches the point $$(1,70)$$ she made on her graph. After day 4, Lin reaches page 190, which does not match the point $$(4,160)$$ she made on her graph. Lin is not sure what went wrong since she knows she followed her reading plan. 1. Sketch a line showing Lin's original plan on the axes. 2. What does the vertical intercept mean in this situation? How do the vertical intercepts of the two lines compare? 3. What does the slope mean in this situation? How do the slopes of the two lines compare? ### Student Response For access, consult one of our IM Certified Partners. ### Student Facing #### Are you ready for more? Jada's grandparents started a savings account for her in 2010. The table shows the amount in the account each year. If this relationship is graphed with the year on the horizontal axis and the amount in dollars on the vertical axis, what is the vertical intercept? What does it mean in this context? year amount in dollars 2010 600 2012 750 2014 900 2016 1050 ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis Ask students: • “How did your graph compare to Lin’s?” (It is steeper but starts off at 30 instead of 40 pages.) • “Which point does your graph have in common with Lin’s?” (The point $$(1,70)$$, 70 pages read after one day.) Invite selected students to share what’s the likely source of Lin’s error (confusing the $$y$$-intercept with the slope). In this context, the $$y$$-intercept is the number of pages Lin read before she starts counting the days (30), and the slope is the number of pages Lin reads per day (40). Emphasize how the $$y$$-intercept and slope influence the graph of a line. • The $$y$$-intercept indicates where the line touches or crosses the $$y$$-axis. • The slope indicates how steep the line is. ## Lesson Synthesis ### Lesson Synthesis Lines have a slope and vertical intercept. The vertical intercept indicates where the line meets the $$y$$-axis. For example, a line represents a proportional relationship when the vertical intercept is 0. Here is a graph of a line showing the amount of money paid for a new cell phone and monthly plan: • “What is the vertical intercept for the graph?” ($$(0,200)$$) • “What does it mean?” (There was an initial cost of \$200 for the phone.)
The slope of the line is 50 (draw a slope triangle connecting the points such as $$(0,200)$$ and $$(2,300)$$). This means that the phone service costs \$50 per month in addition to the initial \$200 for the phone.
## 6.4: Cool-down - Savings (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners.
## Student Lesson Summary
### Student Facing
At the start of summer break, Jada and Lin decide to save some of the money they earn helping out their neighbors to use during the school year. Jada starts by putting \$20 into a savings jar in her room and plans to save \$10 a week. Lin starts by putting \$10 into a savings jar in her room plans to save \$20 a week. Here are graphs of how much money they will save after 10 weeks if they each follow their plans:
The value where a line intersects the vertical axis is called the vertical intercept. When the vertical axis is labeled with a variable like $$y$$, this value is also often called the $$y$$-intercept. Jada's graph has a vertical intercept of \$20 while Lin's graph has a vertical intercept of \$10. These values reflect the amount of money they each started with. At 1 week they will have saved the same amount, \\$30. But after week 1, Lin is saving more money per week (so she has a larger rate of change), so she will end up saving more money over the summer if they each follow their plans. |
# 7. Ratio and Proportion
We need to be a bit careful because lots of people use the words "ratio", "fraction" and "proportion" to mean the same thing in everyday speech.
This makes it difficult when we meet the terms in mathematics, because they are not necessarily used to mean the same thing.
## Ratios and Fractions
### Example 1
Ethanol or methanol (wood-based methyl alcohol) is sometimes added to gasoline to reduce pollution and cost. Car engines can typically run on a petrol-ethanol mixture in a ratio of 9:1. The "9:1" means that for each nine units of petrol, there is 1 unit of ethanol.
For example, if we had 9 L of petrol, we would need 1 L of ethanol.
We can see that altogether we would have 10 L of the mixture.
As fractions, the proportion of each liquid is:
1/10 ethanol
9/10 petrol
You can see a more advanced question involving ratios and gasoline in Applied Verbal Problems, in the algebra chapter.
### Example 2
Concrete
Concrete is a mixture of gravel, sand and cement, usually in the ratio 3:2:1.
We can see that there are 3 + 2 + 1 = 6 items altogether. As fractions, the amount of each component of the concrete is:
Gravel: 3/6=1/2
Sand: 2/6=1/3
Cement: 1/6
### Example 3
One of the most famous ratios is the ratio of the circumference of a circle to its diameter.
The value of that ratio cannot be determined exactly. It is approximately 3.141592654... We call it:
pi (the Greek letter "pi").
See more on Pi.
## Proportion
We can talk about the proportion of one quantity compared to another.
In mathematics, we define proportion as an equation with a ratio on each side.
### Example 4
Considering our ethanol/petrol example above, if we have 54 L of petrol, then we need 6 L of ethanol to give us a 9:1 mix.
We could write this as:
54:6 = 9:1
We could use fractions to write our proportion, as follows:
54/6=9/1
## Rates
### Example 5
A normal walking speed is 1 km in 10 min. This is a rate, where we are comparing how far we can go in a certain amount of time.
Our walking rate is equivalent to 6 km/h.
### Example 6 - Conversion of Units
A bullet leaves a gun travelling at 500 m/s. Convert this speed into km/h.
### A famous ratio: Phi
Now let's move on to an interesting ratio called Phi ("phi"), in Math of Beauty.
### Online Algebra Solver
This algebra solver can solve a wide range of math problems. (Please be patient while it loads.)
### Algebra Lessons on DVD
Easy to understand algebra lessons on DVD. See samples before you commit. |
# NCERT Solutions Class 9th Maths Statistics
In this page we have NCERT Solutions Class 9th Maths Statistics for EXERCISE 3 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
The following number of goals was scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Solution:
Mean= (Sum of all 10 matches) /10
=(2+3+4+5+0+1+3+3+4+3)/10 =2.8
Median:
We need to arrange the goals in increasing order
0,1,2,3,3,3,3,4,4,5
Since the number of terms is 10 (even)
Mode:
We need to arrange the goals in increasing order
0,1,2,3,3,3,3,4,4,5
Since 3 occurs becomes maximum times. It is the mode
Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Solution :
Mean= (Sum of all 15 students) /15
=(41+39+48+52+46+62+54+40+96+52+98+40+42+52+60)/10 =54.8
Median:
We need to arrange the data in increasing order
39,40,40,41,42,46,48,52,52,52,54,60,62,96,98
Since the number of terms is 15(odd)
Median= (n+1)/2=8 term=52
Mode:
52 occurs maximum times.So mode is 52
Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
The number of term is even
So
2x+2=126
x=62
Question 4.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Solution:
14 occurs maximum time,so mode is 14
Question 5.
Find the mean salary of 60 workers of a factory from the following table
Salary( in Rs) Number of Workers 3000 16 4000 12 5000 10 6000 8 7000 6 8000 4 9000 3 10000 1 Total 60
Solution
x f fx 3000 16 48000 4000 12 48000 5000 10 50000 6000 8 48000 7000 6 42000 8000 4 32000 9000 3 27000 10000 1 10000 sum 60 305000
Mean = fx/f= 305000/60=5083.33 |
# Applications of Pascal’s Triangle
Pascal’s triangle has many applications in mathematics and statistics. We can use Pascal’s triangle to find the binomial expansion. Also, Pascal’s triangle is used in probabilistic applications and in the calculation of combinations. Recall that Pascal’s triangle is a pattern of numbers in the shape of a triangle, where each number is found by adding the two numbers above it.
Here, we will look at each of the most important applications of Pascal’s triangle in detail along with some examples to understand their use.
##### ALGEBRA
Relevant for
Learning about some of the applications of Pascal’s triangle.
See applications
##### ALGEBRA
Relevant for
Learning about some of the applications of Pascal’s triangle.
See applications
## Pascal’s triangle in binomial expansion
The following is Pascal’s triangle:
We can use the rows of Pascal’s triangle to facilitate the binomial expansion process. Depending on the power of a binomial, we can use a given row of Pascal’s triangle that represents the coefficients of the expanded values.
We use $latex n + 1$ to determine the row to use, where n represents the power of the binomial. Alternatively, we can consider the first row as 0 and take the row in Pascal’s triangle indicated by the power. For example, let’s look at the following binomial expansion:
$latex {{(a+b)}^2}=1{{a}^2}+2ab+1{{b}^2}$
In this expansion, we have the coefficients 1, 2, 1. This corresponds to the third row of Pascal’s triangle. We see that the power of the binomial is 2, so when using $latex n + 1$, we have 3 that does correspond to the third row used.
### EXAMPLE
• Expand the binomial $latex {{(x+y)}^4}$.
Solution: In this case, the power is 4, so we have to use the row number $latex 4 + 1 = 5$. This row corresponds to the numbers 1, 4, 6, 4, 1. These are the coefficients of the binomial expansion and it tells us that we will have 5 terms in the expansion.
Furthermore, we know that we expand a binomial by starting with each term in the highest power and reducing to 0 each term in the opposite direction:
$latex {{(x+y)}^4}$
$$=1{{x}^4}{{y}^0}+4{{x}^3}{{y}^1}+6{{x}^2}{{y}^2}+4{{x}^1}{{y}^3}+1{{x}^0}{{y}^4}$$
$$=1{{x}^4}(1)+4{{x}^3}{{y}^1}+6{{x}^2}{{y}^2}+4{{x}^1}{{y}^3}+1(1){{y}^4}$$
$$={{x}^4}+4{{x}^3}y+6{{x}^2}{{y}^2}+4x{{y}^3}+{{y}^4}$$
## Pascal’s triangle in probability
Pascal’s triangle can be used in probability to simplify counting the probabilities of some event. For example, Pascal’s triangle can show us in how many ways we can combine heads and tails in a coin toss.
Then, this can show us the probability of any combination. In the following example, T represents tails and H represents heads.
### EXAMPLE
Suppose a coin is tossed 4 times, the probabilities of the combinations are:
• HHHH
• HHHT, HHTH, HTHH, THHH
• HHTT, HTHT, HTTH, THHT, THTH, TTHH
• HTTT, THTT, TTHT, TTTH
• TTTT
Therefore, the observed pattern is 1, 4, 6, 4, 1. If we are looking for the total number of possibilities, we just have to add the numbers. That is, we have $latex 1 + 4 + 6 + 4 + 1 = 16$ possible combinations if we toss a coin 4 times.
## Pascal’s triangle in combinations
Pascal’s triangle can be used to find combinations. The top row of Pascal’s triangle is row 0, and the first number in each row is element zero of that row. For example, if we want to find the combination $latex _{6} C _{4}$, we have to look at row 6 and item 4. The answer is 15:
### EXAMPLE
• We have 5 marbles in a bag, 1 red, 1 blue, 1 green, 1 yellow, and 1 black. How many combinations are possible if we want to choose 2 marbles?
Solutions: The total number of marbles is 5 and we want to choose 2 marbles. Therefore, we have the combination $latex _{5} C _{2}$. This means that we have to look at element 2 in row 5. This means that the answer is 10. |
Courses
Courses for Kids
Free study material
Offline Centres
More
# Find all the zeros of $p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ , if you know that two of its zeros are $\sqrt{2}$and $-\sqrt{2}$.
Last updated date: 29th Feb 2024
Total views: 381k
Views today: 4.81k
Verified
381k+ views
Hint: Given that the function is $p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ and given zeros are $\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)=\left( {{x}^{2}}-2 \right)$ . We will perform the long division method for polynomials using the basic methodology, Dividend = Divisor * Quotient + Remainder, where dividend is the given function p(x), divisor is the product of the given zeros of the function p(x), quotient is the answer we get after division and remainder is any remaining term or number after division.
The given function is$p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ . The roots of the given function is $\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)=\left( {{x}^{2}}-2 \right)$
First arrange the term of dividend and the divisor in the decreasing order of their degrees.
To obtain the first term of the quotient divide the highest degree term of the dividend by the highest degree term of the divisor.
To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor.
Continue this process till the degree of remainder is less than the degree of divisor.
\begin{align} & {{x}^{2}}-2\overset{\,\,2{{x}^{2}}-3x+1}{\overline{\left){2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2}\right.}} \\ & \,\,\,\,\,\,\,\,\,\,-\underline{\left( 2{{x}^{4}} \right)\,\,\,\,\,\,\,+\left( -4{{x}^{2}} \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3{{x}^{3}}+\,\,\,\,{{x}^{2}}+6x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\underline{\left( -\,3{{x}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6x \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\underline{\left( {{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2 \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\ \end{align}
Finding the roots of $2{{x}^{2}}-3x+1$ as,
\begin{align} & 2{{x}^{2}}-3x+1=2{{x}^{2}}-2x-x+1 \\ & =2x\left( x-1 \right)-1\left( x-1 \right) \\ & =\left( 2x-1 \right)\left( x-1 \right) \end{align}
Put the above expression equal to 0 as,
\begin{align} & \left( 2x-1 \right)=0 \\ & 2x=1 \\ & x=\dfrac{1}{2} \end{align}
\begin{align} & \left( x-1 \right)=0 \\ & x=1 \end{align}
Therefore the remaining zeros of the above function $p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ are $\dfrac{1}{2}$ and 1.
Note: The possible error that you may encounter could be that the long division was not done properly. Be careful when calculating the roots and keep in mind the sign of the expression being operated on. |
### 2-1 Power and Radical Functions
```2-1 Power and Radical Functions
Part 1
Chapter 2
Power, Polynomial, and Rational Functions
Objectives
You have already analyzed parent functions and their families
of graphs.
Now we will be
Graphing and analyzing power functions,
Graphing and analyzing radical functions, and
(Solving radical equations next class)
1. Power functions
Power functions previously
studied are
f ( x) x
A power function is any
function of the form
2
f ( x ) ax
n
where a and n are nonzero
constant real numbers.
and
f ( x) x
3
End Behavior: using f(x)= axn
When a > 0
When n is even
When n is odd
When a < 0
As x –∞,
As x –∞,
As x ∞,
As x ∞,
As x –∞,
As x –∞,
As x ∞,
As x ∞,
We should be able to also identify domain, range,
intercepts, continuity, and where the function is increasing
and decreasing.
Example 1a:
f ( x)
1
x
6
2
Graph and analyze the function. Describe the domain, range,
intercepts, end behavior, continuity, and where the function
is increasing or decreasing.
Example 1b:
f ( x) x
5
Graph and analyze each function. Describe the domain,
range, intercepts, end behavior, continuity, and where the
function is increasing or decreasing.
2. Functions with Negative Exponents
Recall that f ( x )
1
or x
1
is undefined at x = 0.
x
2
Similarly, f ( x ) x
and f ( x ) x 3 are undefined at x = 0.
Because power functions can be undefined when n < 0 (negative
exponents), these functions will have discontinuities.
Example 2a:
f ( x) 2 x
4
Graph and analyze the function. Describe the domain, range,
intercepts, end behavior, continuity, and where the function
is increasing or decreasing.
Example 2b: f ( x ) 2 x
3
Graph and analyze the function. Describe the domain, range,
intercepts, end behavior, continuity, and where the function
is increasing or decreasing.
3. Functions with Rational Exponents
1
Recall that f ( x ) x n is a root function.
p
f ( x ) x n indicates the nth root of xp
p
If n is an even integer, then the domain must be restricted to
f (x) x n
nonnegative values.
3
Example 3a:
f (x) 2 x 4
Graph and analyze the function. Describe the domain, range,
intercepts, end behavior, continuity, and where the function
is increasing or decreasing.
5
Example 3b:
f ( x ) 10 x 3
Graph and analyze the function. Describe the domain, range,
intercepts, end behavior, continuity, and where the function
is increasing or decreasing.
An expression with rational exponents can be written in
Exponential form
p
f (x) x n
f ( x)
n
x
p
Example 4a:
f ( x) 5 2 x
3
Graph and analyze the function. Describe the domain, range,
intercepts, end behavior, continuity, and where the function
is increasing or decreasing.
Example 4b:
f ( x)
1
5
3x 4
2
Graph and analyze the function. Describe the domain, range,
intercepts, end behavior, continuity, and where the function
is increasing or decreasing.
Assignment: p. 92
1, 5, 9, 13,19, 23, 27, 35, 37
Part 2
Chapter 2
Power, Polynomial, and Rational Functions
Warm up
Graph and analyze the function. Describe the domain, range,
intercepts, end behavior, continuity, and where the function
is increasing or decreasing.
5
1.
g ( x)
2.
h( x) 3 x
x
8
8
7
3. f ( x ) 3 6 3 x
g ( x)
5
8
x
8
h( x) 3 x
7
f ( x) 3 6 3x
Questions on the homework?
Objectives:
Learn how to use a graphing calculator to find a power
function to model a set of data.
1.
3.
Homework quiz over last night’s homework.
4. Power Regression
The table shows the braking distance in feet at several speeds
in miles per hour for a specific car on a dry, well-paved
Speed
10
20
30
40
50
60
70
Distance
4.2
16.7
37.6
66.9
104.5
150.5
204.9
If we wanted to find a model to represent this relationship, a
graphing calculator should make this an easy task.
Write these steps down in your notes to refer to later, when
you need to do another model.
Speed
10
20
30
40
50
60
70
Distance
4.2
16.7
37.6
66.9
104.5
150.5
204.9
1. Enter the data in L1 (speed) and L2 (distance)
(To access, press STAT, then EDIT. Clear the old lists but DO NOT
DEL(ete).
2. Create a scatter plot of the data
(2nd STATPLOT, Select Plot 1, Turn it on [if it is off]. Select a scatter
plot, and the appropriate lists. Press ZOOM 9)
3. If the plot appears to be a power function, use the use the STAT, CALC,
PwrReg function.
(To place this equation in the Y= menu, go to Y1=, then VARS,
Statistics, EQ, RegEQ)
4. Check your scatter plot again. The regression line should appear on the
screen with the data points.
5. Use the equation to make a prediction of the braking distance of a car
going 80 miles per hour. A table of values starting at 79 would work
just fine.
Then, raise each side of the equation to a power equal to the
Note: This process sometimes produces extraneous solutions,
which are solutions which do not satisfy the original
equation.
Always check your solutions in the original equation to
determine if extraneous solutions exist.
Example 1.
2x
28 x 29 3
Example 2.
12
3
( x 2) 8
2
Example 3.
x 1 1
2 x 12
Assignment: p. 92-93, 33, 45 - 55
odds, 100 - 103.
``` |
# Sampling (statistics)
(Redirected from Target population)
This page provides an introductory overview of statistical sampling. For more detailed information, see Sampling (statistics) on Wikipedia.
## Definition of sampling
Key terms: Sampling, sampling frame, target population
Sampling is a term used in statistics. It is the process of choosing a representative sample from a target population and collecting data from that sample in order to understand something about the population as a whole.
Was this sample slice good? Would you infer from this sample that the rest of the pie is good? (Generalisation).
Here is a simple illustration. You have before you a freshly baked homemade cherry pie (the greater whole) and you are pondering the question, “Does it taste delicious?” It looks delicious, it smells delicious, but is it delicious? You take a bite (sample) of the pie (greater whole of many bites), let your taste buds study it, and then make a generalization, otherwise known as an inference, about the whole pie ... mmm this pie is delicious!
The term used in statistical sampling to describe the greater whole is population. A population could be a group of people or any group of objects you are studying (e.g., rocks containing gold, dog biscuits made by x-brand, or all left handed one-toothed people in the world). Studying populations can be complicated, expensive, and time consuming so researchers have developed several different ways to sample whatever it is they are studying. Broadly, these sampling techniques are either probability-based (random sampling) or non-probability-based (non-random sampling).
Let's say after the cherry pie experiment you decide to become a researcher. Before you can determine which sampling method to use you must first decide what will be your target population. From that you would develop a sampling frame, or list, of the set of people from whom data could be collected. This can be a difficult task at times. Then you would probably want to apply one of the following methods for determining what, or who, will be in the sample.
## Probability sampling
### (Simple) random sampling
Search for Simple random sample on Wikipedia.
This doesn’t mean haphazard. It means every left-handed one-toothed person in the sampling frame (list) has an equal and unbiased chance of being sampled. Usually you would number the names or items on the list and then randomly pick numbers to make up the sample (or use an equivalent electronic process). This method is great for statistical accuracy, but very difficult to do sometimes in practice. What if you have to travel to Guam, Brazil, Canada and Germany to sample your left handed one toothed people?
### Systematic (random) sampling
Search for Systematic sampling on Wikipedia.
Here random sampling is given a little structure. You decide you want to sample 100 of the 1000 left-handed one-toothed people in your sampling frame. First, a random starting point is picked and then the rest of the sample is selected at equal intervals from that starting point. So for our example, you would pick a number from 1-100, say the#8, and then pick every 100th person from that number. You start with person #8, which means 8,108, 208, 308 etc. would make up your sample. This method ensures better coverage of the population, as long as nothing quirky comes up as an underlying pattern, such as every 100th person lost their teeth by eating taffy.
### Stratified sampling
Search for Stratified sampling on Wikipedia.
In this method your sampling frame would be divided into non-overlapping groups and then samples from each of those groups is conducted.
For example, we could group our sampling frame of left-handed one-tooth people according to geographic region. This could give us more valuable precise data if the way they are grouped is relevant to what is being studied. So, we might learn not only how left-handed people lose their teeth but that how they lost them differs between Europe, South America, Asia and North America.
### Clustering sampling
First the sampling frame is broken up into groups. Then a sample of groups is randomly picked out of all the groups. Finally, the people in those groups are randomly sampled. This cuts down on travel, time and expense.
For example, you would end up only sampling people from Paris, Buenos Aires, Beijing, and Chicago instead of traveling to 40 different towns and cities in ten different countries. The one drawback is that the groups need to be as dissimilar as possible or you could have a large sampling error. For instance, if all of your groups end up being in Europe you would lose valuable information and it wouldn’t be representative of your population.
## Non-probability sampling
### Convenience sampling
This is where you pick your sample according to what is available. This is why college students are studied so much…no, truly, it isn’t because they are so strange! Another example is when you see someone on a street corner randomly stopping people to do a survey. Convenience sampling is great because…here it comes…it’s convenient, but it is often difficult to make inferences to the population at large.
### Snowball sampling
In this approach the research gains the trust of some people in the target population (e.g., ecstasy users), gathers data, and then asks these people to introduce the researcher to other potential participants. So, the sample gradually snowballs to become larger and larger. |
# USA 1996 Problem 5
Problem 5 of USA Mathematical Olympiad 1996
Triangle ABC has the following property: there is an interior point P such that ∠PAB = 10°, ∠PBA = 20°, ∠PCA = 30°, and ∠PAC = 40°. Prove that triangle ABC is isosceles.
Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Melvin Drimmer, Ph.D.)
Extend CP to meet AB at S. From A draw a line to meet the extension of BP at R and CP at Q such that ∠QAP = 10°. We have BR = AR and
∠BRQ = 40°, ∠QAC = 30°, ∠AQC = 120°, ∠PQR = 60°, ∠QPR = ∠BSP = 80°, and ∠PSA = ∠QPB = 100°
We wish to prove that the triangles QSA and QPB are similar since, if they are, we have ∠QBP = ∠QAS = 20° and ∠QBA = 40°, ∠BQA = 120° = ∠BQC which cause the two triangles BQC and BQA to be congruent, implying BC = BA making triangle ABC isosceles.
In triangles QSA and QPB, ∠QSA = ∠QPB = 100°. So, to prove them similar, it suffices to show that QP / QS = PB / SA (1)
Since AP is bisector of ∠QAS, we have QP/QA = PS/SA (2)
QP/QA = PS/SA = QS/(QA + SA) or
QP/QS = QA/(QA + SA)
Combine with (1), we now need to prove
QA/(QA + SA) = PB/SA = (QA – PB)/QA = (QR + PR)/QA (since BR = AR),
or we need to prove
``` (QR + PR)/QA = PB/SA (3)
```
Using the law of the sines, we have
QP/sin40° = QR/sin80° = PR/sin60° = (QR + PR)/(sin60° + sin80°) and PS/sin20° = PB/sin80°
or QP = (QR + PR)sin40°/(sin60° + sin80°) and PS = PB sin20°/sin80°
Substitute QP and PS to (2), it becomes
[(QR + PR)/QA ] x [sin40°/(sin60° + sin80°)] = [PB/SA] x [sin20°/sin80°]
so now we have to prove sin40°/(sin60° + sin80°) = sin20°/sin80° (4)
or sin20°/sin80° = (sin40° - sin20°)/sin60° or sin10°/sin30° = sin20°/sin80° or sin10°sin80° = sin30°sin20° or ½ (cos70° - cos90°) = cos60°cos70° or ½ = cos60° which is obvious! |
# Compute 4.659×10^4−2.14×10^4. Round the answer appropriately.
– The answer should be expressed as an integer that is rounded off to a proper number of significant figures.
An Operation is a mathematical process such as addition, subtraction, multiplication, and division to solve an equationPEMDAS Rule is the sequence in which these operations are performed. It is abbreviated as follows:
“P” represents the Parenthesis (brackets).
“E” represents the Exponents (Powers or Roots).
“M & D” represents the Multiplication and Division Operations.
“A & S” represents the Addition and Subtraction Operations.
PEMDAS rule defines that operations are to be solved starting from Parenthesis (brackets), then Exponents (Powers or Roots), then Multiplication and Division (from left to right), and lastly Addition and Subtraction (from left to right).
Significant Figures of a number are defined as the number of digits in the given number that are reliable and indicate the accurate quantity.
In solving equations, the following rules are used:
(a) For Addition and subtraction operations, the numbers are rounded by the lowest number of decimals points.
(b) For Multiplication and division operations, the numbers are rounded by the lowest number of significant figures.
(c) Exponential terms $n^x$ are only rounded by the significant figures in the base of the exponent.
Given numbers are:
$a=4.659\times{10}^4$
$b=2.14\times{10}^4$
We need to compute the number resulting from the subtraction of $a$ and $b$.
$a-b=?$
We will first analyze the significant figures of the decimal numbers. As per the significant rule for addition or subtraction of numbers having different significant figures, we will consider rounding off both numbers to the lowest number of decimal points.
$4.659$ has three digits after the decimal point.
$2.14$ has two digits after the decimal point.
Hence, we will round off $4.659$ till it has two digits after the decimal point:
$a=4.66\times{10}^4$
Now we will check the significant figures for Exponential Terms.
$Exponential\ Term={10}^4$
As for the exponential terms, the number of significant figures in the base of the exponent is considered. In both exponential terms, the number of significant figures in the base of the exponent is two.
Now that significant figures are sorted, we will solve the equation using the PEMDAS Rule.
$a-b=4.66\times{10}^4-2.14\times{10}^4$
Taking the exponential term common:
$a-b=(4.66-2.14)\times{10}^4$
As per the PEMDAS Rule, we will first solve the term in the parenthesis (brackets) as follows:
$4.66-2.14=2.52$
So:
$a-b=2.52\times{10}^4$
It can be expressed as follows:
${10}^4=10000$
$a-b=2.52\times 10000$
$a-b=25200$
## Numerical Result
The result for the subtraction of given two numbers is:
$4.659\times{10}^4-2.14\times{10}^4=2.52\times{10}^4$
In Integer form:
$4.659\times{10}^4-2.14\times{10}^4=25200$
## Example
Compute the result of the given equation as per PEMDAS Rule.
$58\div(4\times5)+3^2$
Solution
As per PEMDAS Rule, we will first solve the parenthesis:
$4\times5=20$
$58\div(4\times5)+3^2=58\div20+3^2$
Secondly, we will solve the exponent:
$3^2=9$
$58 \div 20+3^2=58 \div 20+9$
Thirdly, we will solve division:
$58 \div 20+9=2.9+9$
Finally, we will solve the addition:
$2.9+9=11.9$
So:
$58 \div (4\times 5)+3^2=11.9$ |
MNR = MR – MC = 0 MR = MC TC=w*L+r*K Set marginal revenue equal to marginal cost and solve for q. Finding Maximum Profit To find maximum profit, compare the profit level at each price level. This means that we have a positive marginal profit. How can you be certain that you make the best financial decision when evaluating whether to take a job or invest in a new business opportunity? As you can see this forms a rectangle and the area of the rectangle is the TR. We want to begin by starting with revenue. The TR –TC will be the shaded region below. Thus, the correct choice of output is Q = 65. 5.32 Calculate profit (loss) by using the the equation obtained in 5.31. Your Need to understand how to plot the Total Product of Labor Curve, Average Product of Labor Curve, and the Marginal Product of Labor Curve. $\begin{array}{lll}\text{profit}& =& \text{total revenue}-\text{total cost}\\& =& \left(85\right)\left(\5.00\right)-\left(85\right)\left(\3.50\right)\\& =& \127.50\end{array}$, $\begin{array}{lll}\text{profit}& =& \text{(price}-\text{average cost)}\times \text{quantity}\\ & =& \left(\5.00-\3.50\right) \times 85\\ & =& \127.50\end{array}$, $\begin{array}{lll}\text{profit}& =& \text{total revenue}-\text{total cost}\hfill \\ & =& \left(75\right)\left(2.75\right)-\left(75\right)\left(2.75\right)\hfill \\ & =& 0\hfill \end{array}$, $\begin{array}{lll}\text{profit}& =& \text{(price}-\text{average cost)}\times \text{quantity}\hfill \\ & =& \left(2.75-2.75\right)\times 75\hfill \\ & =& 0\hfill \end{array}$, $\begin{array}{lll}\text{profit}& =& \text{(total revenue}-\text{ total cost)}\hfill \\ & =& \left(65\right)\left(2.00\right)-\left(65\right)\left(2.73\right)\hfill \\ & =& -47.45\hfill \end{array}$, $\begin{array}{lll}\text{profit}& =&\text{(price}-\text{average cost)}\times \text{quantity}\hfill \\ & =& \left(2.00-2.73\right) \times 65\hfill \\ & =& -47.45\hfill \end{array}$. For a firm in perfect competition, demand is perfectly elastic, therefore MR=AR=D. This is aimed toward those who have taken or are currently taking Intermediate Microeconomics. TC = VC + FC The TC and TR are combined. This is how we will derive the MC and AVC curve. Marginal revenue represents the change in total revenue associated with an additional unit of output, and marginal cost is the change in total cost for an additional unit of output. TR = P*Q So we must find where MC =MR and draw a vertical line down to the Quantity axis and find the Quantity which correlates to the Price chosen. In the firm this in the only range in which it will produce output. Δ = the change in Total costs will be the quantity of 85 times the average cost of $3.50, which is shown by the area of the rectangle from the origin to a quantity of 85, up to point C, over to the vertical axis and down to the origin. 2. The firm's marginal cost function is MC = 3 + 0.001Q, and at the profit maximizing level of output the average variable cost (AVC) is$5.50 and the average fixed cost (AFC) is $0.75. Profit Maximisation in the Real World From previous knowledge we know that TVC =wL. Thus, the firm is losing money and the loss (or negative profit) will be the rose-shaded rectangle. It should be noticeable from the graphs that the TC area is larger than the TR area.The Second Graph The firm will continue to produce if Marginal Revenue is greater then the Marginal Cost. Necessary Conditions: To find these values in the calculator, plot the profit function P(x) in the same way as was outlined in part 4) r*K = wage rate * Capital Profit maximization. The firm will continue to produce if Marginal Revenue is greater then the Marginal Cost. Or, we can calculate it as: profit = (price−average cost) ×quantity = ($2.75−$2.75)×75 =$0 profit = (price − average cost) × quantity = ( $2.75 −$ 2.75) × 75 = $0. TC is always above TVC. Profit = Total Revenue – Total Cost And a rational firm will want to maximize its profit. Watch this video for more practice solving for the profit-maximizing point and finding total revenue using a table. Characteristics of Perfect Competition: TVC = Total Variable Cost Since the price is less than average cost, the firm’s profit margin is negative. When the TC = TR the AC = MR. As we stated above when the total revenue is greater then the total cost we have positive profit and when the TC is greater then the TR the profit is negative. TR = P*Q So we must find where MC =MR and draw a vertical line down to the Quantity axis and find the Quantity which correlates to the Price chosen. Jan Hagemejer dvanced Microeconomics. As you can see this forms a rectangle and the Area of the rectangle is the TR. TR is P*Q which is a linear relationship and increases as Price and Quantity increase.Second Graph TR = PQ Remember, however, that the firm has already paid for fixed costs, such as equipment, so it may make sense to continue to produce and incur a loss. Your accounting profit is still$60,000, but now your economic profit is -$10,000. Next we find the slope of the cost curve. Figure 1 illustrates three situations: (a) where at the profit maximizing quantity of output (where P = MC), price is greater than average cost, (b) where at the profit maximizing quantity of output (where P = MC), price equals average cost, and (c) where at the profit maximizing quantity of output (where P = MC), price is less than average cost. Revenue = Price * Quantity Next we combine all of the information we just found. Thus, the profit-maximizing quantity is … As the marginal product of labor increases the MC decreases and when the marginal product of labor decreases the MC increases. MC – Marginal Cost Loss is greater then the variable cost therefor the firm will shut down. To find the Average of the variable cost we must divide by Q. It is as though all the previous actions are ‘sunk’. This means we will have a horizontal line at the chosen price which is shown on the graph. Simply calculate the firm’s total revenue (price times quantity) at each quantity. To double-check your calculations, examine the marginal cost at … MR= ΔTR/ΔQ= (Δ(P*Q))/ΔQ=(P* ΔQ)/ΔQ=P The difference between total revenues and total costs is profits. Background: Pick two very close points to the location of our extrema (t = 1/4). Next we have to find the TC. This will give us our Average Revenue (AR) Then, to find the profit gener ated from this output level, substitute this x-value into P(x). Next we have to find the TC. The firm maximises profit where MR=MC (at Q1). C) TR >TC : profit is positive Profit Maximisation in Perfect Competition. The highest level of profit is the maximum profit and the associated product price is the profit-maximizing price. profit = total revenue−total cost = (75)($2.75)−(75)($2.75) =$0 profit = total revenue − total cost = ( 75) ( $2.75) − ( 75) ($ 2.75) = $0. We have our necessary quantity marked and now we must look at the area under the AC curve. At this price and output level, where the marginal cost curve is crossing the average cost curve, the price the firm receives is exactly equal to its average cost of production. The solutions to the problems are my own work and not necessarily the only way to solve the problems. APL = Average Product of Labor Next we have to find the TC. Now, profit, you are probably already familiar with the term. Target Audience: When AVCTR : profit is negative If the market price that a perfectly competitive firm receives leads it to produce at a quantity where the price is greater than average cost, the firm will earn profits. Homogenous product (perfect substitutes) f (t) = 100 (1/4) 2 – 50 (1/4) + 9 = 2.75. Many producers Next we want to observe the average value of the revenue and to do this we must divide the total revenue by the quantity. Now consider Figure 1(b), where the price has fallen to$2.75 for a pack of frozen raspberries. Share it with us! AR= TR/Q=(P*Q)/Q=P 5)If you choose to find the output level that maximizes profit by hand, use the formula to find the vertex of the profit function, P(x). TR = P*Q Marginal net benefit of the first drink is $13 ($20 – $7), the 2nd is$5 ($12 –$7), and the third is -$1 ($6 – $7). The average cost of producing 65 packs is shown by Point C” which shows the average cost of producing 65 packs is about$2.73. Its demand is estimated that: Q = 100,000 - … At this price, marginal revenue intersects marginal cost at a quantity of 65. MR = MC is a necessary condition for perfect competition Neoclassical economics, currently the mainstream approach to microeconomics, usually models the firm as maximizing profit.. Total costs will be the quantity of 75 times the average cost of $2.75, which is shown by the area of the rectangle from the origin to a quantity of 75, up to point E, over to the vertical axis and down to the origin. The total profit of this firm is then$25, or: T R − T C = 1 0 0 − 7 5 TR - TC = 100 - 75 T R − T C = 1 0 0 − B = Point of Maximum Slope The pattern of costs for the monopoly can be analyzed within the same framework as the costs of a perfectly comp… The curvature of the profit function is consistent with a negative second derivative and results in q* being a quantity of maximum profit. So shift the revenue function parallel downward toward costs until it only touches on one point. This is also the point where our MC = MR. Did you make this project? It never makes sense for a firm to choose a level of output on the downward sloping part of the MC curve, because the profit is lower (the loss is bigger). As we can see from the graph above we can observe profit by looking at the change in TR and TC. or advanced microeconomics course. Play the simulation below multiple times to practice applying these concepts and to see how different choices lead to different outcomes. To find the maximum or minimum value of a quadratic function, start with the general form of the function and combine any similar terms. We want for our marginal net revenue to equal 0. AC=AVC+AFC We want to look at how profit changes with respect to quantity, meaning we want to look at the slope. π = TR - TC Now we shall explain the conditions (10.3) and (10.5) of maximum profit with the help of the firm’s MR and MC curves shown in Fig. Graphically this means the slope of the cost function equals the slope of the revenue function at the maximum profit point. It should be noticeable from the graphs that the TR area is larger than the TC area. Game Theory It’s also important to anticipate the actions of your competitor, even in the nonprofit sector where competitors can also be considered partners. Thus, profits will be the blue shaded rectangle on top. ***It is important to note that between point B and C the MPL is positive and declining. Visual tutorial on production theory. ... Economics AP®︎/College Microeconomics Production, cost, and the perfect competition model Profit maximization. MNR – Marginal Net Revenue To maximize its profit, the firm must its of the product for $20 per unit. Then subtract the firm’s total cost (given in the table) at each quantity. 5.34 Calculate the break-even point Q using the equation obtained in 5.31 and the numbers of 5.32. When Profit is maximized and minimized the MC = MR. For t = 1/4: Total Cost = Variable Cost + Fixed Cost https://cnx.org/contents/vEmOH-_p@4.48:EkZLadKh@7/How-Perfectly-Competitive-Firm, https://www.youtube.com/watch?v=BQvtnjWZ0ig, Use the average cost curve to calculate and analyze a firm’s profits and losses, Identify and explain the firm’s break-even point. Now we can find the profit. Profit maximisation will also occur at an output where MR = MC When MR> MC the firms is increasing its profits and Total Profit is increasing. Substitute q equals 2,000 in order to determine average total cost at the profit-maximizing quantity of output. Total revenues will be the quantity of 85 times the price of$5.00, which is shown by the rectangle from the origin over to a quantity of 85 packs (the base) up to point E’ (the height), over to the price of $5, and back to the origin. The firm is making money, but how much? TPL = Total Product of Labor We’d love your input. TC = Total Cost How to Find the Maximum Profit for a Perfectly Competitive Firm Step 1: Begin With Previous Knowledge of Production Theory. We want to change the equation above to look at the change in profit divided by the change in quantity. Quantity = Q Profit = Total Revenue – Total Costs Therefore, profit maximization occurs at the most significant gap or the biggest difference between the total revenue and the total cost. However, maximizing profit does not necessarily mean that economic profit will be earned. The average product is the TPL/Q and the MPL is the slope of the TPL curve. MNR = MR – MC = 0 The Total Product Curve is shown in the first graph. *Begin with previous knowledge of the Production Theory. First Graph The average cost of producing 85 packs is shown by point C’ or about$3.50. As you can see this forms a rectangle and the Area of the rectangle is the TR. AXES Instead of using the golden rule of profit maximization discussed above, you can also find a firm’s maximum profit (or minimum loss) by looking at total revenue and total cost data. Halloween Pumpkin With a Moving Animatronic Eye | This Pumpkin Can Roll Its Eye. Figure 1. TR was greater than TC and therefor the profit was positive.The Third Graph At this point P =AVC the firm must make decisions as to whether it should continue to produce or shut down. A graph showing a profit curve that has an inverted U-shape and has a peak at the profit maximizing quantity. APL= TPL/Q= Q/L At the inflection point (A) the MPL reaches its maximum and continues to decline from that point and intersects the maximum of the APL. TFC = Totao Fixed Cost The answer depends on firm’s profit margin (or average profit), which is the relationship between price and average total cost. In (c), price intersects marginal cost below the average cost curve. Profit = Total Revenue – Total Cost This occurs when the difference between TR – TC is the greatest. In perfect competition, the same rule for profit maximisation still applies. Pro t maximization problem The formal de nition: (with production set Y ) given a price vectorm p ˛0 and a production vector y 2RL: the pro t is ˇ(p ) = p y = PL l =1 p l y l:(total revenue minus total cost) (1) the pro t maximization problem (PMP): Max y p y ; s.t. We divide the change in Total Cost by the change in Quantity A negative economic profit implies that you could be doing better by pursuing an alternative opportunity. • FC = 240 • AVC = 5 • AR (= Price) = 8 • Q = 70 5.33 Use the equation obtained in 5.31 and the numbers of 5.32 to calculate Q if we target a profit of 60. If the price that a firm charges is higher than its average cost of production for that quantity produced, then the firm’s profit margin is positive and it is earning economic profits. As we have seen when P>AVC the firm continues to produce and when P TC : profit is maximized. We call this the break-even point, since the profit margin is zero. There are three characteristic points that have been pointed out: Price and Average Cost at the Raspberry Farm. AVC= TVC/Q= wL/Q=w/(Q/L)= w/APL Step 2: Derive the Cost Curve From the APL/MPL Curves. The shaded box represents the TR. Previously known information: C = Slope of zero This means that we have a positive marginal profit. From this the ΔQ’s cancel leaving only P. From this we see MR = P In (b), price intersects marginal cost at the minimum point of the average cost curve. These questions allow you to get as much practice as you need, as you can click the link at the top of the first question (“Try another version of these questions”) to get a new set of questions. The firm will continue to operate as long as it covers its variable cost, which is does. Marginal Revenue is the change in total revenueas a result of changing the rate of sales by one unit. P=AVC MNR = MR – MC The calculations are as follows: In Figure 1(c), the market price has fallen still further to \$2.00 for a pack of frozen raspberries. |
# Prove that $A \vartriangle B \subseteq C$ iff $A \cup C = B \cup C$.
This is an exercise from Velleman's "How To Prove It". The end of chapter questions have escalated in difficulty, so I just want to make sure that I am on the right track.
1. Suppose $$A$$, $$B$$, and $$C$$ are sets. Prove that $$A \vartriangle B \subseteq C$$ iff $$A \cup C = B \cup C$$.
Proof: Suppose $$A \vartriangle B \subseteq C$$. Let $$x$$ be arbitrary. Suppose $$x \in A \cup C$$, then either $$x \in A$$ or $$x \in C$$. We consider these two cases:
Case 1. $$x \in A$$. Suppose $$x \notin B \cup C$$. So $$x \notin B$$ and $$x \notin C$$. Since $$x \in A$$ and $$x \notin B$$, $$x \in A\setminus B$$. It follows that $$x \in A \setminus B \cup B \setminus A$$, so $$x \in A \vartriangle B$$. Since $$A \vartriangle B \subseteq C$$ and $$x \in A \vartriangle B$$, $$x \in C$$. But then we have $$x \in C$$ and $$x \notin C$$, which is a contradiction. Thus, $$x \in B \cup C$$
Case 2. $$x \in C$$. It immediately follows that $$x \in B \cup C$$.
In every case, we have shown that $$x \in B \cup C$$. The proof of $$x \in B \cup C \implies x \in A \cup C$$ will be similar, but with the roles of $$A$$ and $$B$$ switched. Therefore, $$A \cup C = B \cup C$$.
Now suppose $$A \cup C = B \cup C$$. Let $$x \in A \vartriangle B$$ be arbitrary. Then $$x \in A \setminus B \cup B \setminus A$$, which means that $$x \in A \setminus B$$ or $$x \in B \setminus A$$. We consider these two cases:
Case 1. $$x \in A \setminus B$$. Then $$x \in A$$ and $$x \notin B$$. Suppose $$x \notin C$$. Then since $$x \notin B$$ and $$x \notin C$$, $$x \notin B \cup C$$. Since $$x \in A$$, $$x \in A \cup C$$. Then since $$A \cup C = B \cup C$$, $$x \in B \cup C$$. But then we have $$x \in B \cup C$$ and $$x \notin B \cup C$$, which is a contradiction. Thus, $$x \in C$$.
Case 2. $$x \in B \setminus A$$. By similar reasoning as case 1 with $$A$$ and $$B$$ switched, we also find that $$x \in C$$.
In every case, we have shown that $$x \in C$$. Since $$x$$ was arbitrary, it follows that $$A \vartriangle B \subseteq C$$. Therefore, $$A \vartriangle B \subseteq C$$ iff $$A \cup C = B \cup C$$. $$\square$$
• A suggested shorter proof, easier (for me at least!) to grasp as a whole: $$(A \cup C) \setminus (B \cup C) = ((A \cup C) \setminus C) \setminus B = (A \setminus C) \setminus B = (A \setminus B) \setminus C,$$ and similarly $(B \cup C) \setminus (A \cup C) = (B \setminus A) \setminus C,$ therefore: $$(A \cup C) \vartriangle (B \cup C) = (A \vartriangle B) \setminus C,$$ whence the result. Apr 7, 2020 at 14:59
• You're welcome, @Iyeeke. I mean starting with $A \vartriangle B\subseteq C$ then using iff statements to get intermediate results until, eventually, you get $A\cup C=B\cup C$. I'm thinking of how to do it myself, so I might update this answer soon. Apr 7, 2020 at 14:14 |
# 高德納箭號表示法
## 簡介
${\displaystyle 3\uparrow \uparrow 2={^{2}3}=3^{3}=27}$
${\displaystyle 3\uparrow \uparrow 3={^{3}3}=3^{3^{3}}=3^{27}=7,625,597,484,987}$
${\displaystyle 3\uparrow \uparrow 4={^{4}3}=3^{3^{3^{3}}}=3^{7625597484987}\approx 1.2580143\times 10^{3638334640024}}$
${\displaystyle 3\uparrow \uparrow 5={^{5}3}=3^{3^{3^{3^{3}}}}=3^{3^{7625597484987}}\approx 3^{1.2580143\times 10^{3638334640024}}}$
${\displaystyle 3\uparrow \uparrow \uparrow 2=3\uparrow \uparrow 3={^{3}3}=3^{3^{3}}=3^{27}=7,625,597,484,987\,\!}$
${\displaystyle 3\uparrow \uparrow \uparrow 3=3\uparrow \uparrow 3\uparrow \uparrow 3={^{^{3}3}3}={^{7625597484987}3}={\begin{matrix}\underbrace {3^{3^{.^{.^{.{3}}}}}} \\7625597484987\end{matrix}}}$
## 使用指數來解釋高德納箭號表示法
${\displaystyle a\uparrow \uparrow b}$代表重複的冪，或迭代冪次，例如： ${\displaystyle a\uparrow \uparrow 4=a\uparrow (a\uparrow (a\uparrow a))=a^{a^{a^{a}}}}$
${\displaystyle a\uparrow \uparrow b=\underbrace {a^{a^{.^{.^{.{a}}}}}} _{b}}$
${\displaystyle a\uparrow \uparrow \uparrow 2=a\uparrow \uparrow a=\underbrace {a^{a^{.^{.^{.{a}}}}}} _{a}}$
${\displaystyle a\uparrow \uparrow \uparrow 3=a\uparrow \uparrow (a\uparrow \uparrow a)=\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{a}}}$
${\displaystyle a\uparrow \uparrow \uparrow 4=a\uparrow \uparrow [a\uparrow \uparrow (a\uparrow \uparrow a)]=\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{a}}}}$
${\displaystyle a\uparrow \uparrow \uparrow b=\left.\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}b}$
${\displaystyle a\uparrow \uparrow \uparrow \uparrow 2=a\uparrow \uparrow \uparrow a=\left.\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}a}$
${\displaystyle a\uparrow \uparrow \uparrow \uparrow 3=a\uparrow \uparrow \uparrow (a\uparrow \uparrow \uparrow a)=\left.\left.\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}a}$
${\displaystyle a\uparrow \uparrow \uparrow \uparrow 4=a\uparrow \uparrow \uparrow [a\uparrow \uparrow \uparrow (a\uparrow \uparrow \uparrow a)]=\left.\left.\left.\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}a}$
${\displaystyle a\uparrow \uparrow \uparrow \uparrow b=\underbrace {\left.\left.\left.\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\cdots \right\}a} _{b}}$
## 一般化
${\displaystyle {\begin{matrix}a\uparrow ^{n}b&=&{\mbox{hyper}}(a,n+2,b)&=&a\to b\to n\\{\mbox{(Knuth)}}&&&&{\mbox{(Conway)}}\end{matrix}}}$
## 定義
${\displaystyle a\uparrow ^{n}b=\left\{{\begin{matrix}1,\\a^{b},\\a\uparrow ^{n-1}(a\uparrow ^{n}(b-1)),\end{matrix}}\right.}$ 若${\displaystyle b=0}$； 若${\displaystyle n=1}$； 其他。
## 參考
• Knuth, Donald E., "Coping With Finiteness", Science vol. 194 n. 4271 (Dec 1976), pp. 1235-1242.
• Robert Munafo, Large Numbers |
What Is 59/60 as a Decimal + Solution With Free Steps
The fraction 59/60 as a decimal is equal to 0.98333.
A Complex Fraction is also a type of fraction which is defined as a Fraction expression that contains a fraction in the numerator, denominator, or either in both. 2/3/4 is a complex fraction in this example the numerator 2/3 is a fraction and the denominator is 4. Whereas 1/2/3 is also a complex fraction in which the denominator 2/3 contains the fraction and the numerator is 1. 1/2/3/4 is also a complex fraction in which the numerator 1/2 and denominator 3/4 both contain fractions.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 59/60.
Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 59
Divisor = 60
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 59 $\div$ 60
This is when we go through the Long Division solution to our problem. The following figure shows the long division:
Figure 1
59/60 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 59 and 60, we can see how 59 is Smaller than 60, and to solve this division, we require that 59 be Bigger than 60.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 59, which after getting multiplied by 10 becomes 590.
We take this 590 and divide it by 60; this can be done as follows:
 590 $\div$ 60 $\approx$ 9
Where:
60 x 9 = 540
This will lead to the generation of a Remainder equal to 590 – 540 = 50. Now this means we have to repeat the process by Converting the 50 into 500 and solving for that:
500 $\div$ 60 $\approx$ 8Â
Where:
60 x 8 = 480
This, therefore, produces another Remainder which is equal to 500 – 480 = 20. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 200.
200 $\div$ 60 $\approx$ 3
Where:
60 x 3 = 180
Finally, we have a Quotient generated after combining the three pieces of it as 0.983=z, with a Remainder equal to 20.
Images/mathematical drawings are created with GeoGebra.
5/5 - (5 votes) |
# Math Snap
## Linear Equations: Tutorial 16 of 22 Save \& ? Question Drag the labels to the table. Each label can be used more than once, but not all labels will be used. Choose the justification for each step in the solution to the equation $6 x-9=45$. \begin{tabular}{|l|l|l|} \hline Step & Statements & \multicolumn{1}{c}{ Reasons } \\ \hline 1 & $6 x-9=45$ & given \\ \hline 2 & $6 x-9+9=45+9$ & addition property of equality \\ \hline 3 & $6 x=54$ & division property of equality \\ \hline 4 & $\frac{6 x}{6}=\frac{54}{6}$ & simplification \\ \hline 5 \end{tabular} addition property of equality subtraction property of equality multiplication property of equality division property of equality simplification
#### STEP 1
Assumptions1. The given equation is $6x -9 =45$ . We can use the properties of equality to solve the equation3. The properties of equality include addition, subtraction, multiplication, and division4. Simplification is also allowed in the steps of solving the equation
#### STEP 2
The given equation is $6x -9 =45$.
#### STEP 3
We can use the addition property of equality to add9 to both sides of the equation. This gives us $6x -9 +9 =45 +9$.
#### STEP 4
implify the equation to get $6x =54$.
#### STEP 5
Now, we can use the division property of equality to divide both sides of the equation by. This gives us $\frac{x}{} = \frac{54}{}$.
##### SOLUTION
implify the equation to get $x =9$. So, the solution to the equation $6x -9 =45$ is $x =9$. |
# How do you simplify (9 1/3+4 1/3)div 6-(-12)?
Apr 30, 2018
$14 \frac{5}{18}$
#### Explanation:
Count the number of terms first and simplify each term separately.
$\textcolor{b l u e}{\left(9 \frac{1}{3} + 4 \frac{1}{3}\right) \div 6} \text{ } \textcolor{red}{- \left(- 12\right)}$
$= \textcolor{b l u e}{\left(13 \frac{2}{3}\right) \div 6} \text{ } \textcolor{red}{+ 12}$
$= \textcolor{b l u e}{\frac{41}{3} \times \frac{1}{6}} \text{ } \textcolor{red}{+ 12}$
$= \textcolor{b l u e}{\frac{41}{18}} \text{ } \textcolor{red}{+ 12}$
$= \textcolor{b l u e}{2 \frac{5}{18}} \text{ } \textcolor{red}{+ 12}$
$= 14 \frac{5}{18}$ |
## Algebra
Showing posts with label distributive property. Show all posts
Showing posts with label distributive property. Show all posts
### Multiplying Polynomials
The distributive property and the product rule for exponents are the keys to multiplication of polynomials.
Remember the product rule,
It says that when we multiply two numbers with the same base we must add the exponents.
Monomial x Polynomial
Multiply.
A common mistake is to distribute when working with multiplication. The distributive property only applies when addition or subtraction separates the terms.
Binomial x Polynomial: Use the distributive property to multiply polynomials by binomials.
We can save a step by distributing each term in the binomial individually. Sometimes this technique is referred to as the FOIL method. (FOIL – Multiply the First, Outer, Inner and then Last terms together.)
Multiply.
When combining like terms, be sure that the variable parts are exactly the same. Go slow and work in an organized fashion because it is easy to make an error when many terms are involved.
It is good practice to recheck your distributive step.
Trinomial x Polynomial
The following problem is one that we will have to be able to do before moving on to the next algebra course. These are tedious, time consuming, and often worked incorrectly. Use caution, take your time and work slowly.
Function notation for multiplication looks like
Given the functions f and g find (f g)(x).
The following special products will simplify things if we memorize them.
This special product is often called difference of squares. Notice that the middle terms cancel because one term will always be positive and the other will be negative.
We may use these formulas as templates when multiplying binomials. It is a good idea to memorize them.
Multiply.
### Simplifying Algebraic Expressions
The properties of real numbers are very important in our study of Algebra. These properties can be applied in Algebra because a variable is simply a letter that represents a real number. The distributive property is one that we apply often when simplifying algebraic expressions. Given real numbers a, b, c:
a(b + c) = ab + ac
When multiplying an expression within parentheses you must multiply everything inside by the number or variable that you are distributing.
Introduction to the Distributive Property
Simplify.
When simplifying, we will often have to combine like terms after we distribute. This step is consistent with the order of operations, multiplication before addition.
Combining Like Terms
Simplify.
Simplifying Algebraic Expressions
To combine like terms, the variable parts have to be exactly the same. But before combining like terms, generally, we will first distribute if necessary. When distributing negative numbers notice that the operations change.
Simplify.
Profit Word Problem: Profit is equal to revenues less cost of production. If the revenue R can be represented by
and the cost C can be represented by
where x represents the number of units produced, find an equation that represents the profit.
Subtracting Variable Expressions: What is the difference between 3x − 4 and −2x + 5? |
LCM the 36 and 45 is the smallest number amongst all common multiples that 36 and also 45. The first couple of multiples the 36 and also 45 room (36, 72, 108, 144, 180, 216, . . . ) and (45, 90, 135, 180, 225, 270, 315, . . . ) respectively. There space 3 commonly used methods to discover LCM that 36 and also 45 - by prime factorization, by division method, and by listing multiples.
You are watching: What is the least common multiple of 36 and 45
1 LCM of 36 and also 45 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM that 36 and also 45 is 180.
Explanation:
The LCM of two non-zero integers, x(36) and y(45), is the smallest confident integer m(180) that is divisible by both x(36) and also y(45) without any type of remainder.
The techniques to discover the LCM that 36 and also 45 are defined below.
By element Factorization MethodBy division MethodBy Listing Multiples
### LCM of 36 and also 45 by prime Factorization
Prime administrate of 36 and 45 is (2 × 2 × 3 × 3) = 22 × 32 and (3 × 3 × 5) = 32 × 51 respectively. LCM that 36 and 45 can be derived by multiply prime determinants raised to their respective highest power, i.e. 22 × 32 × 51 = 180.Hence, the LCM of 36 and 45 by element factorization is 180.
### LCM the 36 and 45 by division Method
To calculate the LCM of 36 and also 45 through the department method, we will certainly divide the numbers(36, 45) by their prime determinants (preferably common). The product of these divisors provides the LCM of 36 and 45.
Step 3: continue the steps until just 1s are left in the last row.
The LCM the 36 and also 45 is the product of every prime number on the left, i.e. LCM(36, 45) by division method = 2 × 2 × 3 × 3 × 5 = 180.
### LCM the 36 and 45 through Listing Multiples
To calculation the LCM that 36 and also 45 by listing out the typical multiples, we can follow the given listed below steps:
Step 1: list a few multiples that 36 (36, 72, 108, 144, 180, 216, . . . ) and also 45 (45, 90, 135, 180, 225, 270, 315, . . . . )Step 2: The common multiples from the multiples that 36 and also 45 room 180, 360, . . .Step 3: The smallest typical multiple that 36 and 45 is 180.
∴ The least common multiple of 36 and also 45 = 180.
## FAQs top top LCM that 36 and 45
### What is the LCM that 36 and also 45?
The LCM the 36 and 45 is 180. To uncover the LCM of 36 and 45, we require to discover the multiples that 36 and also 45 (multiples of 36 = 36, 72, 108, 144 . . . . 180; multiples the 45 = 45, 90, 135, 180) and choose the the smallest multiple that is precisely divisible by 36 and also 45, i.e., 180.
### If the LCM that 45 and also 36 is 180, discover its GCF.
LCM(45, 36) × GCF(45, 36) = 45 × 36Since the LCM of 45 and 36 = 180⇒ 180 × GCF(45, 36) = 1620Therefore, the greatest typical factor (GCF) = 1620/180 = 9.
### What is the least Perfect Square Divisible by 36 and also 45?
The least number divisible by 36 and also 45 = LCM(36, 45)LCM that 36 and 45 = 2 × 2 × 3 × 3 × 5 ⇒ the very least perfect square divisible by each 36 and also 45 = LCM(36, 45) × 5 = 900 Therefore, 900 is the compelled number.
### Which that the complying with is the LCM that 36 and 45? 27, 5, 180, 16
The worth of LCM the 36, 45 is the smallest usual multiple the 36 and also 45. The number to solve the given problem is 180.
See more: What Is The Average Penis Size For 16 Yr.Old, When Does Your Penis Stop Growing
### What is the Relation between GCF and LCM the 36, 45?
The adhering to equation can be provided to express the relation in between GCF and also LCM that 36 and 45, i.e. GCF × LCM = 36 × 45. |
# Graph (mathematics)
Graph (mathematics)
A drawing of a labeled graph on 6 vertices and 7 edges.
In mathematics, a graph is an abstract representation of a set of objects where some pairs of the objects are connected by links. The interconnected objects are represented by mathematical abstractions called vertices, and the links that connect some pairs of vertices are called edges. Typically, a graph is depicted in diagrammatic form as a set of dots for the vertices, joined by lines or curves for the edges. Graphs are one of the objects of study in discrete mathematics.
The edges may be directed (asymmetric) or undirected (symmetric). For example, if the vertices represent people at a party, and there is an edge between two people if they shake hands, then this is an undirected graph, because if person A shook hands with person B, then person B also shook hands with person A. On the other hand, if the vertices represent people at a party, and there is an edge from person A to person B when person A knows of person B, then this graph is directed, because knowing of someone is not necessarily a symmetric relation (that is, one person knowing of another person does not necessarily imply the reverse; for example, many fans may know of a celebrity, but the celebrity is unlikely to know of all their fans). This latter type of graph is called a directed graph and the edges are called directed edges or arcs.
Vertices are also called nodes or points, and edges are also called lines. Graphs are the basic subject studied by graph theory. The word "graph" was first used in this sense by J.J. Sylvester in 1878.[1]
## Definitions
Definitions in graph theory vary. The following are some of the more basic ways of defining graphs and related mathematical structures.
### Graph
A general example of a graph (actually, a pseudograph) with three vertices and six edges.
In the most common sense of the term,[2] a graph is an ordered pair G = (VE) comprising a set V of vertices or nodes together with a set E of edges or lines, which are 2-element subsets of V (i.e., an edge is related with two vertices, and the relation is represented as unordered pair of the vertices with respect to the particular edge). To avoid ambiguity, this type of graph may be described precisely as undirected and simple.
Other senses of graph stem from different conceptions of the edge set. In one more generalized notion,[3] E is a set together with a relation of incidence that associates with each edge two vertices. In another generalized notion, E is a multiset of unordered pairs of (not necessarily distinct) vertices. Many authors call this type of object a multigraph or pseudograph.
All of these variants and others are described more fully below.
The vertices belonging to an edge are called the ends, endpoints, or end vertices of the edge. A vertex may exist in a graph and not belong to an edge.
V and E are usually taken to be finite, and many of the well-known results are not true (or are rather different) for infinite graphs because many of the arguments fail in the infinite case. The order of a graph is | V | (the number of vertices). A graph's size is | E | , the number of edges. The degree of a vertex is the number of edges that connect to it, where an edge that connects to the vertex at both ends (a loop) is counted twice.
For an edge {uv}, graph theorists usually use the somewhat shorter notation uv.
The edges E of an undirected graph G induce a symmetric binary relation ~ on V that is called the adjacency relation of G. Specifically, for each edge {uv} the vertices u and v are said to be adjacent to one another, which is denoted u ~ v.
## Types of graphs
### Distinction in terms of the main definition
As stated above, in different contexts it may be useful to define the term graph with different degrees of generality. Whenever it is necessary to draw a strict distinction, the following terms are used. Most commonly, in modern texts in graph theory, unless stated otherwise, graph means "undirected simple finite graph" (see the definitions below).
#### Undirected graph
A simple undirected graph with three vertices and three edges. Each vertex has degree two, so this is also a regular graph.
An undirected graph is one in which edges have no orientation. The edge (a, b) is identical to the edge (b, a), i.e., they are not ordered pairs, but sets {uv} (or 2-multisets) of vertices.
#### Directed graph
A directed graph
A directed graph or digraph is an ordered pair D = (VA) with
• V a set whose elements are called vertices or nodes, and
• A a set of ordered pairs of vertices, called arcs, directed edges, or arrows.
An arc a = (xy) is considered to be directed from x to y; y is called the head and x is called the tail of the arc; y is said to be a direct successor of x, and x is said to be a direct predecessor of y. If a path leads from x to y, then y is said to be a successor of x and reachable from x, and x is said to be a predecessor of y. The arc (yx) is called the arc (xy) inverted.
A directed graph D is called symmetric if, for every arc in D, the corresponding inverted arc also belongs to D. A symmetric loopless directed graph D = (VA) is equivalent to a simple undirected graph G = (VE), where the pairs of inverse arcs in A correspond 1-to-1 with the edges in E; thus the edges in G number |E| = |A|/2, or half the number of arcs in D.
A variation on this definition is the oriented graph, in which not more than one of (xy) and (yx) may be arcs.
#### Mixed graph
A mixed graph G is a graph in which some edges may be directed and some may be undirected. It is written as an ordered triple G = (VEA) with V, E, and A defined as above. Directed and undirected graphs are special cases.
#### Multigraph
A loop is an edge (directed or undirected) which starts and ends on the same vertex; these may be permitted or not permitted according to the application. In this context, an edge with two different ends is called a link.
The term "multigraph" is generally understood to mean that multiple edges (and sometimes loops) are allowed. Where graphs are defined so as to allow loops and multiple edges, a multigraph is often defined to mean a graph without loops,[4] however, where graphs are defined so as to disallow loops and multiple edges, the term is often defined to mean a "graph" which can have both multiple edges and loops,[5] although many use the term "pseudograph" for this meaning.[6]
#### Simple graph
As opposed to a multigraph, a simple graph is an undirected graph that has no loops and no more than one edge between any two different vertices. In a simple graph the edges of the graph form a set (rather than a multiset) and each edge is a pair of distinct vertices. In a simple graph with n vertices every vertex has a degree that is less than n (the converse, however, is not true - there exist non-simple graphs with n vertices in which every vertex has a degree smaller than n).
#### Weighted graph
A graph is a weighted graph if a number (weight) is assigned to each edge. Such weights might represent, for example, costs, lengths or capacities, etc. depending on the problem at hand.
#### Half-edges, loose edges
In exceptional situations it is even necessary to have edges with only one end, called half-edges, or no ends (loose edges); see for example signed graphs and biased graphs.
### Important graph classes
#### Regular graph
A regular graph is a graph where each vertex has the same number of neighbors, i.e., every vertex has the same degree or valency. A regular graph with vertices of degree k is called a k‑regular graph or regular graph of degree k.
#### Complete graph
A complete graph with 5 vertices. Each vertex has an edge to every other vertex.
Complete graphs have the feature that each pair of vertices has an edge connecting them.
#### Finite and infinite graphs
A finite graph is a graph G = (VE) such that V and E are finite sets. An infinite graph is one with an infinite set of vertices or edges or both.
Most commonly in graph theory it is implied that the graphs discussed are finite. If the graphs are infinite, that is usually specifically stated.
#### Graph classes in terms of connectivity
In an undirected graph G, two vertices u and v are called connected if G contains a path from u to v. Otherwise, they are called disconnected. A graph is called connected if every pair of distinct vertices in the graph is connected; otherwise, it is called disconnected.
A graph is called k-vertex-connected or k-edge-connected if no set of k-1 vertices (respectively, edges) exists that, when removed, disconnects the graph. A k-vertex-connected graph is often called simply k-connected.
A directed graph is called weakly connected if replacing all of its directed edges with undirected edges produces a connected (undirected) graph. It is strongly connected or strong if it contains a directed path from u to v and a directed path from v to u for every pair of vertices uv.
## Properties of graphs
Two edges of a graph are called adjacent (sometimes coincident) if they share a common vertex. Two arrows of a directed graph are called consecutive if the head of the first one is at the nock (notch end) of the second one. Similarly, two vertices are called adjacent if they share a common edge (consecutive if they are at the notch and at the head of an arrow), in which case the common edge is said to join the two vertices. An edge and a vertex on that edge are called incident.
The graph with only one vertex and no edges is called the trivial graph. A graph with only vertices and no edges is known as an edgeless graph. The graph with no vertices and no edges is sometimes called the null graph or empty graph, but the terminology is not consistent and not all mathematicians allow this object.
In a weighted graph or digraph, each edge is associated with some value, variously called its cost, weight, length or other term depending on the application; such graphs arise in many contexts, for example in optimal routing problems such as the traveling salesman problem.
Normally, the vertices of a graph, by their nature as elements of a set, are distinguishable. This kind of graph may be called vertex-labeled. However, for many questions it is better to treat vertices as indistinguishable; then the graph may be called unlabeled. (Of course, the vertices may be still distinguishable by the properties of the graph itself, e.g., by the numbers of incident edges). The same remarks apply to edges, so graphs with labeled edges are called edge-labeled graphs. Graphs with labels attached to edges or vertices are more generally designated as labeled. Consequently, graphs in which vertices are indistinguishable and edges are indistinguishable are called unlabeled. (Note that in the literature the term labeled may apply to other kinds of labeling, besides that which serves only to distinguish different vertices or edges.)
## Examples
A graph with six nodes.
• The diagram at right is a graphic representation of the following graph:
V = {1, 2, 3, 4, 5, 6}
E = {{1, 2}, {1, 5}, {2, 3}, {2, 5}, {3, 4}, {4, 5}, {4, 6}}.
## Important graphs
Basic examples are:
• In a complete graph, each pair of vertices is joined by an edge; that is, the graph contains all possible edges.
• In a bipartite graph, the vertex set can be partitioned into two sets, W and X, so that no two vertices in W are adjacent and no two vertices in X are adjacent. Alternatively, it is a graph with a chromatic number of 2.
• In a complete bipartite graph, the vertex set is the union of two disjoint sets, W and X, so that every vertex in W is adjacent to every vertex in X but there are no edges within W or X.
• In a linear graph or path graph of length n, the vertices can be listed in order, v0, v1, ..., vn, so that the edges are vi−1vi for each i = 1, 2, ..., n. If a linear graph occurs as a subgraph of another graph, it is a path in that graph.
• In a cycle graph of length n ≥ 3, vertices can be named v1, ..., vn so that the edges are vi−1vi for each i = 2,...,n in addition to vnv1. Cycle graphs can be characterized as connected 2-regular graphs. If a cycle graph occurs as a subgraph of another graph, it is a cycle or circuit in that graph.
• A planar graph is a graph whose vertices and edges can be drawn in a plane such that no two of the edges intersect (i.e., embedded in a plane).
• A tree is a connected graph with no cycles.
• A forest is a graph with no cycles (i.e. the disjoint union of one or more trees).
More advanced kinds of graphs are:
## Operations on graphs
There are several operations that produce new graphs from old ones, which might be classified into the following categories:
• Elementary operations, sometimes called "editing operations" on graphs, which create a new graph from the original one by a simple, local change, such as addition or deletion of a vertex or an edge, merging and splitting of vertices, etc.
• Graph rewrite operations replacing the occurrence of some pattern graph within the host graph by an instance of the corresponding replacement graph.
• Unary operations, which create a significantly new graph from the old one. Examples:
• Binary operations, which create new graph from two initial graphs. Examples:
## Generalizations
In a hypergraph, an edge can join more than two vertices.
An undirected graph can be seen as a simplicial complex consisting of 1-simplices (the edges) and 0-simplices (the vertices). As such, complexes are generalizations of graphs since they allow for higher-dimensional simplices.
Every graph gives rise to a matroid.
In model theory, a graph is just a structure. But in that case, there is no limitation on the number of edges: it can be any cardinal number, see continuous graph.
In computational biology, power graph analysis introduces power graphs as an alternative representation of undirected graphs.
In geographic information systems, geometric networks are closely modeled after graphs, and borrow many concepts from graph theory to perform spatial analysis on road networks or utility grids.
## Notes
1. ^ Gross, Jonathan L.; Yellen, Jay (2004). Handbook of graph theory. CRC Press. p. 35. ISBN 9781584880905
2. ^ See, for instance, Iyanaga and Kawada, 69 J, p. 234 or Biggs, p. 4.
3. ^ See, for instance, Graham et al., p. 5.
4. ^ For example, see Balakrishnan, p. 1, Gross (2003), p. 4, and Zwillinger, p. 220.
5. ^ For example, see. Bollobás, p. 7 and Diestel, p. 25.
6. ^ Gross (1998), p. 3, Gross (2003), p. 205, Harary, p.10, and Zwillinger, p. 220.
## References
• Balakrishnan, V. K. (1997-02-01). Graph Theory (1st ed.). McGraw-Hill. ISBN 0-07-005489-4.
• Berge, Claude (1958) (in French). Théorie des graphes et ses applications. Dunod, Paris: Collection Universitaire de Mathématiques, II. pp. viii+277. Translation: . Dover, New York: Wiley. 2001 [1962].
• Biggs, Norman (1993). Algebraic Graph Theory (2nd ed.). Cambridge University Press. ISBN 0-521-45897-8.
• Bollobás, Béla (2002-08-12). Modern Graph Theory (1st ed.). Springer. ISBN 0-387-98488-7.
• Bang-Jensen, J.; Gutin, G. (2000). Digraphs: Theory, Algorithms and Applications. Springer.
• Diestel, Reinhard (2005). Graph Theory (3rd ed.). Berlin, New York: Springer-Verlag. ISBN 978-3-540-26183-4 .
• Graham, R.L., Grötschel, M., and Lovász, L, ed (1995). Handbook of Combinatorics. MIT Press. ISBN 0-262-07169-X.
• Gross, Jonathan L.; Yellen, Jay (1998-12-30). Graph Theory and Its Applications. CRC Press. ISBN 0-8493-3982-0.
• Gross, Jonathan L., & Yellen, Jay, ed (2003-12-29). Handbook of Graph Theory. CRC. ISBN 1-58488-090-2.
• Harary, Frank (January 1995). Graph Theory. Addison Wesley Publishing Company. ISBN 0-201-41033-8.
• Iyanaga, Shôkichi; Kawada, Yukiyosi (1977). Encyclopedic Dictionary of Mathematics. MIT Press. ISBN 0-262-09016-3.
• Zwillinger, Daniel (2002-11-27). CRC Standard Mathematical Tables and Formulae (31st ed.). Chapman & Hall/CRC. ISBN 1-58488-291-3.
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Graph — may refer to:* A graphic (such as a chart or diagram) depicting the relationship between two or more variables used, for instance, in visualising scientific data.In mathematics:* Graph (mathematics), a set of vertices connected with edges * Graph … Wikipedia
• Graph theory — In mathematics and computer science, graph theory is the study of graphs : mathematical structures used to model pairwise relations between objects from a certain collection. A graph in this context refers to a collection of vertices or nodes and … Wikipedia
• Graph pebbling — is a mathematical game and area of interest played on a graph with pebbles on the vertices. Game play is composed of a series of pebbling moves. A pebbling move on a graph consists of taking two pebbles off one vertex and placing one on an… … Wikipedia
• Graph paper — Regular graphing paper (upper); Logarithmic graphing paper (lower). Graph paper, graphing paper, grid paper or millimeter paper is writing paper that is printed with fine lines making up a … Wikipedia
• Graph coloring — A proper vertex coloring of the Petersen graph with 3 colors, the minimum number possible. In graph theory, graph coloring is a special case of graph labeling; it is an assignment of labels traditionally called colors to elements of a graph… … Wikipedia
• Graph property — In graph theory a graph property is any inherently graph theoretical property of graphs (formal definitions follow), distinguished from properties of graphs described in terms of various graph representations: graph drawings, data structures for… … Wikipedia
• mathematics — /math euh mat iks/, n. 1. (used with a sing. v.) the systematic treatment of magnitude, relationships between figures and forms, and relations between quantities expressed symbolically. 2. (used with a sing. or pl. v.) mathematical procedures,… … Universalium
• Graph factorization — Not to be confused with Factor graph. 1 factorization of Desargues graph: each color class is a 1 factor … Wikipedia
• Mathematics — Maths and Math redirect here. For other uses see Mathematics (disambiguation) and Math (disambiguation). Euclid, Greek mathematician, 3r … Wikipedia
• Graph drawing — This article is about the general subject of graph drawing. For the annual research symposium, see International Symposium on Graph Drawing. Graphic representation of a minute fraction of the WWW, demonstrating hyperlinks. Graph drawing is an… … Wikipedia |
Lesson Video: Finding the Prime Factorization of a Number | Nagwa Lesson Video: Finding the Prime Factorization of a Number | Nagwa
Lesson Video: Finding the Prime Factorization of a Number Mathematics • 6th Grade
With a series of examples using factor trees and the method of dividing by prime numbers, we carefully walk you through the process of finding the prime factorization of a composite number then leave you with a few tips at the end.
09:53
Video Transcript
Let’s look at finding the prime factorization of a number. Before we get started, there are two words we need to know here. And that would be “factor” and “prime number.”
Factors are numbers we multiply together to produce another number. Here’s an example. Two times three is six. Two and three are factors of six. And prime number is a whole number greater than one that has exactly two factors: one and itself. For example, the number seven, the only two numbers that multiply together to equal seven are one and seven. So seven is a prime number. Then, we say that prime factorization is writing a composite number as the product of only prime numbers. Just a reminder, a composite number is a whole number greater than one that has more than two factors. For example, 10, you can find 10 by multiplying one by 10 or two by five. 10 has more than two factors.
Okay, back to prime factorization. We’re going to take these composite numbers and write them as a product of prime numbers. Remember that product means numbers that are multiplied together. We’re gonna look at two methods for finding prime factorization: the first one is by using a factor tree and the second one is dividing by prime numbers.
Find the prime factorization of 60.
Let’s start by using the factor tree method to factor 60. What you wanna do here is think of two numbers that you know that multiply together to equal 60. I’m gonna choose six and 10. Six times 10 equals 60. Now, we look at the numbers six and 10 and we see what multiplies together to equal each of those. For six, I know that two times three equals six. And for 10, two times five equals 10.
The factor tree method is finished when each of your branches is a prime number. So here, the only things that multiply together to equal two are two and one. Two is a prime number. Three is a prime number for the same reason: one times three is three. That’s the only factors. When we find that all of our branches have prime numbers, we can stop and write down the prime factorization. Two times two times three times five is the prime factorization of 60. 60 equals two times two times three times five. It’s often helpful to write it with exponents. So we say two squared times three times five for clarity. And we call this method the factor tree method. Let’s solve the same problem again using divide by prime numbers method.
For this method, I need to think of a prime number that 60 is divisible by. I’m going to start with two. I know that 60 is divisible by two. And I know that two is a prime number. 60 divided by two is 30. Now, I need a prime number here that 30 is divisible by. I chose three. 30 divided by three is 10. We need a prime number that 10 is divisible by. I’m gonna choose two. 10 divided by two is five. This method is finished when the number in the box is a prime number. Since five is a prime number, we’re finished with this step. We take all the prime numbers we’ve been using to divide and the bottom number. And this is the prime factorization.
You would say 60 equals two times three times two times five. Or more simply, two squared times three times five. Both methods lead us to the prime factorization of sixty. Though the methods are different, there is only one prime factorization of 60. Only one set of prime numbers multiply together to equal 60.
Our next example says, “Find all the prime factors of 28.”
This time we’re gonna try and divide by prime factors to solve the problem. I noticed that 28 is an even number. So I’m gonna start by dividing it by two. 28 divided by two leaves us with 14. Another even number! I’m gonna divide by two again. 14 divided by two is seven. We now recognize that seven is a prime number. And so that’s the end of this step. 28 equals two times two times seven. And we prefer to write it with exponents which leaves us with 28 equals two squared times seven.
Our next example: Find the prime factorization of 468.
You might be thinking that 468 is a really big number. Is it going to be super hard to find the prime factorization for 468? But the answer is no. We’re going to follow the same procedure. And it will be no different than finding the prime factorization for other numbers, other smaller numbers. Okay! Then let’s use the factor tree method and find the prime factorization here. Again, we recognize that this is an even number. So you can automatically start with two. Two times 234 equals 468. Two is a prime number. So this branch is finished.
234 is an even number. Let’s divide 234 by two. Two times 117 equals 234. Two is a prime number. This branch is finished. Now we need some factors of 117. If you don’t immediately recognize what the factors of 117 are, the best thing to do is try to check common prime factors. For example, we know that this is an odd number and not divisible by two. The sum of the digits one, one, seven is nine, which makes this divisible by three.
So our next step would be to divide 117 by three. Three times 39 equals 117. Three is a prime number. This branch is finished. I’m gonna move the thirty-nine over here to give us a little bit more space. What two factors multiply together to equal 39? It’s pretty easy to spot that 39 is divisible by three. 39 divided by 13 is three. Three is a prime number. This branch is finished. 13 is also prime, which means we’re finally down to all prime factors.
The end of each branch is a prime factor. We’re gonna put them all together to make the prime factorization. We get 468 is equal to two times two times three times three times 13, for a final answer with exponents of two squared times three squared times 13 equals 468. And it’s not so bad after all.
And finally, here are just a few tips for finding prime factorization. Number one: use the method that works for you. If you prefer dividing by prime factors, that’s great! You can find the answer by dividing by prime factors. If you prefer using the factor tree and that works for you, use that method. And secondly, if you get stuck and you’re not sure what factors to check for, check for larger prime numbers. Try dividing the number you’re factoring by 13, 17, 19, or other prime numbers. And third, just practice. Recognizing what factors are in numbers comes with practice. You’ll get faster and more accurate at checking for factorization with practice. |
# Dividing Fractions
Part One
This game is intended to help you visualize what is going on with fraction division . If you get lost at
any point in the program, you can scroll down the window for instructions.
You will be shown a point of the number line for which you must enter the corresponding fraction. You must enter
the fraction in the form shown in the pictures below. The picture on the left shows how you would enter and
the other picture shows 5. (Why is this 5?)
The number you have entered re presents the dividend of your division problem. Next, you will see a second
number line showing a new fraction- the divisor. You must now enter this number. Finally, you must decide on
the answer (the quotient) and enter it. Click OK. Click DRAW to get a new problem.
Play the game for at least 5 complete division problems. Think about what it is that you are doing when you divide.
Write down your solution process for two of the problems you solved. Include number line drawings like the ones
on the computer. Explain why the solution to each problem makes sense.
Part Two
Pretend that you were given these problems on the computer:
Create drawings for each problem to imitate what the computer screen
would show. Explain in a sentence why the answer makes sense.
For example, the picture at right illustrates This
makes sense because six one-fourths fit into .
Part Three
Most of you remember the “flip and multiply” algorithm for dividing fractions:
Why does this method work ? The purpose of this part of the lab is to develop an understanding of why “flip and
multiply” works.
Example:
Wanda the Witch had 3 pounds of candy corn to give away at her Haunted House. She decided to make 1/4 pound
bags to give away. How many bags of candy corn could she make?
(a) Let’s represent this story with a picture and use the picture to de termine the
.
Each rectangle represents 1 pound of candy corn. I divided each pound into 4 equal regions to represent
1/4 pound bags. Now I can solve the problem by counting the number of 1/4ths in the 3 pounds.
There are 12 1/4-pound bags in the 3 pounds of candy corn.
(b) Let’s write a division statement for this problem and connect each number to the
problem.
Division statement: 3 ÷ 1/4. The 3 represents the whole: 3 pounds. The 1/4 represents the size of the
part that I am repeatedly taking away: 1/4 pound bags. So 3 pounds ÷ 1/4 pounds per bag = 12 bags.
(c) Let’s write a multiplication statement for this problem and connect each number to
the problem.
Multiplication statement: 3 x 4. The 3 represents how many pounds of candy corn Wanda has. The 4
represents the number of 1/4 pound bags that are in each pound of candy corn.
So 3 pounds x 4 bags per pound = 12 bags.
Explore the fraction division problems be low by completing steps (a)-(c) as illustrated in the
example above.
(1) A turtle undertakes a journey of 2 miles. Each day, the turtle travels 1/5 of a mile. How many
days will it take the turtle to finish the journey?
(2) The Habitat for Humanity team can build 2/3 of a house in one day . They plan to build 4
houses. How long will it take to finish all 4 houses?
(3) The vet has a new bottle containing 4 ounces of special medicine. She will be administering
several doses of this medicine to her cats and dogs. Each dose is ounces. How many doses
are in the bottle?
Look back at your solutions to the problems above. Write an explanation for why it makes
sense to “flip and multiply” when you do fraction division.
Prev Next
Start solving your Algebra Problems in next 5 minutes!
2Checkout.com is an authorized reseller
of goods provided by Sofmath
Attention: We are currently running a special promotional offer for Algebra-Answer.com visitors -- if you order Algebra Helper by midnight of February 19th you will pay only \$39.99 instead of our regular price of \$74.99 -- this is \$35 in savings ! In order to take advantage of this offer, you need to order by clicking on one of the buttons on the left, not through our regular order page.
If you order now you will also receive 30 minute live session from tutor.com for a 1\$!
You Will Learn Algebra Better - Guaranteed!
Just take a look how incredibly simple Algebra Helper is:
Step 1 : Enter your homework problem in an easy WYSIWYG (What you see is what you get) algebra editor:
Step 2 : Let Algebra Helper solve it:
Step 3 : Ask for an explanation for the steps you don't understand:
Algebra Helper can solve problems in all the following areas:
• simplification of algebraic expressions (operations with polynomials (simplifying, degree, synthetic division...), exponential expressions, fractions and roots (radicals), absolute values)
• factoring and expanding expressions
• finding LCM and GCF
• (simplifying, rationalizing complex denominators...)
• solving linear, quadratic and many other equations and inequalities (including basic logarithmic and exponential equations)
• solving a system of two and three linear equations (including Cramer's rule)
• graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions)
• graphing general functions
• operations with functions (composition, inverse, range, domain...)
• simplifying logarithms
• basic geometry and trigonometry (similarity, calculating trig functions, right triangle...)
• arithmetic and other pre-algebra topics (ratios, proportions, measurements...)
ORDER NOW! |
# Ch. 8.2-3: Solving Systems of Equations Algebraically.
## Presentation on theme: "Ch. 8.2-3: Solving Systems of Equations Algebraically."— Presentation transcript:
Ch. 8.2-3: Solving Systems of Equations Algebraically
What is a System of Linear Equations? If the system of linear equations is going to have a solution, then the solution will be an ordered pair (x (x, y) where x and y make both equations true at the same time. We will only be dealing with systems of two equations using two variables, x and y.y. We will be working with the graphs of linear systems and how to find their solutions graphically. A system of linear equations is simply two or more linear equations using the same variables.
inconsistent consistent dependent independent
Definitions Consistent: has at least one solution Dependent: has an infinite amount of solutions
Inconsistent: has no solutions
Independent: has exactly one solution Consistent: has at least one solution
x y Consider the following system: x – y = –1 x + 2y = 5 Using the graph to the right, we can see that any of these ordered pairs will make the first equation true since they lie on the line. We can also see that any of these points will make the second equation true. However, there is ONE coordinate that makes both true at the same time… (1, 2) The point where they intersect makes both equations true at the same time. How to Use Graphs to Solve Linear Systems
x – y = –1 x + 2y = 5 How to Use Graphs to Solve Linear Systems x y Consider the following system: (1, 2) We must ALWAYS verify that your coordinates actually satisfy both equations. To do this, we substitute the coordinate (1, 2) into both equations. x – y = –1 (1) – (2) = –1 Since (1, 2) makes both equations true, then (1, 2) is the solution to the system of linear equations. x + 2y = 5 (1) + 2(2) = 1 + 4 = 5
Graphing to Solve a Linear System While there are many different ways to graph these equations, we will be using the slope – intercept form. To put the equations in slope intercept form, we must solve both equations for y. Start with 3x + 6y = 15 Subtracting 3x from both sides yields 6y = –3x + 15 Dividing everything by 6 gives us… Similarly, we can add 2x to both sides and then divide everything by 3 in the second equation to get Now, we must graph these two equations Solve the following system by graphing: 3x + 6y = 15 –2x + 3y = –3
Graphing to Solve a Linear System Solve the following system by graphing: 3x + 6y = 15 –2x + 3y = –3 Using the slope intercept forms of these equations, we can graph them carefully on graph paper. x y Start at the y – intercept, then use the slope. Label the solution! (3, 1) Lastly, we need to verify our solution is correct, by substituting (3, 1). Since and, then our solution is correct!
Graphing to Solve a Linear System Let's summarize! There are 4 steps to solving a linear system using a graph. Step 1: Put both equations in slope – intercept form Step 2: Graph both equations on the same coordinate plane Step 3: Estimate where the graphs intersect. Step 4: Check to make sure your solution makes both equations true. Solve both equations for y, so that each equation looks like y = mx + b. Use the slope and y – intercept for each equation in step 1. Be sure to use a ruler and graph paper! This is the solution! LABEL the solution! Substitute the x and y values into both equations to verify the point is a solution to both equations.
Substitution 1. Given two equations. Solve one equation for a variable. 2. Plug this expression in for the variable into the other equation. 3. Solve. 4. Plug this value into the first equation and solve it as well. 5. Write the answer as an ordered pair.
Example: Solve 1. Solve for p 2. Plug into other equation and solve for q. 3. Plug value of q into initial equation. 4. Write the ordered pair (ABC order)
Example: Solve 1. Solve for n 2. Plug into other equation and solve for m. 3. Plug value of m into initial equation. 4. Write the ordered pair (ABC order)
Elimination 1. Given two equations. Make sure variables are lined vertically 2. Choose a variable to eliminate. It must become the opposite value of the same variable in the other equation. 3. Multiply the entire equation to create the needed values. 4. Add the two equations together 5. Solve for variable left. 6. Plug value into initial equation and solve. 7. Write the answer as an ordered pair
Example: Solve 1. Eliminate n since they are opposites. 2. Add the two equations 3. Solve for m 4. Plug m back into the initial equation 5. Solve for n 6. Write the ordered pair (ABC order)
Example: Solve 1. Eliminate h since they are opposites. 2. Multiply the second equation by 2. 3. Add the two equations 4. Solve for g 5. Plug g back into the initial equation 6. Solve for h 7. Write the ordered pair (ABC order)
Ex#1 Solve the system by substitution. Check by graphing – x + 2(-2x + 1) = 2 – x – 4x + 2 = 2 -5x + 2 = 2 -2 -2 -5x = 0 -5 x = 0 y = -2x + 1 y = -2(0) + 1 y = 0 + 1 y = 1 (0, 1) (x, y)
(0, 1)
Ex#2 Solve the system by substitution. 4x – 3(-2x + 13) = 11 4x + 6x – 39 = 11 10x – 39 = 11 +39 +39 10x = 50 10 x = 5 y = -2x + 13 y = -2(5) + 13 y = -10 + 13 y = 3 (5, 3) (x, y)
Try These: Solve the system by substitution. 3(2y – 7) + 4y = 9 6y – 21 + 4y = 9 10y – 21 = 9 +21 +21 10y = 30 10 y = 3 x = 2y – 7 x = 2(3) – 7 x = 6 – 7 x = -1 (-1, 3) (x, y)
Ex#2 Solve the system by substitution. 6x – 3(2x – 5) = 15 6x – 6x + 15 = 15 15 = 15 Infinite solutions
Ex#2 Solve the system by substitution. 6x + 2(-3x + 1) = 5 6x – 6x + 2 = 5 2 5 No Solution
To eliminate a variable line up the variables with x first, then y. Afterward make one of the variables opposite the other. You might have to multiply one or both equations to do this.
Example #3: Find the solution to the system using elimination. 2 6x – 2y = 16 x + 2y = 5 7x + 0 = 21 7x = 21 7 x = 3 3 + 2y = 5 -3 -3 2y = 2 y = 1(x, y) (3, 1) 3x – y = 8 x + 2y = 5
Example #3: Find the solution to the system using elimination. -3 -6x – 3y = –18 4x + 3y = 24 –2x = 6 x = – 3 2(–3) + y = 6 -6 + y = 6 y = 12 (x, y) (–3, 12) 2x + y = 6 4x + 3y = 24
3 3x – 4y = 16 5x + 6y = 14 9x – 12y = 48 2 19x = 76 x = 4 (x, y) (4, -1) 5(4) + 6y = 14 -20 6y = -6 y = -1 20 + 6y = 14 10x + 12y = 28 Example #3: Find the solution to the system using elimination.
5 -3x + 2y = -10 5x + 3y = 4 -15x + 10y = -50 3 19y = -38 y = -2 (x, y) (2, -2) 5x + 3(-2) = 4 5x = 10 x = 2 5x – 6 = 4 15x + 9y = 12
Example #3: Find the solution to the system using elimination. 12x – 3y = -9 -4x + y = 3 12x – 3y = -9 3 0 = 0 Infinite solutions -12x + 3y = 9
Example #3: Find the solution to the system using elimination. 6x + 15y = -12 -2x – 5y = 9 6x + 15y = -12 3 0 = 15 No solution -6x – 15y = 27 |
# Class 12 RD Sharma Solutions – Chapter 18 Maxima and Minima – Exercise 18.2
### Question 1. f(x) = (x – 5)4
Solution:
Given function
f(x) = (x – 5)4
Now, differentiate the given function w.r.t. x
f ‘(x) = 4(x-5)3
Now, for local maxima and minima
Put f ‘(x) = 0
⇒ 4(x – 5)3 = 0
⇒ x – 5 = 0
⇒ x = 5
So, at x = 5, f'(x) changes from negative to positive. Hence, x = 5 is the point of local minima
So, the minimum value is f(5) = (5 – 5)4 = 0
### Question 2. f(x) = x3– 3x
Solution:
Given function
f(x) = x3– 3x
Now, differentiate the given function w.r.t. x
f ‘(x) = 3x2– 3
Now, for local maxima and minima
Put f ‘(x) = 0
⇒ 3x2– 3 = 0
⇒ x = ±1
Now, again differentiating f'(x) function w.r.t. x
f “(x) = 6x
Put x = 1 in f”(x)
f “(1)= 6 > 0
So, x = 1 is point of local minima
Put x = -1 in f”(x)
f “(-1)= -6 < 0
So, x = -1 is point of local maxima
So, the minimum value is f(1) = x3– 3x = 13 – 3 = -2
and the maximum value is f(-1) = x3– 3x = (-1)3 – 3(-1) = 2
### Question 3. f(x) = x3 (x – 1)2
Solution:
Given function
f(x) = x3(x – 1)2
Now, differentiate the given function w.r.t. x
f ‘(x) = 3x2 (x- 1)2 + 2x3(x- 1)
= (x – 1) (3x2(x – 1) + 2x3)
= (x – 1) (3x3 – 3x2 + 2x3)
= (x – 1) (5x3 – 3x2)
= x2(x – 1) (5x – 3)
Now, for all maxima and minima,
Put f ‘(x) = 0
= x2(x – 1) (5x- 3) = 0
x = 0, 1, 3/5
So, at x = 3/5, f'(x) changes from negative to positive. Hence, x = 3/5 is a point of minima
So, the minimum value is f(3/5) = (3/5)3(3/5 – 1)2 = 108/3125
At x = 1, f'(x) changes from positive to negative. Hence, x = 1 is point of maxima.
So, the maximum value is f(1) = (1)3(1 – 1)2 = 0
### Question 4. f (x) = (x – 1) (x + 2)2
Solution:
Given function
f(x) = (x – 1)(x + 2)2
Now, differentiate the given function w.r.t. x
f ‘(x) = (x + 2)2 + 2(x – 1)(x + 2)
= (x+ 2) (x+ 2 + 2x – 2)
=(x + 2) (3x)
Now, for all maxima and minima,
Put f ‘(x) = 0
⇒ (x + 2) (3x) = 0
x = 0,-2
So, at x = -2, f(x) changes from positive to negative. Hence, x = -2 is a point of Maxima
So, the maximum value is f(-2) = (-2 – 1)(-2 + 2)2 = 0
At x = 0, f ‘(x) changes from negative to positive. Hence, x = 0 is point of Minima.
So, the minimum value is f(0) = (0 – 1)(0 + 2)2 = -4
### Question 5. f(x) = (x – 1)3 (x + 1)2
Solution:
Given function
f(x) = (x – 1)3 (x + 1)2
Now, differentiate the given function w.r.t. x
f ‘(x) = 3(x – 1)2 (x + 1)2 + 2(x – 1)3 (x + 1)
= (x – 1)2 (x + 1) {3(x + 1) + 2(x – 1)}
= (x – 1)2 (x + 1) (5x + 1)
Now, for local maxima and minima,
Put f ‘(x) = 0
⇒ (x – 1)2 (x + 1) (5x + 1) = 0
⇒ x = 1, -1, -1/5
So, at x = -1, f ‘(x) changes from positive to negative. Hence, x = -1 is point of maxima
So, the maximum value is f(-1) = (-1 – 1)3 (-1 + 1)2 = 0
At x = -1/5, f ‘(x) changes from negative to positive so x= -1/5 is point of minima
So, the minimum value is f(-1/5) = (-1/5 – 1)3 (-1/5 + 1)2 = -3456/3125
### Question 6. f(x) = x3 – 6x2 + 9x +15
Solution:
Given function
f(x) = x3 – 6x2 + 9x + 15
Now, differentiate the given function w.r.t. x
f ‘(x) = 3x2 – 12x + 9
= 3 (x2 – 4x + 3)
= 3 (x – 3) (x – 1)
Now, for local maxima and minima,
Put f ‘(x) = 0
⇒ 3 (x – 3) (x – 1) – 0
⇒ x = 3, 1
At x = 1, f'(x) changes from positive to negative. Hence, x = 1 is point of local maxima
So, the maximum value is f(1) = (1)3 – 6(1)2 + 9(1) + 15 = 19
At x = 3, f'(x) changes from negative to positive. Hence, x = 3 is point of local minima
So, the minimum value is f(x) = (3)3 – 6(3)2 + 9(3) + 15 = 15
### Question 7. f(x) = sin2x, 0 < x < π
Solution:
Given function
f(x) = sin2x, 0 < x, π
Now, differentiate the given function w.r.t. x
f'(x) = 2 cos 2x
Now, for local maxima and minima,
Put f'(x) = 0
⇒ 2x =
⇒ x =
At x = π​/4, f'(x) changes from positive to negative. Hence, x = π​/4, Is point of local maxima
So, the maximum value is f(π​/4) = sin2(π​/4) = 1
At x = 3π​/4, f'(x) changes from negative to positive. Hence, x = 3π​/4 is point of local minima,
So, the minimum value is f(3π​/4) = sin2(3π​/4) = -1
### Question 8. f(x) = sin x – cos x, 0 < x < 2Ï€
Solution:
Given function
f(x) = sin x – cos x, 0 < x < 2Ï€
Now, differentiate the given function w.r.t. x
f'(x)= cos x + sin x
Now, for local maxima and minima,
Put f'(x) =0
cos x = -sin x
tan x = -1
x = ∈ (0, 2π)
Now again differentiate the given function w.r.t. x
f”(x) = -sin x + cos x
<0
>0
Therefore, by second derivative test, is a point of local maxima
Hence, the maximum value is
However, is a point of local minima
Hence, the minimum value is
### Question 9. f(x) = cos x, 0< x < π
Solution:
Given function
f(x) = cos x, 0< x < π
Now, differentiate the given function w.r.t. x
f'(x) = – sin x
Now, for local maxima and minima,
Put f ‘(x) – 0
⇒ – sin x = 0
⇒ x = 0, and π
But, these two points lies outside the interval (0, π)
So, no local maxima and minima will exist in the interval (0, π).
### Question 10. f(x) = sin 2x – x, -Ï€/2 ≤ x ≤ Ï€/2
Solution:
Given function
f(x) = sin2x – x
Now, differentiate the given function w.r.t. x
f ‘(x) = 2 cos 2x – 1
Now, for local maxima and minima,
Put f'(x) = 0
⇒ 2cos 2x – 1 = 0
⇒ cos 2x = 1/2 = cos π/3
⇒ 2x = π/3, -π/3
⇒ x =
At x = -π/6, f'(x) changes from negative to positive. Hence, x = π/6 is point of local minima.
So, the minimum value is
At x = π/6, f'(x) changes from positive to negative. Hence, x = π/6 is point of local maxima
The maximum value is
### Question 11. f(x) = 2sin x – x, -Ï€/2 ≤ x ≤ Ï€/2
Solution:
Given function
f(x) = 2sin x – x, -Ï€/2≤ x ≤ Ï€/2
Now, differentiate the given function w.r.t. x
f ‘(x) = 2cos x – 1 = 0
Now, for local maxima and minima,
Put f'(x) = 0
⇒ cos x = 1/2 = cos π/3
⇒ x = -π/3, π/3
So, at x = -Ï€/3, f'(x) changes from negative to positive. Hence, x = -Ï€/3 is point of local minima
So, the minimum value is f(-Ï€/3) = 2sin(-Ï€/3) – (-Ï€/3) = -√3 – Ï€/3
At x = π/3, f'(x) changes from positive to negative. Hence, x = π/3 is point of local minima
The maximum value is f(Ï€/3) = 2sin(Ï€/3) – (Ï€/3) = √3 – Ï€/3
### Question 12. f(x) = x, x > 0
Solution:
Given function
f(x) = x, x > 0
Now, differentiate the given function w.r.t. x
f'(x) =
Now, for local maxima and minima,
Put f'(x) = 0
⇒
⇒ 2 – 3x = 0
⇒ x = 2/3
f “(x) =
=
= (3x – 4)/4(1 – x)2
f “(2/3) =
Therefore, x = 2/3 is a point of local maxima and the local maximum value of f at x = 2/3 is
f(2/3) = 2/3(√1/3) = (2√3)/9
### Question 13. f(x) = x3(2x – 1)3
Solution:
Given function
f(x) = x3(2x – 1)3
Now, differentiate the given function w.r.t. x
f'(x) = 3x2(2x – 1)2 + 6x3(2x – 1)2
= 3x2(2x – 1)2(2x – 1 + 2x)
= 3x2(4x – 1)
Now, for local maxima and minima,
Put f'(x) = 0
⇒ 3x2(4x – 1) = 0
⇒ x = 0, 1/4
At x = 1/4, f'(x) changes from negative to positive. Hence, x = 1/4 is the point of local minima,
So, the minimum value is f(1/4)= (1/4)3(2(1/4) – 1)3= -1/512
### Question 14. f(x) = x/2 + 2/x, x > 0
Solution:
Given function
f(x) = x/2 + 2/x, x > 0
Now, differentiate the given function w.r.t. x
f'(x) = 1/2 – 2/x2, x > 0
Now, for local maxima and minima,
Put f'(x) = 0
⇒ 1/2 – 2/x2 = 0
⇒ x2 – 4 = 0
⇒ x = 2, -2
At x = 2, f'(x) changes from negative to positive. Hence, x = 2 is point of local minima
So, the local minimum value is f(2) = 2/2 + 2/2 = 2
### Question 15. f(x) = 1/(x2 + 2)
Solution:
Given function
f(x) = 1/(x2 + 2)
Now, differentiate the given function w.r.t. x
f'(x) = -(2x)/(x2 + 2)2
Now, for local maxima and minima,
Put f'(x) = 0 f'(x) = 0
f'(x) = -(2x)/(x2 + 2)2 = 0
⇒ x = 0
At x = 0, f'(x) > 0
At x = 0+, f'(x) < 0
Therefore, local minimum and maximum value of f(0) = 1/2
Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!
Previous
Next |
Home » Physics » What is graph in physics?
# What is graph in physics?
In this article learn about the what is graph in physics and how to plot a basics of graphs in physics. Here you will learn about
1. Why we need graphs
2. What are graphs and
3. How to plot graphs
After reading this article students would be able to define graphs and plot graphs using data on dependent and independent variables.
Let us now start with why we need graphs. We need graphs in physics because a graph is a very powerful method of presenting the information.
We can use both tables and graphs to represent the same information but graphs are a lot easier to read and interpret information than tables. Having established the need for graphs let us now define what is a graph.
## What is graph in physics?
A graph is a line straight or curved that shows relation between two quantities out of which one varies as a result of change in other.
To make it clear let us consider we have two variables x and y where,
$y=2x$
The variable that is made to alter or change is called the independent variable. Here in our case x is an independent variable. Another variable that varies as a result of the change in the independent variable is called the dependent variable.
Here $y$ is the dependent variable whose value changes with change in the value of $x$.
This relation between the independent variable and the dependent variable can be shown by means of a graph.
## Plotting a basic graph in physics
We already know that a graph is plotted to display the relationship between two quantities. To understand how a graph can be plotted, consider a car that is moving along a straight line. The distance moved by car every 6 minutes is given in this table.
let us now use the example of the above table and start looking at the steps involved in plotting a graph. Before starting it is important to note that here time t is the independent quantity and distance traveled depends on it.
### Step 1: Choosing the axes
Step 1 in plotting the graph is Choosing the axes. In this step first, mark the axis by drawing two perpendicular lines crossing each other at a point. Now mark the x-axis and y-axis.
We will take time along the x-axis because generally independent variables are taken along the x-axis. We will take distance which is the dependent variable along the y-axis. So the line OX represents time and line OY represents distance. These lines OX and OY are named after the quantities they represent.
### Step 2: Choose the scale
In our second step, we would choose the scale. We have to choose the scale because the size of the paper on which the graph is drawn is limited. Values are marked at equal distances on the available length of the axis. You have to do it in such a way that all the values of the quantity represented on the axis can be accommodated in the available length.
### Step 3: Plotting the points
In the third step, we would be plotting the points. Now each set of values of the two quantities is represented by a point on the graph. For example, the set 6 minutes and 10 kilometers is represented by point A on the graph. To get this point we mark 6 minutes on the x-axis and draw a perpendicular at the point on the x-axis. Similarly, locate 10 kilometers on the y-axis and draw a perpendicular on the y-axis at the point.
The point of intersection of these perpendiculars is point A. In this way, we will mark all the information given in the table on the graph.
### Step 4: Joining the points
Our fourth step would be joining the points. Once all these points corresponding to available information in the table are plotted they are joined by a smooth curve to get the graph.
Here, in this case, we get a straight line that represents a uniform motion. I think now you have an idea about the basics of the graph and how to plot them.
So if you have knowledge of how distance varies with time for any object you should be able to plot a distance-time graph or position-time graph. Similarly, you can also plot the speed-time graph using this procedure. So, we can easily plot motion of object with the help of these steps.
Related Articles
### 3 thoughts on “What is graph in physics?”
1. glad to hear that it helped.
This site uses Akismet to reduce spam. Learn how your comment data is processed. |
# Perpendicular bisectors of a triangle meet
### Proof of the three perpendicular bisectors of the sides of a triangle are concurrent.
Perpendicular Bisector of a Triangle As we can see that all the perpendicular bisectors of a triangle meet at one single point which is called the circumcenter of . Perpendicular Bisectors of Triangle Meet at Point By definition of perpendicular bisector: From Triangle Side-Angle-Side Equality. The perpendicular bisector of a line segment can be constructed using a compass by drawing A triangle's three perpendicular bisectors meet (Casey , p.
And then we know that the CM is going to be equal to itself. And so this is a right angle. We have a leg, and we have a hypotenuse.
We know by the RSH postulate, we have a right angle. We have one corresponding leg that's congruent to the other corresponding leg on the other triangle. We have a hypotenuse that's congruent to the other hypotenuse, so that means that our two triangles are congruent. Well, if they're congruent, then their corresponding sides are going to be congruent. So that tells us that AM must be equal to BM because they're their corresponding sides.
So this side right over here is going to be congruent to that side. So this really is bisecting AB. So this line MC really is on the perpendicular bisector. And the whole reason why we're doing this is now we can do some interesting things with perpendicular bisectors and points that are equidistant from points and do them with triangles. So just to review, we found, hey if any point sits on a perpendicular bisector of a segment, it's equidistant from the endpoints of a segment, and we went the other way.
If any point is equidistant from the endpoints of a segment, it sits on the perpendicular bisector of that segment. So let's apply those ideas to a triangle now. So let me draw myself an arbitrary triangle. I'll try to draw it fairly large. So let's say that's a triangle of some kind. Let me give ourselves some labels to this triangle. That's point A, point B, and point C. You could call this triangle ABC.
## Bisectors of Triangles
Now, let me just construct the perpendicular bisector of segment AB. So it's going to bisect it. So this distance is going to be equal to this distance, and it's going to be perpendicular. So it looks something like that.
And it will be perpendicular. Actually, let me draw this a little different because of the way I've drawn this triangle, it's making us get close to a special case, which we will actually talk about in the next video.
Let me draw this triangle a little bit differently. This one might be a little bit better. And we'll see what special case I was referring to. So this is going to be A.
This is going to be B. This is going to be C. Now, let me take this point right over here, which is the midpoint of A and B and draw the perpendicular bisector.
So the perpendicular bisector might look something like that. And I don't want it to make it necessarily intersect in C because that's not necessarily going to be the case.
But this is going to be a degree angle, and this length is equal to that length. And let me do the same thing for segment AC right over here.
• Perpendicular Bisector of a Triangle
• Perpendicular Bisector
• Circumcenter of a triangle
Let me take its midpoint, which if I just roughly draw it, it looks like it's right over there. And then let me draw its perpendicular bisector, so it would look something like this.
So this length right over here is equal to that length, and we see that they intersect at some point. Just for fun, let's call that point O.
And now there's some interesting properties of point O. We know that since O sits on AB's perpendicular bisector, we know that the distance from O to B is going to be the same as the distance from O to A.
That's what we proved in this first little proof over here. So we know that OA is going to be equal to OB. Well, that's kind of neat. But we also know that because of the intersection of this green perpendicular bisector and this yellow perpendicular bisector, we also know because it sits on the perpendicular bisector of AC that it's equidistant from A as it is to C.
So we know that OA is equal to OC. Now, this is interesting. OA is equal to OB. So we also know that OC must be equal to OB. OC must be equal to OB. Well, if a point is equidistant from two other points that sit on either end of a segment, then that point must sit on the perpendicular bisector of that segment. That's that second proof that we did right over here.
### Perpendicular Bisector of a Triangle - Definition & Examples | [email protected]
So it must sit on the perpendicular bisector of BC. So if I draw the perpendicular bisector right over there, then this definitely lies on BC's perpendicular bisector.
And what's neat about this simple little proof that we've set up in this video is we've shown that there's a unique point in this triangle that is equidistant from all of the vertices of the triangle and it sits on the perpendicular bisectors of the three sides. Or another way to think of it, we've shown that the perpendicular bisectors, or the three sides, intersect at a unique point that is equidistant from the vertices.
And this unique point on a triangle has a special name. We call O a circumcenter. And because O is equidistant to the vertices, so this distance-- let me do this in a color I haven't used before. This distance right over here is equal to that distance right over there is equal to that distance over there. An important type of segment, ray, or line that can help us prove congruence is called an angle bisector. Understanding what angle bisectors are and how they affect triangle relationships is crucial as we continue our study of geometry.
Let's investigate different types of bisectors and the theorems that accompany them. Segment CD is the perpendicular bisector to segment AB. We derive two important theorems from the characteristics of perpendicular bisectors. We can use these theorems in our two-column geometric proofs, or we can just use them to help us in geometric computations.
Perpendicular Bisector Theorem If a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment. These theorems essentially just show that there exist a locus of points which form the perpendicular bisector that are equidistant from the endpoints of a given segment which meet at the midpoint of the segment at a right angle.
An illustration of this concept is shown below. Together, they form the perpendicular bisector of segment AB. The perpendicular bisectors of a triangle have a very special property. Let's investigate it right now. Circumcenter Theorem The perpendicular bisectors of the sides of a triangle intersect at a point called the circumcenter of the triangle, which is equidistant from the vertices of the triangle.
Point G is the circumcenter of? Angle Bisectors Now, we will study a geometric concept that will help us prove congruence between two angles. Any segment, ray, or line that divides an angle into two congruent angles is called an angle bisector. We will use the following angle bisector theorems to derive important information from relatively simple geometric figures.
### Perpendicular Bisector -- from Wolfram MathWorld
Angle Bisector Theorem If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle. If a point in the interior of an angle is equidistant from the sides of the angle, then it lies on the bisector of the angle. The points along ray AD are equidistant from either side of the angle. Together, they form a line that is the angle bisector. Similar to the perpendicular bisectors of a triangle, there is a common point at which the angle bisectors of a triangle meet.
Let's look at the corresponding theorem. Incenter Theorem The angle bisectors of a triangle intersect at a point called the incenter of the triangle, which is equidistant from the sides of the triangle. Point G is the incenter of? Summary While similar in many respects, it will be important to distinguish between perpendicular bisectors and angle bisectors.
We use perpendicular bisectors to create a right angle at the midpoint of a segment. Any point on the perpendicular bisector is equidistant from the endpoints of the given segment.
Perpendicular Bisectors and the Circumcenter |
# Integration by partial fractions with complex numbers
I learned you should use integration by partial fractions when the discriminant $D\ge0$ but are you allowed to use it in other cases?
For example take the integral $I=\int{2x-1\over x^2+1}dx$
The traditional way to solve it is: $$I=\int{d(x^2+1)\over x^2+1}-\int{dx\over x^2+1}=\ln(x^2+1)-\arctan(x)+c_1$$
With partial fractions: $${2x-1\over x^2+1}=\frac{A}{x+i}+\frac{B}{x-i}=\frac{(A+B)x+(B-A)i}{x^2+1} \\ \Leftrightarrow \left\{ \begin{array}{c} A+B=2 \\ (B-A)i=-1 \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}{c} B=2-A \\ (B-A)=i \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}{c} B={2+i\over2} \\ A={2-i\over2} \\ \end{array} \right. \\ \Leftrightarrow I=A\int{dx\over x+i}+B\int{dx\over x-i}={2-i\over2}\ln|x+i|+{2+i\over2}\ln|x-i|+c_2$$
Is this correct?
How does ${2-i\over2}\ln|x+i|+{2+i\over2}\ln|x-i|+c_2$ simplify to $\ln(x^2+1)-\arctan(x)+c_1$? I checked for $x=0$ and the constants aren't equal.
The formula $\int \frac 1 {x+i} \, dx =\ln |x+i|$ is not valid. In fact the left side is a complex valued function and the right side is real valued. You have borrowed a formula valid for real the real case and applied it to the case.
You made a mistake at the point where you said $(B-A)i=-1$. It should be $(B-A)i=1$. This implies that $B=1-i/2$ and $A=1+i/2$.
• No it should be $(B-A)i=-1$. The mistake I made was ${2x+1\over x^2+1}$ instead of ${2x-1\over x^2+1}$. It should be fixed now. – fejfo Jun 26 '18 at 10:32
• The next comment I have is that the primitive of $\frac{1}{x+i}$ is given by $\log(x+i)$ although you should think about on which branch you define the logarithm. Moreover, you should also think about on which contour you are integrating, because every time you circle around $-i$ or $+i$, the integral will increase by $2\pi i$. – Stan Tendijck Jun 26 '18 at 10:49
• I would specify that you are integrating over real positive $x$ (it is then not too hard to generalize it for all real $x$, I think) – Stan Tendijck Jun 26 '18 at 10:50 |
# What is the probability that the person picks up all three red balls?
Let's say there are $10$ orange balls and $3$ red balls in a bag.
And a person picks a ball at random and puts it on the side. So for example if he picked up an orange ball then there would be $9$ orange balls and $3$ red balls left.
What is the probability that the person picks up all three red balls in three tries?
Would be $\frac1{14}\times\frac1{13}\times\frac1{12}$
• Are there three or four red balls in the bag – Plus Twenty Jun 16 '18 at 1:03
• There are three red balls – Bas bas Jun 16 '18 at 1:04
• If they picks up 3 balls and they all happen to be red – Bas bas Jun 16 '18 at 1:05
The probability of getting all red balls would be, if he is only grabbing balls three times:
$\frac3{13}\times\frac2{12}\times\frac1{11} = \frac1{286}$
Since the balls are being removed you must subtract 1 from both the numerator and denominator as the number of red balls and total amount of balls are decreasing every time a ball is removed.
This probability is $$\frac{\text{# ways of choosing 3 red balls (order matters)}}{\text{# ways of choosing 3 balls (order matters)}} = \frac{3 \times 2 \times 1}{13 \times 12 \times 11}.$$
A proper way to denote exactly what is described in the accepted answer of Conal.
For $i=1,2,3$ let $E_i$ denote the event that the $i$-th ball picked is red.
Then you are looking for:$$P(E_1\cap E_2\cap E_3)=P(E_1)P(E_2\mid E_1)P(E_3\mid E_1\cap E_2)=\frac3{13}\frac2{12}\frac1{11}$$
Another way to solve this is to realize that the $13\choose3$ subsets of $3$ balls of the urn’s $13$ balls are equally likely outcomes of drawing three balls without replacement and these subsets comprise all possible outcomes. Only $1$ of these subsets consists of three red balls, so the probability of obtaining this subset is $1\big/{13\choose3}$.
• Mmmm, are you sure there are 11 balls? – Tony Hellmuth Jun 16 '18 at 2:48
• Oops, and thanks! I fixed it. – Steve Kass Jun 16 '18 at 18:59 |
# How to Round Numbers Up: Methods and Examples
Methods and examples for rounding numbers up with our easy-to-follow guide. From rounding to the nearest ten to rounding to the nearest hundredth, we've got you covered.
By Brian Dean- publish on march 08, 2023
When working with numbers, sometimes it is necessary to round them up to a certain degree of accuracy. Rounding numbers up can be useful in many different scenarios, including calculating budgets, estimating costs, and predicting results. However, it can be challenging to know which method of rounding up to use, and how to do it correctly. In this article, we will explore different methods of rounding up numbers and provide examples to help you understand how to apply them correctly.
## Understanding Rounding Up
Before diving into the different methods of rounding up numbers, it is essential to understand what rounding up means. Rounding up is a mathematical technique used to simplify the numbers by eliminating decimals or reducing the precision of numbers to a specific degree. Rounding up is necessary when a more straightforward number is needed for calculations or analysis.
## Different Methods of Rounding Up Numbers
There are several methods for rounding up numbers, and each method has its specific use case. The most common methods include:
## Round Up to the Nearest Integer
Rounding up to the nearest integer is the simplest method of rounding up numbers. This method involves rounding the number to the nearest whole number. If the number is a fraction, it is rounded up to the next highest whole number. For example:
2.3 rounds up to 3
4.7 rounds up to 5
8.0 remains 8
## Round Up to the Nearest Tenth
Rounding up to the nearest tenth is used when more precision is required than rounding up to the nearest integer. This method involves rounding the number to the nearest tenth decimal place. If the number is already at the tenth decimal place or higher, it remains unchanged. For example:
3.43 rounds up to 3.5
2.68 rounds up to 2.7
6.0 remains 6.0
## Round Up to the Nearest Hundredth
Rounding up to the nearest hundredth is used when even more precision is needed. This method involves rounding the number to the nearest hundredth decimal place. If the number is already at the hundredth decimal place or higher, it remains unchanged. For example:
1.234 rounds up to 1.24
5.678 rounds up to 5.68
7.0 remains 7.0
## Round Up to a Specific Decimal Place
Rounding up to a specific decimal place is used when a particular level of precision is required. This method involves rounding the number to a specific decimal place. For example:
3.14159 rounded up to the 3rd decimal place becomes 3.142
2.71828 rounded up to the 4th decimal place becomes 2.7183
6.0 rounded up to the 2nd decimal place becomes 6.00
## Tips for Rounding Up Numbers
When rounding up numbers, there are a few tips that can help ensure you are doing it correctly:
• Always be aware of the rounding rules for your specific use case
• Always round up or down consistently throughout your calculations to avoid errors
• When rounding to the nearest whole number, remember that .5 rounds up to the next highest whole number
• When rounding to a specific decimal place, make sure you know the correct place value to round up to 5
## Examples of Rounding Up Numbers
To help you better understand how to round up numbers, here are some examples:
Example 1:
Rounding up to the nearest integer
Suppose you need to round the number 4.6 up to the nearest integer. In this case, the answer is 5.
Example 2:
Rounding up to the nearest tenth
Suppose you need to round the number 7.42 up to the nearest tenth. In this case, the answer is 7.5.
Example 3:
Rounding up to the nearest hundredth
Suppose you need to round the number 9.876 up to the nearest hundredth. In this case,the answer is 9.88.
Example 4:
Rounding up to a specific decimal place
Suppose you need to round the number 3.1415926535 up to the 4th decimal place. In this case, the answer is 3.1416.
## Conclusion
Rounding up numbers can be a useful tool for simplifying calculations and making predictions. By understanding the different methods of rounding up numbers and following some tips for doing it correctly, you can avoid errors and make more accurate calculations. Remember to consider the precision needed for your specific use case and to always round up or down consistently throughout your calculations.
## FAQs
### What is the difference between rounding up and rounding down?
Rounding up involves increasing a number to the nearest whole number or decimal place, while rounding down involves decreasing it to the nearest whole number or decimal place.
### Why is it important to round up numbers?
Rounding up numbers can simplify calculations and make predictions more accurate by reducing the precision of numbers to a specific degree.
### Are there situations where rounding up should not be used?
Yes, in some situations where high levels of precision are required, rounding up may not be appropriate and may lead to errors.
### How can I determine the appropriate level of precision for rounding up numbers?
The appropriate level of precision for rounding up numbers will depend on your specific use case and the level of accuracy required.
### What should I do if I am unsure how to round up a number?
If you are unsure how to round up a number, it is always best to consult a reliable source or seek guidance from a trusted expert in the field. |
# Compound Interest
Quantitative Aptitude Questions and Answers section on “Compound Interest” with solution and explanation for competitive examinations such as CAT, MBA, SSC, Bank PO, Bank Clerical and other examinations.
1.
The compound interest on Rs. 10000 in 2 years at 4% per annum, the interest being compounded half-yearly, is:
[A]Rs. 912.86
[B]Rs. 828.82
[C]Rs. 824.32
[D]Rs. 636.80
Rs. 824.32
$A=10000\left ( 1+\frac{2}{100} \right )^{4}$
$=>10000\left ( \frac{51}{50} \right )^{4}$
$=>10000\left ( 1.08243216 \right )=10824.3216$
Since Interest = 10824.3216 – 10000 = Rs.824.32
Hence option [C] is the right answer.
2.
In what time will Rs. 1000 becomes Rs. 1331 at 10% per annum compounded annually?
[A]2 years
[B]3 years
[C]3.5 years
[D]2.5 years
3 years
Let the required time be n years, Then,
$1331=1000\left ( 1+\frac{10}{100} \right )^{n}$
$=>\frac{1331}{1000}=\left ( \frac{10+1}{10} \right )^{n}$
$=>\left ( \frac{11}{10} \right )^{n}=\left ( \frac{11}{10} \right )^{3}$
$=>n=3$
Hence option [B] is the right answer.
3.
A sum becomes Rs. 1352 in 2 years at 4% per annum compound interest. The sum is :
[A]Rs. 1250
[B]Rs. 1270
[C]Rs. 1245
[D]Rs. 1225
Rs. 1250
Let the sum be Rs. x
$\therefore 1352=x\left ( 1+\frac{4}{100} \right )^{2}$
$=>1352=x\left ( 1+\frac{1}{25} \right )^{2}$
$=>1352=x\left ( \frac{26}{25} \right )^{2}$
$=>x=\frac{1352\times 25\times 25}{26\times 26}$
$=>x=Rs. 1250$
Hence option [A] is the right answer.
4.
In how may years will Rs. 2000 amounts to Rs. 2420 at 10% per annum compound interest?
[A]2.5 years
[B]1.5 years
[C]2 years
[D]3 years
2 years
According to question,
$2420=2000\left ( 1+\frac{10}{100} \right )^{t}$
$=>\frac{2420}{2000}=\left ( \frac{11}{10} \right )^{t}$
$=>\frac{121}{100}=\left ( \frac{11}{10} \right )^{t}$
$=>\left ( \frac{11}{10} \right )^{2}=\left ( \frac{11}{10} \right )^{t}$
∴ t = 2 years
Hence option [C] is the right answer.
5.
The principal, which will amount to Rs. 270.40 in 2 years at the rate of 4% per annum compound interest, is
[A]Rs. 220
[B]Rs. 250
[C]Rs. 225
[D]Rs. 200
Rs. 250
Let the principal be Rs.P.
$\therefore 270.40=P\left ( 1+\frac{4}{100} \right )^{2}$
$=>270.40=P\left ( 1+0.04 \right )^{2}$
$=>P=\frac{270.40}{1.04\times 1.04}=Rs. 250$
Hence option [B] is the right answer.
6.
The compound interest on Rs. 2000 in 2 years if the rate of interest is 4% per annum for the first year and 3% per annum for the second year, will be
[A]Rs. 143.40
[B]Rs. 140.40
[C]Rs. 141.40
[D]Rs. 142.40
Rs. 142.40
$Amount = 2000\left ( 1+\frac{4}{100} \right )\left ( 1+\frac{3}{100} \right )$
$=>2000\times 1.04\times 1.03$
$=>Rs. 2142.40$
$\therefore CI = Rs. (2142.40 - 2000)$
$=> Rs. 142.40$
Hence option [D] is the right ansewr.
7.
At what percent per annum will Rs. 3000/- amounts to Rs. 3993/- in 3 years if the interest is compounded annually?
[A]13%
[B]10%
[C]9%
[D]11%
10%
P = Rs. 3000 , A = Rs. 3993 , n = 3 years
$A = p \left ( 1+\frac{r}{100} \right )^{n}$
$\therefore \left ( 1 + \frac{r}{100} \right )^{n} = \frac{A}{P}$
$=>\left ( 1+\frac{r}{100} \right )^{3} = \frac{3993}{3000} = \frac{1331}{1000}$
$=>\left ( 1+\frac{r}{100} \right )^{3} = \left ( \frac{11}{10} \right )^{3}$
$=>1+\frac{r}{100} = \frac{11}{10}$
$=>\frac{r}{100} = \frac{11}{10}-1$
$=> \frac{r}{100} = \frac{1}{10}$
$=> r = \frac{100}{10} = 10\%$
Hence option [B] is the right ansewr.
8.
The compound interest on Rs. 8,000 at 15% per annum for 2 years 4 months, compounded annually is :
[A]Rs. 3109
[B]Rs. 3091
[C]Rs. 2980
[D]Rs. 3100
3109 Rs.
$Amount = p\left ( 1+\frac{R}{100} \right )^{t}$
$= 8000\left ( 1+\frac{15}{100} \right )^{2\frac{1}{3}}$
$= 8000\left ( 1+\frac{3}{20} \right )^{2}\left ( 1+\frac{3}{20\times 3} \right )$
$= 8000\times \frac{23}{20}\times\frac{23}{20}\times\frac{21}{20}$
$= Rs. 11109$
∴ Compound Interest = Rs. (11109 – 8000) = Rs. 3109
Hence option [A] is the right ansewr.
9.
A sum of money on compound interest amounts to Rs. 10648 in 3 years and Rs. 9680 in 2 years. The rate of interest per annum is :
[A]20%
[B]5%
[C]10%
[D]15%
10%
Let the sum be Rs. P and rate of interest be R% per annum. Then,
$P\left ( 1+\frac{R}{100} \right )^{2} = 9680 .........(1)$
$P\left ( 1+\frac{R}{100} \right )^{3} = 10648 ......(2)$
On dividing equation (2) by (1) :
$1+\frac{R}{100} = \frac{10648}{9680}$
$=> \frac{R}{100} = \frac{10648}{9680} - 1$
$=> \frac{R}{100} =\frac{10648 - 9680}{9680}$
$=> \frac{R}{100} = \frac{968}{9680} = \frac{1}{10}$
$=> R = \frac{1}{10}\times 100 = 10\%$
Hence option [C] is the right ansewr.
10.
At what rate per cent per annum will Rs. 2304 amount to Rs. 2500 in 2 years at compound interest?
[A]$4\frac{1}{3}\%$
[B]$4\frac{1}{2}\%$
[C]$4\frac{1}{6}\%$
[D]$4\frac{1}{5}\%$
$\mathbf{4\frac{1}{6}\%}$
Let the rate per cent per annum be r. Then,
$2500 = 2304 \left ( 1 + \frac{r}{100} \right )^{2}$
$=> \left ( 1 + \frac{r}{100} \right )^{2} = \frac{2500}{2304} = \left ( \frac{50}{48} \right )^{2}$
$=> 1+\frac{r}{100} = \frac{50}{48} = \frac{25}{24}$
$=> \frac{r}{100} = \frac{25}{24} - 1 = \frac{1}{24}$
$=> r = \frac{100}{24} = \frac{25}{6} = 4\frac{1}{6}\%$
Hence option [C] is the right ansewr. |
# Converting Roman Numerals to Standard Numbers
Page content
## Prior Learning
Before students begin this lesson, they should be taught the rules of Roman numerals as instructed in the previous lesson plan. Once the students understand which each Roman numerals means and understands the rules that accompany them, explain the method below to your students. Then they too can begin converting Roman numerals to standard numbers.
Reading and writing Roman numerals means understanding what each symbol represents as well as what math function should be performed. Look at the Roman numeral below.
MDCXVI
When you look at this number, the first thing to look for is to see if there are any digits representing smaller digits on the left side of the larger digits. The answer to this question will tell you what math function to perform. In this case, the digits descend from highest to lowest where M=1000, D=500, C=100, X=10, V=5 and I=1. This tells you that you will add the numbers together, which is something you will do as the last step in deciphering every Roman numeral. The number is 1616 by the formula of 1000+500+100+10+5+1
Have students practice using the following numbers.
• CVII (107)
• MDX (1510)
• MC (1100)
Sometimes you will have to subtract before you can add when converting Roman numerals to standard numbers. Use the number below as an example.
MCMXCIV
There is a C before the M, an X before the C and a I before the V. Before taking the last step in converting the Roman numerals by adding them, you must first subtract the smaller numbers on the left from the larger ones on the right. The formula for figuring out this Roman numeral would look like this; 1000 + (1000-100) + (100- 10) + (5-1) because the way the Roman numeral is written is actually the formula listed after this sentence. M + (M-C) + (C-X) + (V-I) We simply translated it! The answer to the formula is 1994.
Have the students convert the Roman numerals below.
• MM (2000)
• MCMVI (1906)
• XXIII (23)
• LVI (56)
• DIII (503)
• MCDIV (1404)
Now that students have learned about converting Roman numerals to standard numbers, they are ready to move on to the next lesson plan in this series which will teach them how to write Roman numerals.
## This post is part of the series: Roman Numerals Today
This series is dedicated to learning about Roman Numerals. They are more than just pretty markings on a clock! |
# Lesson Worksheet: Introduction to Hypothesis Testing Mathematics
In this worksheet, we will practice using hypothesis testing to assess claims about population parameters.
Q1:
Suppose the average SAT score of graduating seniors is greater than 1,100. What are the null and alternative hypotheses?
• A,
• B,
• C,
• D,
• E,
Q2:
A package of gum claims that the flavor lasts more than 39 minutes. What would the null hypothesis of a test to determine the validity of the claim be?
• A
• B
• C
• D
• E
What sort of test is this?
• ARight-tailed test
• BTwo-sided test
• CLeft-tailed test
Q3:
Which of the following statements is not true for hypothesis testing?
• AThe null hypothesis must contain the equal symbol.
• BType II error is the probability of accepting false null hypothesis given that the alternative hypothesis is true.
• CType I error is the probability of rejecting true null hypothesis.
• DType II error is the probability of accepting false null hypothesis.
Q4:
Some teachers at a school say that high-school juniors use computers for an average of 3.2 hours per day. The school principal wants to test if this is true. What are the null and alternative hypotheses?
• A,
• B,
• C,
• D,
• E,
Q5:
The school nurse thinks the average height of 7th graders has increased. The average height of a 7th grader 5 years ago was 145 cm with a standard deviation of 20 cm. She takes a random sample of 200 students and finds that the average height of her sample is 147 cm. What are the null and alternative hypotheses?
• A,
• B,
• C,
• D,
• E,
Conduct a single-tailed hypothesis test using a .05-significance level to test the null and alternative hypotheses. What is your conclusion?
• AThere is insufficient evidence to reject the alternative hypothesis at the given significance level.
• BThere is insufficient evidence to reject the null hypothesis at the given significance level.
• CThere is sufficient evidence to reject the null hypothesis at the given significance level.
Are 7th graders now taller than they were before?
• AYes
• BNo
Q6:
A manufacturer specifies that the mean lifetime of a certain type of batteries is at least 273 hours. A sample of 49 batteries has an average lifetime of 270.5 hours and a standard deviation of 9 hours.
State the null and the alternative hypotheses.
• A and
• B and
• C and
• D and
• E and
Test the manufacturerβs claim at the significance level of 0.05.
• AThe claim fails to be rejected.
• BThe claim is rejected.
Q7:
True or False: Significant level is the probability of rejecting when is true.
• AFalse
• BTrue
Q8:
The water temperature in a storeroom is normally distributed. The mean temperature should not exceed . The standard deviation of water temperature is . Measurements on 9 randomly selected days produce a mean of . Should the water temperature be regarded as acceptable with a significant level of 0.05?
• AYes
• BNo
Q9:
A rod used in a machine application must have a diameter of 1.2 mm. A random sample of 25 rods produces a mean diameter of 1.194 cm. The diameter is known to be normally distributed with a standard deviation of 0.03 cm. Test this hypothesis with a significant level of 0.02.
• A
• B
Q10:
A machine produces rods used in car engines. A random sample of 25 rods is selected and their diameters are measured. The resulting mean and standard deviation are 7.212 and 0.005 respectively. Test the hypotheses and at a significant level of 0.01.
• A
• B |
# Frank Solutions for Chapter 20 Mensuration II Class 10 ICSE Mathematics
### Exercise 20.1
1. Find the curved surface area, the total surface area and the volume of a cone if its:
(i) Height = 12 cm, radius = 5 cm
(ii) Height = 15 cm, radius = 8 cm
(iii) Height = 16 cm, diameter = 24 cm
(iv) Height = 8 cm, diameter = 12 cm
(i) From the question it is given that,
Height = 12 cm, radius = 5 cm
We know that, curved surface area = (Ï€r√(h2 + r2))
= (22/7) × 5 × √(122 + 52)
= (22/7) × 5 × √(169)
= (22/7) × 5 × 13
= 204.29 cm2
Then, total surface area = area of circular base + curved surface area
= Ï€r2 + (Ï€r√(h2 + r2))
= (22/7) × 52 + 204.29
= 78.57 + 204.29
Therefore, total surface area = 282.86 cm2
Volume = 1/3 × (Ï€r2) × h
= 1/3 × ((22/7) × 52) × 12
= 314.29 cm3
(ii) From the question it is given that,
Height = 15 cm, radius = 8 cm
We know that, curved surface area = (Ï€r√(h2 + r2))
= (22/7) × 8 × √(152 + 82)
= (22/7) × 8 × √(289)
= (22/7) × 8 × 17
= 427.43 cm2
Then, total surface area = area of circular base + curved surface area
= Ï€r2 + (Ï€r√(h2 + r2))
= (22/7) × 82 + 427.43
= 201.14 + 427.43
Therefore, total surface area = 628.57 cm2
Volume = 1/3 × (Ï€r2) × h
= 1/3 × ((22/7) × 82) × 15
= 1005.71 cm3
(iii) From the question it is given that,
Height = 16 cm, diameter = 24 cm
= 24/2
= 12 cm
We know that, curved surface area = (Ï€r√(h2 + r2))
= (22/7) × 12 × √(162 +122)
= (22/7) × 12 × √(400)
= (22/7) × 12 × 20
= 754.29 cm2
Then, total surface area = area of circular base + curved surface area
= Ï€r2 + (Ï€r√(h2 + r2))
= (22/7) × 122 + 754.29
= 452.57 + 754.29
Therefore, total surface area = 1206.86 cm2
Volume = 1/3 × (Ï€r2) × h
= 1/3 × ((22/7) × 122) × 16
= 2413.71 cm3
(iv) From the question it is given that,
Height = 8 cm, diameter = 12 cm
= 12/2
= 6 cm
We know that, curved surface area = (Ï€r√(h2 + r2))
= (22/7) × 6 × √(82 +62)
= (22/7) × 6 × √(100)
= (22/7) × 6 × 10
= 188.57 cm2
Then, total surface area = area of circular base + curved surface area
= Ï€r2 + (Ï€r√(h2 + r2))
= (22/7) × 62 + 188.57
= 113.14 + 188.57
Therefore, total surface area = 301.71 cm2
Volume = 1/3 × (Ï€r2) × h
= 1/3 × ((22/7) × 62) × 8
= 301.71 cm3
2. Find the radius of the circular base of the cone, if its volume is 154 cm3and the perpendicular height is 12 cm.
From the question it is given that,
Volume of circular based cone = 154 cm3
Perpendicular height of the circular based cone = 12 cm
We know that, volume of the cone = 1/3 × (Ï€r2) × h
So,
1/3 × (Ï€r2) × h = 154
⇒ 1/3 × ((22/7) × r2) × 12 = 154
⇒ r2 = (154 × 3 × 7)/(12 × 22)
⇒ r2 = 12.25
⇒ r = √12.25
⇒ r = 3.5 cm
Therefore, radius of the circular base of the cone is 3.5 cm.
3. Find the volume of the right circular cone whose height is 12 cm and slant height is 15 cm. (Ï€ = 3.14)
From the question it is given that,
Height of the right circular cone, h = 12 cm
Slant height of the right circular cone, l = 15 cm
Let us assume radius of the base = r
So, we know that l2 = h2 + r2
r2 = l2 – h2
⇒ r2 = 152 – 122
⇒ r2 = 225 – 144
⇒ r2 = 81
⇒ r = √81
⇒ r = 9 cm
Then, volume = 1/3 × (Ï€r2) × h
= 1/3 × (3.14 × 92) × 12
= 1017.36 cm3
Therefore, volume of the cone is 1017.36 cm3.
4. Find the curved surface area of a cone whose height is 8 cm and base diameter is 12 cm.
From the question it is given that,
Height of the cone = 8 cm
Base diameter of the cone = 12 cm
We know that, curved surface area = (Ï€r√(h2 + r2))
= (22/7) × 6 × √(82 +62)
= (22/7) × 6 × √(100)
= (22/7) × 6 × 10
= 188.57 cm2
Therefore, curved surface area cone is 188.57 cm2.
5. The diameter of a right circular is 12 m and the slant height is 10 m. Find its curved surface area and the total surface area.
From the question it is given that,
Diameter of a right circular = 12 m
Slant height of a right circular = 6 m
We know that, curved surface area = circumference of the base × height
= 2Ï€r × h
= 2 × (22/7) × 6 × 10
= 377.14 m2
Therefore, area of curved surface = 377.14 m2
Then, total surface area = curved surface area + (2 × base area)
= 2Ï€rh + 2Ï€r2
= 2Ï€r (h + r)
= 2 × (22/7) × 6 × (10 + 6)
= 2 × (22/7) × 6 × 16
= 603.42
Therefore, total surface area = 603.42 m2
6. The heights of two cones are in the ratio 1: 3 and their base radii are in the ratio 3: 1. Find the ratio of their volumes.
Let us assume radius of the first cone be 3r and height be h, then radius of second cone will be r and height will be 3h.
We know that, volume of cone = 1/3 × (Ï€r2) × h
Ratio of volumes of cone = (volume of first cone)/(Volume of second cone)
= [1/3 × (Ï€(3r)2) × h]/[1/3 × (Ï€r)2 × 3h]
= ((1/3) π9r2h)/((1/3) πr23h)
= 3
Therefore, ratio of volume of cone = 3 : 1.
7. The base circumference of two cones are the same. If their slant heights are in the ratio 5: 4, find the ratio of their curved surface areas.
From the question it is given that,
Ratio of slant heights of two cones = 5: 4
Let us assume radius be r,
Let us assume slant height of first cone = 5x
Slant height of second cone = 4x
Then, curved surface area of cone = πrl
So, ratio of curved surface areas = (Ï€r × 5x)/(Ï€r × 4x)
= 5/4
Therefore ratio of curved surface areas = 5 : 4.
8. Find the height of the cone whose base radius is 5 cm and volume is 75Ï€ cm3.
From the question it is given that,
Base radius of the cone = 5 cm
Volume of cone = 75Ï€ cm3
We know that,
Volume of cone = 1/3 × (Ï€r2) × h
⇒ 75Ï€ = 1/3 × (Ï€ × 52) × h
⇒ h = 225/25
⇒ h = 9 cm
Therefore, height of the cone is 9 cm.
9. The curved surface area of a right circular cone of radius 11.3 cm is 710 cm2. What is the slant height of the cone?
From the question it is given that,
Radius of right circular cone, r = 11.3 cm
Curved surface area = 710 cm2
Let us assume slant height be l,
Then, πrl = 710
⇒ 22/7 × 11.3 × l = 710
By cross multiplication we get,
l = (710 × 7)/(11.3 × 22)
⇒ l = 19.99 cm
Therefore, slant height of cone is approximately equal to 20 cm.
10. A conical tent requires 264 mof canvas. If the slant height is 12 m, find the vertical height of the cone.
From the question it is given that,
Curved surface area of the tent = 264 m2
Slant height of tent, l = 12 m
Then, πrl = 264
⇒ 22/7 × r × 12 = 264
By cross multiplication we get,
r = (264 × 7)/(22 × 12)
⇒ r = 7 m
Let us assume h be the vertical height,
Then,
l2 = r2 + h2
⇒ h = √(l2 – r2)
⇒ h = √(122 – 72)
⇒ h = √(144 – 49)
⇒ h = √(95)
⇒ h = 9.75 m
Therefore, vertical height of cone = 9.75 m.
11. A conical tent with a capacity of 600 m3stands on a circular base of area 160 m2. Find in m2the area of the canvas.
From the question it is given that,
Volume of the conical tent = 600 m3
Area of circular base cone = 160 m2
Ï€r2 = 160
By cross multiplication we get,
r2 = 160/Ï€
⇒ r = √(160 × 7)/22
⇒ r = √50.909
⇒ r = 7.134 m
We know that,
Volume of cone = 1/3 × (Ï€r2) × h = 600
⇒ 1/3 × (Ï€ × 7.1342) × h = 600
⇒ h = (600 × 7 × 3)/(7.1342 × 22)
⇒ h = 11.265 m
Then, we know that slant height, l = √(r2 + h2)
l = √(7.1342 + 11.2652)
⇒ l = √(177.624)
⇒ l = 13.327 m
Therefore, curved surface area = πrl
= 22/7 × 7.134 × 13.327
= 298.9 m2
Therefore, the area of the canvas is 298.9 m2
12. A hallow metallic cylindrical tube has an internal radius of 3.5 cm and height 21 cm. The thickness of the metal tube is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place.
From the question it is given that,
Internal radius of hallow metallic cylindrical tube, r = 3.5 cm
Height of hallow metallic cylindrical tube, h = 21 cm
Thickness of the metal = 0.5 cm
Then, outer radius of the tube, R = (3.5 + 0.5)cm = 4 cm
We know that volume of hallow metallic cylindrical tube = Ï€h(R2 – r2)
= (22/7) × 21 × (42 – 3.52)
= (22/7) × 21 × (16 – 12.25)
= (22/7) × 21 × 3.75
= 247.5 cm3
Therefore, volume of metal used = 247.5 cm3
Tube is melted and cast into a right circular cone of height 7 cm.
Now, let us assume r1 be the radius of right circular cone,
So, Volume = 1/3Ï€r12h = 247.5
1/3 × 22/7 × r12 × 7 = 247.5
By cross multiplication we get,
r12 = (247.5 × 3 × 7)/(22 × 7)
⇒ r12 = 33.75
⇒ r1 = √(33.75)
⇒ r1 = 5.8 cm
Therefore, radius of the cone is 5.8 cm.
13. A canvas tent is in the shape of a cylinder surrounded by a conical roof. The common diameter of the cone and the cylinder is 14 m. The height of the cylindrical part is 8 m and the height of the conical roof is 4 m. Find the area of the canvas used to make the tent.
From the question it is give that,
The common diameter of the cone and the cylinder = 14 m
The common radius of the cone and the cylinder = 14/2 = 7 m
Height of the cylindrical part, H = 8 m
Height of the conical part, h = 4 m
We know that, slant height of the cone l = √(r2 + h2)
l = √(72 + 42)
⇒ l = √65
⇒ l = 8.06 m
Then,
Area of the canvas used = curved surface area of cylinder + curved surface area of cone
= 2πrH + πrl
= (2 × 22/7 × 7 × 8) + (22/7 × 7 × 8.06)
= 352 + 177.32
= 529.32 m2
14. A circus tent is cylindrical to a height of 5 m and conical above it. If its diameter is 42 m and slant height of the cone is 53 m, calculate the total area of the canvas required.
From the question it is given that,
Height of circle tent, h = 5m
Diameter of circular tent = 42 m
Radius of circular tent = 42/2 = 21 m
Then,
Area of the canvas used = curved surface area of cylinder + curved surface area of cone
= 2πrh + πrl
= (2 × 22/7 × 21 × 5) + (22/7 × 21 × 53)
= 660 + 3498
= 4158 m2
15. Sand from a cylindrical bucket 32 cm in height and 18 cm in radius is poured onto the ground making a conical heap 24 cm high. Find the radius of the conical heap.
From the question it is given that,
Height of the cylinder, h1 = 32 cm
Radius of bucket, r1 = 18 cm
Height of the conical heap, h2 = 24 cm
Let us assume radius of conical heap be r2,
Then,
Volume of sand in the bucket = volume of sand in conical heap
Ï€ × r12 × h1 = 1/3 × Ï€ × r22 × h2
⇒ 18 × 18 × 32 = 1/3 × r2 × 24
⇒ r22 = (10368 × 3)/24
⇒ r22 = 1296
⇒ r2 = √(1296)
⇒ r2 = 36 cm
Therefore, radius of the conical heap is 36 cm.
### Exercise 20.2
1. Find the volume and the surface area of the spheres in the following :
(iii) Diameter = 6.3 cm
2. Find the diameter of the sphere in each of the following :
(i) Volume = 523.17/21 cm3
(ii) Volume = 72Ï€ cm3
(iii) Surface Area = 221. 76 cm2
(iv) Surface area = 576 π cm2
4. Find the radius of the sphere whose surface area is equal to its volume.
5. Find the radius of a sphere whose surface area is equal to the area of the circle of diameter 2.8 cm.
6. Find the length of the wire of diameter 4 m that can be drawn from a solid sphere of radius 9 m.
Radius of solid sphere = 9 m
7. The radius of a sphere is 9 cm. It is melted and drawn into aa wire of diameter 2mm. Find the length of the wire in metre.
8. A sphere and a cone have the same radii. If their volumes are also eqyual, prove that the height of the cone is twice its radius.
Let r be the radii of sphere and cone.
9. The radius and height of a cylinder, a cone and a sphere are same. Calculate the ratio of their volumes.
Let r, h be the radius and height of cylinder, Cone and sphere.
10. A cylindrical beaker of 7 cm diameter is a partly filled with water. Determine the number of spherical marbles of diameter 1.4 cm that are to be submerged in it to raise the water level by 5.6 cm.
11. A cylindrical bucket, whose base is 20 cm, is filled with water to a height of 25 cm. A heavy iron spherical ball of radius 10 cm is dropped to submerge completely in water in the bucket. Find the increase in the level in water.
12. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the top which is open is 5 cm. It is filled with water. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the flower out. Find the number of leadshots dropped in the water.
13. Find the total surface area and volume of a hemisphere whose radius is 10 cm.
14. Find the cost of painting a hemispherical dome of diameter 10 m at the rat of Rs 1.40 per square metre.
15. A solid sphere metal is cut through its centre into 2 equal parts. If the diameter of the sphere is 3.1/3 cm, find the total surface of each part, correct to two decimal places.
16. A circular hall, surmounted by a hemispherical roof, contains 5236 m3 of air. If the internal diameter of the room is equal to the height of the highest point of the roof from the floor, find the height of the hall.
17. Find the volume of the hollow sphere whose inner diameter is 8 cm and the thickness of the material of which it is made is 1 cm.
18. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm, Find the height of the cone.
19. A hollow metallic sphere is 2 cm thick all around and has an external diameter of 12 cm. Find the radius of the solid sphere made by recasting this hollow sphere.
20. A buoy is made in the form of hemisphere surmounted by a right circular cone whose circular base coincides with the plate surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third the volume of hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places. |
# Probability in Terms of Odds for and against the Event
Odds for and against an event represent a ratio of the desired outcomes versus the field. In other words, the odds for an event are the ratio of the number of ways the event can occur to the number of ways the event does not occur. Thus:
Given the probability of an event ‘E’ i.e. P(E),
\begin{align*} \text{Odds for E} & =\cfrac {P(E)}{ \left\{ 1 – P(E) \right\} } \\ \text{Odds against E} & = \cfrac { \left\{ 1 – P(E) \right\} }{ P(E) } \\ \end{align*}
### Example: Odds for and against an event
A box contains five blue balls, two green balls, and six yellow balls. What are the odds of drawing a blue ball from the box?
First, we have to establish the probability of drawing a blue ball:
Let P(B) represent the event that a blue ball is drawn from the box. Thus,
$$P(B) = \cfrac {5}{13}$$
\begin{align*} \text{The odds for a blue ball} & =\cfrac {5}{13} ÷ \left(1 – \cfrac {5}{13} \right)\\ & = \cfrac {5}{13} ÷ \left( \cfrac {8}{13} \right) \\ & =\cfrac {5}{13} * \cfrac {13}{8} \\ & =\cfrac {5}{8} \\ \end{align*}
Therefore, the odds for a blue ball are 5:8 (pronounced as ‘5 to 8’).
Similarly, we can calculate the odds against drawing a blue ball:
\begin{align*} \text{Odds against a blue ball} &= \left\{1 – \cfrac{5}{13} \right\} ÷ \cfrac{5}{13} \\ &= \cfrac{8}{13} ÷ \cfrac{5}{13} \\ &= \cfrac{8}{13} * \cfrac{13}{5} \\ & = \cfrac{8}{5} \end{align*}
Therefore, the odds against are 8:5 (pronounced as ‘8 to 5’).
You should notice that the odds against an event = reciprocal of odds for the same event.
## Question
Suppose you toss a fair coin. What are the odds against obtaining a head?
A. 2:2
B. 1:2
C. 1:1
Solution
The correct answer is C.
The probability of obtaining a head, P(H) = 1/2
Thus,
\begin{align*} \text{Odds against a head} & = \left\{ 1 – P(H) \right\} ÷ P(H) \\ & = \left\{ 1 – \cfrac{1}{2} \right\} ÷ \cfrac{1}{2} \\ & = \cfrac{1}{2} * \cfrac{2}{1} \\ & = \cfrac{1}{1} \end{align*}
Therefore, the odds against a head are 1:1, pronounced as ‘1 to 1.’
Note that in this case, the odds for a head are also 1:1.
Reading 8 LOS 3c: describe the probability of an event in terms of odds for and against the event;
Shop CFA® Exam Prep
Offered by AnalystPrep
Featured Shop FRM® Exam Prep Learn with Us
Subscribe to our newsletter and keep up with the latest and greatest tips for success
Shop Actuarial Exams Prep Shop Graduate Admission Exam Prep
Sergio Torrico
2021-07-23
Excelente para el FRM 2 Escribo esta revisión en español para los hispanohablantes, soy de Bolivia, y utilicé AnalystPrep para dudas y consultas sobre mi preparación para el FRM nivel 2 (lo tomé una sola vez y aprobé muy bien), siempre tuve un soporte claro, directo y rápido, el material sale rápido cuando hay cambios en el temario de GARP, y los ejercicios y exámenes son muy útiles para practicar.
diana
2021-07-17
So helpful. I have been using the videos to prepare for the CFA Level II exam. The videos signpost the reading contents, explain the concepts and provide additional context for specific concepts. The fun light-hearted analogies are also a welcome break to some very dry content. I usually watch the videos before going into more in-depth reading and they are a good way to avoid being overwhelmed by the sheer volume of content when you look at the readings.
Kriti Dhawan
2021-07-16
A great curriculum provider. James sir explains the concept so well that rather than memorising it, you tend to intuitively understand and absorb them. Thank you ! Grateful I saw this at the right time for my CFA prep.
nikhil kumar
2021-06-28
Very well explained and gives a great insight about topics in a very short time. Glad to have found Professor Forjan's lectures.
Marwan
2021-06-22
Great support throughout the course by the team, did not feel neglected
Benjamin anonymous
2021-05-10
I loved using AnalystPrep for FRM. QBank is huge, videos are great. Would recommend to a friend
Daniel Glyn
2021-03-24
I have finished my FRM1 thanks to AnalystPrep. And now using AnalystPrep for my FRM2 preparation. Professor Forjan is brilliant. He gives such good explanations and analogies. And more than anything makes learning fun. A big thank you to Analystprep and Professor Forjan. 5 stars all the way!
michael walshe
2021-03-18
Professor James' videos are excellent for understanding the underlying theories behind financial engineering / financial analysis. The AnalystPrep videos were better than any of the others that I searched through on YouTube for providing a clear explanation of some concepts, such as Portfolio theory, CAPM, and Arbitrage Pricing theory. Watching these cleared up many of the unclarities I had in my head. Highly recommended. |
# Difference between revisions of "2018 AMC 12A Problems/Problem 19"
## Problem
Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots$$of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}$
## Solution
It's just $$\sum_{a=0}^\infty\frac1{2^a}\sum_{b=0}^\infty\frac1{3^b}\sum_{c=0}^\infty\frac{1}{5^c} =\sum_{a=0}^\infty\sum_{b=0}^\infty\sum_{c=0}^\infty\frac1{2^a3^b5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.$$ since this represents all the numbers in the denominator. (athens2016)
## Solution 2
Separate into 7 separate infinite series's so we can calculate each and find the original sum:
The first infinite sequence shall be all the reciprocals of the powers of $2$, the second shall be reciprocals of the powers of $3$, and the third will consist of reciprocals of the powers of 5. We can easily calculate these to be $1, \frac{1}{2}, \frac{1}{4}$ respectively.
The fourth infinite series shall be all real numbers in the form $\frac{1}{2^a3^b}$, where $a$ and $b$ are greater than or equal to 1.
The fifth is all real numbers in the form $\frac{1}{2^a5^b}$, where $a$ and $b$ are greater than or equal to 1.
The sixth is all real numbers in the form $\frac{1}{3^a5^b}$, where $a$ and $b$ are greater than or equal to 1.
The seventh infinite series is all real numbers in the form $\frac{1}{2^a3^b5^c}$, where $a$ and $b$ and $c$ are greater than or equal to 1.
Let us denote the first sequence as $a_{1}$, the second as $a_{2}$, etc. We know $a_{1}=1$, $a_{2}=\frac{1}{2}$, $a_{3}=\frac{1}{4}$, let us find $a_{4}$. factoring out $\frac{1}{6}$ from the terms in this subsequence, we would get $a_{4}=\frac{1}{6}(1+a_{1}+a_{2}+a_{4})$.
Knowing $a_{1}$ and $a_{2}$, we can substitute and solve for $a_{4}$, and we get $\frac{1}{2}$. If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them $\frac{1}{4}$ and $\frac{1}{8}$.
Finally, for the seventh sequence, we see $a_{7}=\frac{a_{8}}{30}$, where $a_{8}$ is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}$, but when we separated the sequence into its parts, we ignored the $1/1$, so adding in the $1$, we get $1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}$, which when we solve for, we get $\frac{29}{8}=\frac{29a_{8}}{30}$, $\frac{1}{8}=\frac{a_{8}}{30}$, $\frac{30}{8}=(a_{8})$, $\frac{15}{4}=(a_{8})$. So our answer is $\frac{15}{4}$, but we are asked to add the numerator and denominator, which sums up to $19$, which is the answer.
~~Edited by mprincess0229~~
## Solution 3
Clearly this is just summing over the reciprocals of the numbers of the form $2^i3^j5^k$, where $i,j,k\in [0,\infty)$. SO our desired sum is $\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}$. By the infinite geometric series formula, $\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}$ is just $\frac{\frac{1}{3^j5^k}}{1-\frac{1}{2}}=\frac{2}{3^j5^k}$. Applying the infinite geometric series formula again gives that $\sum_{j=0}^{\infty}\frac{2}{3^j5^k}=\frac{\frac{2}{5^k}}{1-\frac{1}{3}}=\frac{3}{5^k}$. Applying the infinite geometric series formula again yields $\sum_{k=0}^{\infty}\frac{3}{5^k}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}$. Hence our final answer is $15+4=\boxed{19}$.
-vsamc |
• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon
Suggested languages for you:
Americas
Europe
Q12E
Expert-verified
Found in: Page 39
### Linear Algebra With Applications
Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974
# There exists a${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrix such that ${\mathbit{A}}\left[\begin{array}{c}1\\ 2\end{array}\right]{\mathbf{=}}\left[\begin{array}{c}3\\ 4\end{array}\right]$
True, there exist a$2×2$ matrix A such that role="math" localid="1659642050851" $A\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$
See the step by step solution
## Step 1:Taking the matrix
Suppose the matrix A is.
$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Now according to the given equation,
$A\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$
## Step 2: Justification of answer
Put the value of A matrix in the above equation we get.
$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$
After solving the equation we get
$\left[\begin{array}{c}a+2b\\ c+2d\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$
Now on equating the value of matrix we get
$a+2b=3\phantom{\rule{0ex}{0ex}}c+2d=4\phantom{\rule{0ex}{0ex}}$
On solving the equations we get the value of variables
$a=3-2b\phantom{\rule{0ex}{0ex}}c=4-2d\phantom{\rule{0ex}{0ex}}$
Where, are free variables which can have any value. For instance let the value of $b,d=1$
It implies $a=1,c=2$
Thus, we get the matrix
$A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$
Hence, it is truethat there exist a $2×2$matrix A such that $A\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$ |
# Calculator circle radio, formulas, examples and definitions
To calculate the radius of a circle, three different formulas can be used, depending on the Area, the diameter or the length of the circumference.
But do not worry, with this calculator you can know the radio automatically, if you want more details we also have the formulas and some examples.
## Formulas to calculate the radius of a circle:
1. Formula to calculate the radius from the area. The following formula is used to calculate the radius, when you have the area of the circle:Where:
• A: Area.
• π: Constant Number equivalent to 3.1415
2. Formula for calculating the radius when you have the diameter. When we have the diameter we can calculate the radius by means of the following conversion formula:Where:
• D: Diameter.
3. Formula to perform the calculation of the radius from the length of the circumference. If for some reason you only have the length of the circumference, with this data and the following formula you can find the radius:
Where:
• LC: Length of the circumference.
• π: Constant Number equivalent to 3.1415
## Steps to calculate the radius of a circle:
The above formulas can be used to calculate the radius easily and quickly, the formula you use depends on the data you have (Diameter, area or circumference length):
### – How is the radius of a circle calculated when you have the area ?:
Step 1. Once you have the value of the area you should divide it between the constant pi (π), giving you as a result a number. Example: you have an area of 34, divide between (π) and the result will be: 10,822
Step 2. Then you should take the number from step 1 and take the square root as follows: √10,822 = 3,289.
### – How to calculate the radius of a circle if you have the diameter ?:
Step 1. This is the simplest formula, because it only requires one step, because it is as simple as dividing in diameter between 2. Example: If you have a diameter of 23, you only have to divide 23 between two and the result will be the radius : 23/2 = 11.5
### – How is the radius of a circle calculated if I have the length of the circumference ?:
Step 1. Multiply first 2xπ, the result will always be: 6,283
Step 2. In this case it is essential to have the length of the circumference. Because you must divide the length of the circumference between the value of step 1. Example: You have a length of the circumference of 42, this value is divided as follows: 42 / 6,283 = 6,684 |
Associated Topics || Dr. Math Home || Search Dr. Math
### Finding Pythagorean Triplets
```Date: 02/02/2006 at 17:07:24
From: Faizan
Subject: Pythagorean triplets
Without using the standard a = n^2 - m^2, b = 2nm, c = n^2 + m^2, how
can you work out primitive Pythagorean triplets?
For example, you can work out an equation for cases when c and b
differ by 1 using (2n+1)^2 + b^2 = (b+1)^2.
I cannot work it out when c and b differ by 2. How would you do it?
```
```
Date: 02/04/2006 at 14:21:28
From: Doctor Ricky
Subject: Re: Pythagorean triplets
Hey Faizan,
Thanks for writing Dr. Math!
We should be able to find a Pythagorean triplet where the hypotenuse
and a leg differ by 2. Think of the most basic triplet - 3, 4, 5 -
and note that 5 - 3 = 2.
Assume we are trying to find a triplet where the hypotenuse and a
leg differ by 1.
You had the right idea with your work, but let's look more closely at
the steps that lead to your answer. We would begin by assuming that
one leg and the hypotenuse differ by 1, or in other words:
b (leg) = n
c (hypotenuse) = n + 1
So, using the Pythagorean theorem and plugging in these assumptions,
we get:
a^2 + (n^2) = (n+1)^2. Simplifying, we get:
a^2 + n^2 = n^2 + 2n + 1, and again we simplify to get:
a^2 = 2n + 1
So in other words,
a = sqrt(2n + 1)
Therefore, for any integer value of n that makes sqrt(2n + 1) an
integer, we have a Pythagorean triplet where:
a = sqrt(2n + 1)
b = n
c = n + 1
For example, if n = 4, we get the triplet 3, 4, 5.
If n = 12, we get the triplet 5, 12, 13.
We can keep going with this for all integer values of a.
So now let's attack the situation of when a leg and the hypotenuse
differ by 3 and hopefully you will be able to see how to prove it for
a difference of 2.
So we'll assume:
b (leg) = n
c (hypotenuse) = n + 3
Using the Pythagorean theorem, we get:
a^2 + (n^2) = (n+3)^2
Simplifying, we get:
a^2 + n^2 = n^2 + 6n + 9
Simplifying again, we get:
a^2 = 6n + 9
and
a = sqrt(6n + 9)
Therefore, for any integer value of n that makes sqrt(6n + 9) an
integer, we have a Pythagorean triplet where:
a = sqrt(6n + 9)
b = n
c = n + 3
For example, if we choose n = 12, we get the triplet: 9, 12, 15.
If n = 36, we get the triplet: 15, 36, 39.
Finding values for n that make this work won't be very easy as we
continue to increase the difference between a leg and the hypotenuse,
which is why the general formulas you mentioned in your questions are
great guides for creating Pythagorean triplets.
If you have any more questions or if you were confused about any of
this, please let me know!
- Doctor Ricky, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search |
# Square Root of 9
The square root of 9 is equal to 3. It is represented as √9 in radical form and 91/2 in exponential form. When we multiply a number by itself we get the square of the number. The square root of any number is a value that is multiplied by itself to get the original value. So, the square root of 9, denotes to reverse of the square of the number 9. If x is the square root of 9 then x2 = 9.
## What is Square root of 9?
Square root of 9 is a whole number that is multiplied by itself to get the original number. The actual value of square root of 9 is 3.
√9 = ±3
If we square +3 or -3, the resulting value will be equal to 9.
• (+3)2 = 9
• (-3)2 = 9
Note:
Since, the square root of 9 results in a whole number, therefore, 9 is a perfect square.
Also, the root of 9 is a rational number, because we can represent √9 = ±3 in the form P/Q, such as 3/1.
## How to Find Square Root of 9?
Number 9 is an odd number and not a prime number. Prime numbers have only two factors, 1 and the number itself. But as we know, 9 have three multiple factors, 1, 3 and 9 itself. Thus, we can write the number 9 in the form of multiple as;
1 × 9 = 9
3 × 3 = 9
9 × 1 = 9
You know that number 9 is a multiple of 3 or when 3 is multiplied by itself, we get the number 9. Therefore, we can write the square root of 9 as;
$$\sqrt{9}$$ = $$\sqrt{3 × 3}$$
= $$\sqrt{3^2}$$
The square cancels the square root of a number. Therefore, if we cancel the square root with a square in the above expression, we get,
$$\sqrt{9}$$ = ±3
Basically, the square root of a number gives two root values, one with +ve and another with -ve symbol. Therefore, the value of the square root of 9 can be expressed as +3 and -3 or we can say the roots of 9 are +3 and -3.
Number 9 is the perfect square as it fulfils the criteria which we discussed earlier in the introduction.
## Square Root of 9 by Long Division Method
Long division method is helpful to find the square root of non-perfect squares. Since, 9 is a perfect square, thus, it is very easy to derive its square using division method. Remember the following points:
• 9 is a single-digit number, thus its square root will be less than 9
• 9 is an odd number, thus it is not divisible by any even number less than 9 (i.e.,2,4,6,8 is rejected)
• We have to multiply the rest of the numbers 3, 5, and 7 with themselves to find the value equal to 9
• Hence, 3 x 3 = 9
• So, the square root of 9 is equal to 3
## Important Points
• If the number ends with 2, 3, 7 and 8, then the number is not a perfect square.
• A perfect square number ends with 1, 4, 5, 6 and 9.
• It is easy to find the square root of a perfect square number as it results in the whole number, whereas imperfect squares result in fractional or decimal numbers. We can easily square a number by just multiplying the number to itself but taking the square root of the number is a little complicated, especially in the case of imperfect squares.
• The square root is represented by ‘√’ symbol. This symbol is named a radical symbol or radix. And the number underneath the radical symbol or radix is called the radicand. Here, we will find out the square root of 9 by a simple method, where 9 is the radicand.
## Related Articles
Find some other square roots here Square Root Of 1 Square Root Of 2 Square Root Of 3 Square Root Of 4 Square Root Of 5 Square Root Of 6 Square Root Of 7 Square Root Of 8
## Square Root of 9 Solved Examples
Q.1: What is the value of square root of 9/25?
Solution: Finding the root of 9 and 25 individually we have;
√9 = 3
And,
√25 = 5
Thus, √(9/25) = 3/5
Q.2: Find the sum of √9 + 2√49.
Solution: Since,
√9 = 3
√49 = 7
Therefore,
√9 + 2√49 = 3 + 2 (7) = 3 + 14 = 17
## Frequently Asked Questions – FAQs
### What is actual value of square root of 9?
The value of square root of 9 is 3.
### Is root 9 a rational or irrational?
Root 9 is a rational number because √9 = 3 can be expressed as 3/1.
### What is the square root of -9?
The square root of -9 is 3i, since √-1 = i, where i is the imaginary number.
Download BYJU’S – The Learning App and learn to find the squares and roots of numbers with the help of interactive videos. |
# Representative Values and Stem and Leaf Plot
In this article, we are going to talk about some important representative values and one of the main representative methods called Stem and Leaf Plot.
## Mode
Let’s take the idea about the mode by using an example.
Example:
Marks obtained by 10 students for a mathematics test is given below.
78 93 45 67 82 54 82 76 80 97
The number of data in this collection is 10.
Let’s arrange the above marks in ascending order.
45, 54, 67, 76, 78, 80, 82, 82, 93, 97
The marks that have been obtained by the most is 82.
In a collection of data, some of the values could be identical.
The value which occurs most of 10 is called the Mode of that collection of data.
It is not necessary to write the data in ascending order to find the mode.
The collection of data may have two modes. Such a distribution of data area called Bi-Modal distribution.
Example : 4, 3, 5, 2, 3, 8, 4
Modes are 3 and 4.
In this case, we have to mention both modes as the answer.
There are distributions also with three modes. Those areas called Trimodal Distribution.
Example: 23, 65, 46, 32, 65, 84, 23, 16, 56, 32
Modes are 23, 65 and 32
If a collection have more than 1 mode, those distributions are called Multi-modal distribution.
## Median
Median has to be discussed under two categories.
01. Collection of data with an odd number of values.
Example:
8, 7, 3, 9, 4, 5, 2
Arrange the data in ascending order.
2, 3, 4, 5, 7, 8, 9
There are 7 data. The 4th datum is in the center.
Its value is 5.
So the Median is 5.
So the number of values in a collection of data is an odd number. The median is the value in the center when the values are arranged in ascending order.
How to find the place of the median in a collection of odd number of data easily?
According to the above example,
So, the place of the median should be 4th place.
Exercise:
Find the median of the collection of data.
47, 23, 85, 72, 64, 53, 31, 94, 66
02. Collection of data with an even number of values.
Example:
8, 6, 8, 2, 4, 7, 3, 9
Arrange the data in ascending order.
2, 3, 4, 6, 7, 8, 8, 9
There are 8 values in this collection of data.
The two data in the center are 4th and 5th data. Those values are 6 and 7 respectively.
If there is an even number of in the collection of data, the median is half the sum of the values the two data in the center.
Its a must to arrange data in ascending order.
In the above example, the median is,
## Mean
Mean is also called as the average value in daily life.
Mean is obtained when,
The sum of all the values of the collection of data is divided by the number of values.
Let’s make it clear through an example.
Example:
Marks obtained by Kevin for his mid-term test for 10 different subjects are as follows.
75, 82, 54, 98, 81, 69, 96, 74, 66, 98
His average mark or the mean value is 79.3
It is not necessary to arrange data in ascending order to find mean.
Except for these three representative values mode, median and mean, there are another things which will help you on the way in statistics.
It is Range.
## What is Range?
Range is the difference between the greatest value and the least value of a collection of data.
Example:
Let’s take the previous example which we have taken to discuss about mean, about Kevin’s marks.
To find the range, you have to arrange data in ascending order.
54, 66, 69, 74, 75, 81, 82, 96, 98, 98
54 – Least value
98 – Greatest value
Range = Greatest value – Least value
= 98 – 54
= 44
## Stem and Leaf Plot
Stem and Leaf plots are one of the main data representation methods.
There are few rules that you have to follow when you are entering data to a stem and leaf plot.
• Only the digit in the unit place is indicated as the leaf.
• We don’t use commas to separate leaf values. We just leave a gap.
• It is easier to make a stem and leaf plot if we arrange our data in ascending order before we enter our data. Otherwise, you can just enter your data to the given order and arrange the data in ascending order in another stem and leaf plot.
Let’s discuss an example.
The marks obtained by 20 students for a mathematics test is given below.
17 20 15 10 23 13 19 25 9 11 24 23 15 18 17 8 19 25 20 14
You can arrange your data in ascending order and then enter.
8 9 10 11 13 14 15 15 17 17 18 19 19 20 20 23 23 24 25 25
Key 2 | 3 means 23
The above method is easier when you have small number of data.
Usually, when we are entering data to stem and leaf diagram, we don’t arrange the data in ascending order instead of that we just enter the value according to the given order in a separate stem and leaf diagram and then in another diagram you have to rearrange the data in ascending order.
17 20 15 10 23 13 19 25 9 11 24 23 15 18 17 8 19 25 20 14
Now draw another stem and leaf diagram and enter the values belongs to each stem in ascending order.
Key 1 | 5 means 15
Now let’s move on to another example with decimal values as data.
The height of 15 different trees in meters given below.
4.7 3.6 1.2 2.9 6.8 3.2 5.1 2.3 1.4 3.3 5.6 1.1 4.8 3.9 4.5
Key 4 | 5 means 4.5
## Exercise
01. Find the mode, the median, the mean and the range of each collection of data.
1. 9, 10, 13, 11, 13, 8, 9, 7, 11, 6, 9
2. 34, 33, 19, 34, 46, 24, 54, 33, 34
3. 79, 79, 81, 71, 79, 66, 70, 71
4. 4.5, 3.5, 5.8, 2.3, 4.9
5. 13.5, 33.4, 23.6, 9.3
02. The number of candies in 10 jars is given below.
50, 51, 49, 48, 50, 51, 50, 51, 48, 52
For these jars
1. Find the mode
2. Find the median
3. The mean number of candies in a jar.
03. The lengths of the calls received on a certain day by a person who uses a mobile phone are given below to the nearest minute.
4, 3, 6, 11, 2, 4, 8, 4, 5, 7, 3, 5, 4, 9, 12, 5, 4, 3
1. Write the range of the given set of data.
2. What is the mode?
3. Write the median.
4. Using the mean, estimate the time in hours and minutes that could be expected to be spent on 100 calls that are received by this person.
04. The height of 30 students in a certain class is given in centimeters.
116 124 112 115 110 123 123 115 113 119 114 125 126 121 131 125 117 121 135 112 127 128 129 128 132 115 110 132 121 131
1. Represent the above data in a stem and leaf plot.
2. What is the height of the shortest student?
3. What is the height of the tallest student?
4. Find the range of their heights.
5. Find the mode.
## FAQ
What is mode in math?
The value which occurs most in a collection of data called the Mode.
What is median in math?
The middle of the sorted collection of numbers.
How to find median of collection of data with an odd number of values?
How to find median of collection of data with an even number of values?
If there is an even number of in the collection of data, the median is half the sum of the values the two data in the center.
What is the mean in math?
Mean is the average number in a collection of numbers.
How to find mean in math?
What is Range?
Range is the difference between the greatest value and the least value of a collection of data.
What is stem and leaf plot?
Stem and Leaf plot is a data representation method. |
When autocomplete results are available use up and down arrows to review and enter to select. Touch device users, explore by touch or with swipe gestures.
# Online Math Classes
## Collection by TEL Gurus - Meet Your Divine Masters • Last updated 11 days ago
53
Pins
•
16
Followers
Learning maths online offers a number of advantages. Since students can learn at their own pace and cover a topic as many times as required to completely learn it, knowledge retention is higher than in a traditional classroom environment.
1st Step – 0.125 can be represented as 125 / 1000. 2nd Step – Divide 125 / 1000 by 5 3rd Step – Results are as below: 125 / 1000 = 25 25 / 200 = 5 5 / 40 = 1/8 The simplest form fraction for 0.125 is 1 / 8.
We will calculate and show you how to factorize x2 – 9 by difference of squares method. Step 1 – x2 – 9 Step 2 – We use difference of squares method to factorize x2 – 9 = a2 – b2 = (a + b) (a – b) = So, considering a = x and b = 3 = x2 – 32 get the next step on the website
As we know that 60 is a whole number and a composite number (divisible by more than 2 numbers). Let us find the prime factors of 60 by the following method. 1st step – Divide 60 with its smallest prime factor, which is 2 60/2 = 30 continuethis solution on the website
An arithmetic progression is a series in which each term is derived by addition or subtraction of a common numeral to its previous term. The Arithmetic sequence formula is used for determining the nth term of an arithmetic progression.
0:30
A Rational Number is a number that can be in presented in the form of P/Q where P and Q are Integers and Q is not equal to Zero ("0"). Learn With Example
0:08
An arithmetic progression is a series in which each term is derived by addition or subtraction of a common numeral to its previous term. The Arithmetic sequence formula is used for determining the nth term of an arithmetic progression.
0:08
We will calculate and show you how to factorize x2 – 9 by difference of squares method.
We’ve all heard someone ask what the temperature is outside. Perhaps your doctor has advised you to rest on diagnosing you with a temperature. Maybe you’ve complained that the temperature was too hot in the car. So, what is the temperature? TEMPERATURE - Measure of degree of hotness or coldness measured on a definite scale. UNITS - °C and °F How to convert 10°C into °F. Using formula -> (°C × 9/5) + 32 = °F & substituting °C value i.e. 10°C °F = 50 10°C = 50°F
Well, Yes! Winning the lottery is a dream for many people in this world. But do you know the logic behind how that one person in millions gets this jackpot? It's nothing but just a real-life mathematics concept - PROBABILITY. Just imagine, 10 million people buy lottery tickets, and so you do as well. Your probability of winning the lottery is just 1/10,000,000, and so does everyone else. Interactively learn Maths and ensure a strong Maths foundation for your child. Book A Free Demo Class
Linear equation in one variable An equation which is expressed in the form of ax + b = 0, where a and b are two real numbers and x is a variable that has only one solution. For example, 2x - 10 = 8 is a linear equation having a single variable in it. So, this equation will have only one solution, which is x = 9.
Telguru has expert maths tutor in London that provides online mathematics lessons for all primary and secondary mathematics tuition needs of students.
Multiplication is an important tool of mathematics. It is used to solve algebra, calculus, equations and more. If you make your child learn mathematics' multiplication tricks at an early age it will be easy for him in future to solve complex mathematical problems.
Whole Number in Mathematics The Whole Numbers starts from the 0,1,2....... and so on. Whole Numbers are also known as Positive Integers and non-negative integers. It also includes natural numbers which start from 1,2,3 and so on Make your child fear-free from Mathematics with online learning at TEL Gurus.
3:10
You need to be careful while calculating the value of the given thing in a given question. What this means is: When you start your calculation for a given question, make it fast. You need to have fast calculation, which will help you make a decision of whether you should avoid a question or not.
1:33
Our online mathematics lessons cover the academic needs of our students, including the educational needs from Primary to GCSE levels, helping them simplify complex ideas and giving them the confidence to dominate mathematics. |
Go Math Answer Key for Grade 1 aids teachers to differentiate instruction, building, and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Go Math primary school Grade 1 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.
To achieve high scores in Grade 1, students need to solve all questions and exercises included in Go Math book considers both the content and problems. Students of Grade 1 can get a strong foundation on mathematics concepts by referring to the Go Math course Book. It was highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping.
Curious George
Look at the picture. Make up a subtraction story problem.
The number of dogs = 2.
Explanation:
In the above-given question,
given that,
there are 3 dogs.
out of 3 dogs, 1 take away.
3 – 1 = 2.
The number of dogs = 2.
### Subtraction Concepts Show What You Know
Explore Numbers 1 to 4
Show the number with .
Draw the .
Question 1.
The number is 4.
Explanation:
In the above-given question,
given that,
the number is 4.
so we have to draw the 4 buttons.
Question 2.
The number is 2.
Explanation:
In the above-given question,
given that,
the number is 2.
so we have to draw the 2 buttons.
Numbers 1 to 10
How many objects are in each set?
Question 3.
_10_ butterflies
There are 10 butterflies.
Explanation:
In the above-given question,
given that,
there are 10 butterflies.
so the number of butterflies = 10.
Question 4.
_1_ cat
There is 1 cat.
Explanation:
In the above-given question,
given that,
there is 1 cat.
so the number of cats = 1.
Question 5.
_8_ leaves
There are 8 leaves.
Explanation:
In the above-given question,
given that,
there are 8 leaves.
so the number of leaves = 8.
Use Pictures to Subtract
How many are left?
Question 6.
5 – 4 = _1_
The number of apples = 1.
Explanation:
In the above-given question,
given that,
there are 5 apples.
5 – 4 = 1 apple.
Question 7.
4 – 2 = _2_
The number of fishes = 2.
Explanation:
In the above-given question,
given that,
there are 4 fishes.
4 – 2 = 2 fishes.
### Subtraction Concepts Vocabulary Builder
Visualize It
Sort the review words from the box.
Understand Vocabulary
Circle the part you take from the group. Then cross it out
Question 1.
The number of oranges = 3.
Explanation:
In the above-given question,
given that,
There are 5 oranges.
2 oranges are eaten.
5 – 2 = 3 oranges.
The number of oranges = 3.
Question 2.
The number of balloons = 1.
Explanation:
In the above-given question,
given that,
there are 4 ballons.
out of 4 ballons, 3 fly away.
4 – 3 = 1.
The number of ballons = 1
Question 3.
The number of toy cars = 2.
Explanation:
In the above-given question,
given that,
there are 3 toy cars.
out of 3 toy cars, 1 rolls away.
3 – 1 = 2.
The number of toy cars = 2.
### Subtraction Concepts Game: Subtraction Slide
Subtraction Slide
Materials
Play with a partner.
Take turns.
1. Toss the
2. Take away that number from . Use to find what is left.
3. If you see the answer on your slide, cover it with a .
5. The first player to cover a whole slide wins.
### Subtraction Concepts Vocabulary Game
Bingo
Materials
• 1 set of word cards
• 18
How to Play
Play with a partner.
1. Mix the cards. Put the cards in a pile with the blank side up.
2. Take a card. Read the word.
3. Find the matching word on your Bingo board. Cover the word with a . Put the card at the bottom of the pile.
4. The other player takes a turn.
5. The first player to cover 3 spaces in a line wins. The line may go across or down.
The Write Way
Reflect
Choose one idea. Draw and write about it.
• Think about what you did during math today. Complete one of the sentences:
I learned _____.
Sophie has 5 stickers.
She gives 2 to Olivia.
Sophie has 2 stickers left.
Is the answer correct? Draw and write to explain. Use another piece of paper for your drawing.
### Lesson 2.1 Use Pictures to Show Taking From
Essential Question
How can you show taking from with pictures?
__ children now
Math Talk
MATHEMATICAL PRACTICES
Model How did you find how many are in the sandbox now?
Share and Shhow MATH BOARD
Circle the part you are taking from the group. Then cross it out. Write how many there are now.
Question 1.
6 bugs 2 bugs fly away. _4_ bugs now
The number of bugs= 4.
Explanation:
In the above-given question,
given that,
there are 6 bugs.
out of 6 bugs, 2 fly away.
6 – 2 = 4.
The number of bugs = 4.
Question 2.
3 dogs 1 dog walks away. _1_ dogs now
The number of dogs= 2.
Explanation:
In the above-given question,
given that,
there are 3 dogs.
out of 3 dogs, 1 walks away.
3 – 1 = 2.
The number of dogs = 2.
Circle the part you are taking from the group. Then cross it out. Write how many there are now.
Question 3.
7 chicks 2 chicks walk away. __ chicks now
The number of chicks = 7.
Explanation:
In the above-given question,
given that,
there are 7 chicks.
out of 7 chicks, 2 walk away.
7 – 2 = 5.
The number of dogs = 2.
Question 4.
6 ducks 3 ducks walk away. __ ducks now
The number of ducks = 3.
Explanation:
In the above-given question,
given that,
there are 6 ducks.
out of 6 ducks 3 walk away.
6 – 3 = 3.
The number of ducks = 3.
Question 5.
THINK SMARTER
Ellie and Sara see 8 birds in the tree. Ellie sees 2 fewer birds than Sara. Sara sees 5 birds. How many birds does Ellie see?
_3_ birds
Ellie sees 3 birds.
Explanation:
In the above-given question,
given that,
Ellie and Sara see 8 birds in the tree.
Sara sees 5 birds.
Ellie sees 2 fewer birds than Sara.
5 – 2 = 3.
Ellie sees 3 birds.
Question 6.
GO DEEPER
Choose numbers to complete the story. Write the numbers. Draw to show the problem.
__5_ worms _2_ worms wiggle away. 3 worms now
The number of worms = 3.
Explanation:
In the above-given question,
given that,
there are 5 worms.
out of 5 worms 2 wiggle away.
5 – 2 = 3.
The number of worms = 3.
Problem Solving • Applications
MATHEMATICAL PRACTICE
Use Diagrams Solve.
Question 7.
There are 6 cats. 1 cat runs away. How many cats are there now? Draw to show your work.
__ cats
The number of cats = 5.
Explanation:
In the above-given question,
given that,
there are 6 cats.
out of 6 cats 1 cat runs away.
6 – 1 = 5.
The number of cats = 5.
Question 8.
GO DEEPER
There are 6 dogs. 3 dogs run away. Then 1 more dog runs away. How many dogs are there now?
_2_ dogs now
The number of dogs= 2.
Explanation:
In the above-given question,
given that,
there are 6 dogs.
out of 6 dogs, 4 runs away.
3 + 1 = 4
6 – 4 = 2.
The number of dogs = 2.
Question 9.
THINK SMARTER
Use the picture. Write the numbers.
8 birds _5_ birds fly away. _3_ birds now
The number of birds= 8.
Explanation:
In the above-given question,
given that,
there are 8 birds.
out of 8 birds, 5 fly away.
8 – 5 = 3.
The number of birds = 8.
Question 10.
THINK SMARTER
Circle and cross out the part that goes away. Write how many there are now.
10 fish 6 fish swim away. _4_ fish now
The number of fish= 4.
Explanation:
In the above-given question,
given that,
there are 10 fish.
out of 10 fish, 6 swim away.
10 – 6 = 6.
The number of fish = 4.
TAKE HOME ACTIVITY
• Have your child draw and solve the subtraction problem: There are 6 cows. 3 cows walk away. How many cows are there now?
The number of cows = 3.
Explanation:
In the above-given question,
given that,
there are 6 cows.
out of 6 cows, 3 walk away.
6 – 3 = 3.
The number of cows = 3.
### Use Pictures to Show Taking From Homework & Practice 2.1
Use the picture. Circle the part you take from the whole group. Then cross it out. Write how many there are now.
Question 1.
3 cats 1 cat walks away. _2_ cats now
The number of cats = 2.
Explanation:
In the above-given question,
given that,
there are 3 cats.
out of 3 cats 1 cat runs away.
3 – 1 = 2.
The number of cats = 2.
Question 2.
5 horses 2 horses walk away. _3_ horses now
The number of horses = 3.
Explanation:
In the above-given question,
given that,
there are 5 horses.
out of 5 horses, 2 horses walk away.
5 – 2 = 3.
The number of horses = 3
.
Problem Solving
Solve.
Question 3.
There are 7 birds. 2 birds fly away. How many birds are there now?
_5_ birds
The number of birds= 5.
Explanation:
In the above-given question,
given that,
there are 7 birds.
out of 7 birds, 2 fly away.
7 – 2 = 5.
The number of birds = 5.
Question 4.
WRITE Math
Draw a picture to show the problem. There are 9 turtles. 3 turtles walk away. How many turtles are there now?
The number of turtles= 6.
Explanation:
In the above-given question,
given that,
there are 9 turtles.
out of 9 turtles, 3 walk away.
9 – 3 = 6.
The number of turtles = 6.
Lesson Check
Question 1.
There are 4 ducks. 2 ducks swim away. How many ducks are there now?
_2_ ducks
The number of ducks = 2.
Explanation:
In the above-given question,
given that,
there are 4 ducks.
out of 4 ducks 2 walk away.
4 – 2 = 2.
The number of ducks = 2.
Spiral Review
Question 2.
What is the sum for 2 + 0?
Draw to show each addend. Write the sum.
2 + 0 = _2_
The sum for 2 + 0 = 2.
6 + 2 = 8.
Explanation:
In the above-given question,
given that,
The number 2 + 0 = 2.
The sum for 2 + 0 = 2.
6 + 2 = 8.
Question 3.
How many birds? Write how many.
5 birds and 2 birds _7_ birds.
The number of birds = 7.
Explanation:
In the above-given question,
given that,
There are 5 birds and 2 birds = 7 birds.
5 + 2 = 7.
The number of birds = 7.
Question 4.
What is the sum? Write the sum.
The sum for 6 + 2 = 8.
Explanation:
In the above-given question,
given that,
the sum of numbers 6 and 2.
6 + 2 = 8.
the sum for 6 + 2 = 8.
Lesson 2.2 Model Taking From
Essential Question
How do you model taking from a group?
Listen and Draw
Use to show taking from. Draw to show your work.
Math Talk
MATHEMATICAL PRACTICES
Reasoning Are there more than or fewer than 9 butterflies now?
Share and Show MATH BOARD
Use to show taking from. Draw the . Circle the part you take away from the group. Then cross it out. Write the difference.
Question 1.
8 dogs 3 dogs run away.
8 – 3 = _5_
The number of dogs = 5.
Explanation:
In the above-given question,
given that,
there are 8 dogs.
out of 8 dogs, 3 runs away.
8 – 3 = 5.
The number of dogs = 5.
Question 2.
6 frogs 4 frogs hop away.
6 – 4 = _2_
The number of frogs = 2.
Explanation:
In the above-given question,
given that,
there are 6 frogs.
out of 6 frogs 4 hop away.
6 – 4 = 2.
The number of frogs = 2.
Use to show taking from. Draw the . Circle the part you take away from the group. Then cross it out. Write the difference.
Question 3.
5 seals 4 seals swim away.
5 – 4 = _1_
The number of seals = 1.
Explanation:
In the above-given question,
given that,
there are 5 seals.
out of 5 seals 4 swim away.
5 – 4 = 1.
The number of seals = 1.
Question 4.
9 bears 6 bears run away.
9 – 6 = _3_
The number of bears = 3.
Explanation:
In the above-given question,
given that,
there are 9 bears.
out of 9 bears 6 run away.
9 – 6 = 3.
The number of bears = 3.
Question 5.
THINK SMARTER
Kelly sees some fish at the pet shop. A man buys 2 fish. A boy buys 4 fish. Now there are 3 fish. How many fish did Kelly see at first?
_9_ fish
Kelly sees 9 fish at first.
Explanation:
In the above-given explanation,
given that,
Now there are 3 fish.
3 + 4 + 2 = 9.
Kelly sees 9 fish at first.
Question 6.
GO DEEPER
Use the picture. Circle a part to take from the group. Then cross it out. Write the subtraction sentence.
The number of cubes = 6.
Explanation:
In the above-given question,
given that,
there are 9 cubes.
out of 9 cubes 3 take away.
9 – 3 = 6.
The number of cubes = 6.
Problem Solving • Applications
MATHEMATICAL PRACTICE
Make Sense of Problems
Draw to solve. Complete the subtraction sentence.
Question 7.
There are 5 boys. 3 boys go home. How many are there now?
_5_ – _3_ = __2_
_2_ boys
The number of boys = 2.
Explanation:
In the above-given question,
given that,
there are 5 boys.
out of 5 boys 3 boys go home.
5 – 3 = 2.
The number of boys = 2.
Question 8.
There are 8 bunnies. 2 bunnies hop away. How many bunnies are there now?
_8_ – _2_ = _6_
_6_ bunnies
The number of bunnies = 6.
Explanation:
In the above-given question,
given that,
there are 8 bunnies.
out of 8 bunnies 2 bunnies hop away.
8 – 2 = 6.
The number of bunnies = 6.
Question 9.
THINK SMARTER
Draw a picture to show a subtraction sentence. Write the subtraction sentence.
_3_ – _1_ = _2_
The number of cats = 2.
Explanation:
In the above-given question,
given that,
there are 3 cats.
out of 3 cats 1 cat runs away.
3 – 1 = 2.The number of cats = 2.
Question 10.
THINK SMARTER
Take away to show the problem. Write the difference. There are 7 mice. 5 run away.
7 – 5 = _2_
The number of mice = 2.
Explanation:
In the above-given question,
given that,
there are 7 mice.
out of 7 mice 5 runs away.
7 – 5 = 2.
The number of mice = 2.
TAKE HOME ACTIVITY
• Use small objects of the same kind to model a subtraction situation for numbers within 10. Ask your child to write the subtraction sentence for it. Then switch roles and repeat the activity.
### Model Taking From Homework & Practice 2.2
Use to show taking from. Draw the . Circle the part you take from the group. Then cross it out. Write the difference.
Question 1.
4 turtles 1 turtle walks away.
4 – 1 = _3_
The number of turtles = 3.
Explanation:
In the above-given question,
given that,
there are 4 turtles.
out of 4 turtles, 1 walks away.
4 – 1 = 3.
The number of turtles = 3.
Question 2.
8 birds 7 birds fly away.
8 – 7 = _1_
The number of birds = 1.
Explanation:
In the above-given question,
given that,
There are 8 birds, 7 birds fly away.
8 – 7 = 1.
The number of birds = 1.
Problem Solving
Draw to solve. Complete the subtraction sentence.
Question 3.
There are 8 fish. 4 fish swim away. How many fish are there now?
_8_ – _4_ = _4_
_4_ fish
The number of fish = 4.
Explanation:
In the above-given question,
given that,
there are 8 fish.
out of 8 fish, 4 fish swim away.
8 – 4 = 4.
The number of fish = 4.
Question 4.
WRITE Math
Use pictures and numbers to model 9 − 2.
The number of fish = 7.
Explanation:
In the above-given question,
given that,
there are 9 fish.
out of 9 fish, 2 fish swim away.
9 – 2 = 7.
The number of fish = 7.
Lesson Check
Question 1.
Show taking from. Circle the part you take from the group. Then cross it out. Write the difference.
5 – 2 = _3_
The number of cubes = 3.
Explanation:
In the above-given question,
given that,
there are 5 cubes.
out of 5 cubes, 2 take away.
5 – 2 = 3.
The number of cubes = 3.
Spiral Review
Question 2.
How many snails? Write the number.
7 snails and 1 snail _6_ snails
The number of snails = 6.
Explanation:
In the above-given question,
given that,
there are 7 snails.
out of 7 snails 1 runs away.
7 – 1 = 6.
The number of snails = 6.
Question 3.
Circle the number sentences that show the same addends in a different order.
The number sentences that show the same addends = 6 + 2 = 8 and 2 + 6 = 8.
Explanation:
In the above-given question,
given that,
The number sentences are
7 + 1 = 8, 6 + 2 = 8 and 2 + 6 = 8.
The number sentences that show the same addends = 6 + 2 = 8 and 2 + 6 = 8.
### Lesson 2.2 Model Taking From
Essential Question
How do you model taking from a group?
Listen and Draw
Use to show taking from. Draw to show your work.
Math Talk
MATHEMATICAL PRACTICES
Reasoning Are there more than or fewer than 9 butterflies now?
Share and Show MATH BOARD
Use to show taking from. Draw the . Circle the part you take away from the group. Then cross it out. Write the difference.
Question 1.
8 dogs 3 dogs run away.
8 – 3 = _5_
The number of dogs = 5.
Explanation:
In the above-given question,
given that,
there are 8 dogs.
out of 8 dogs, 3 runs away.
8 – 3 = 5.
The number of dogs = 5.
Question 2.
6 frogs 4 frogs hop away.
6 − 4 = _2_
The number of frogs = 2.
Explanation:
In the above-given question,
given that,
there are 4 frogs.
out of 6 frogs, 4 hop away.
6 – 4 = 2.
The number of frogs = 2.
Use to show taking from. Draw the . Circle the part you take away from the group. Then cross it out. Write the difference.
Question 3.
5 seals 4 seals swim away.
5 – 4 = _1_
The number of seals = 1.
Explanation:
In the above-given question,
given that,
there are 5 seals.
out of 5 seals, 4 swim away.
5 – 4 = 1.
The number of seals = 1.
Question 4.
9 bears 6 bears run away.
9 – 6 = __
The number of bears = 3.
Explanation:
In the above-given question,
given that,
there are 9 bears.
out of 9 bears, 6 runs away.
9 – 6 = 3.
The number of bears = 3.
Question 5.
THINK SMARTER
Kelly sees some fish at the pet shop. A man buys 2 fish. A boy buys 4 fish. Now there are 3 fish. How many fish did Kelly see at first?
_9_ fish
Kelly sees 9 fish at first.
Explanation:
In the above-given explanation,
given that,
Now there are 3 fish.
3 + 4 + 2 = 9.
Kelly sees 9 fish at first.
Question 6.
GO DEEPER
Use the picture. Circle a part to take from the group. Then cross it out. Write the subtraction sentence.
The number of cubes = 5.
Explanation:
In the above-given question,
given that,
there are 9 cubes.
out of 9 cubes, 4 take away.
9 – 4 = 5.
The number of cubes = 5.
Problem Solving • Applications WRITE Math
MATHEMATICAL PRACTICE
Make Sense of Problems
Draw to solve. Complete the subtraction sentence.
Question 7.
There are 5 boys. 3 boys go home. How many are there now?
_5_ – _3_ = _2_
__ boys
The number of boys = 2.
Explanation:
In the above-given question,
given that,
there are 5 boys.
out of 5 boys 3 boys go home.
5 – 3 = 2.
The number of boys = 2.
Question 8.
There are 8 bunnies. 2 bunnies hop away. How many bunnies are there now?
__ – __ = __
__ bunnies
The number of bunnies = 6.
Explanation:
In the above-given question,
given that,
there are 8 bunnies.
out of 8 bunnies 2 bunnies hop away.
8 – 2 = 6.
The number of bunnies = 6.
Question 9.
THINK SMARTER
Draw a picture to show a subtraction sentence. Write the subtraction sentence.
__ – __ = __
The number of cats = 2.
Explanation:
In the above-given question,
given that,
there are 3 cats.
out of 3 cats 1 cat runs away.
3 – 1 = 2.
The number of cats = 2.
Question 10.
THINK SMARTER
Take away to show the problem. Write the difference. There are 7 mice. 5 run away.
7 – 5 = _2_
The number of mice = 2.
Explanation:
In the above-given question,
given that,
there are 7 mice.
out of 7 mice 5 runs away.
7 – 5 = 2.
The number of mice = 2.
TAKE HOME ACTIVITY
• Use small objects of the same kind to model a subtraction situation for numbers within 10. Ask your child to write the subtraction sentence for it. Then switch roles and repeat the activity.
### Model Taking From Homework & practice 2.2
Use to show taking from. Draw the . Circle the part you take from the group. Then cross it out. Write the difference.
Question 1.
4 turtles 1 turtle walks away.
4 – 1 = __
The number of turtles = 3.
Explanation:
In the above-given question,
given that,
there are 4 turtles.
out of 4 turtles, 1 walks away.
4 – 1 = 3.
The number of turtles = 3.
Question 2.
8 birds 7 birds fly away.
8 – 7 = _1_
The number of birds = 1.
Explanation:
In the above-given question,
given that,
There are 8 birds, 7 birds fly away.
8 – 7 = 1.
The number of birds = 1.
Problem Solving
Draw to solve. Complete the subtraction sentence.
Question 3.
There are 8 fish. 4 fish swim away. How many fish are there now?
_8_ – _4_ = _4__
_4_ fish
The number of fish = 4.
Explanation:
In the above-given question,
given that,
there are 8 fish.
out of 8 fish, 4 fish swim away.
8 – 4 = 4.
The number of fish = 4.
Question 4.
WRITE Math
Use pictures and numbers to model 9 − 2.
The number of fish = 7.
Explanation:
In the above-given question,
given that,
there are 9 fish.
out of 9 fish, 2 fish swim away.
9 – 2 = 7.
The number of fish = 7.
Lesson Check
Question 1.
Show taking from. Circle the part you take from the group. Then cross it out. Write the difference.
5 – 2 = _3_
The number of cubes = 3.
Explanation:
In the above-given question,
given that,
there are 5 cubes.
out of 5 cubes, 2 take away.
5 – 2 = 3.
The number of cubes = 3.
Spiral Review
Question 2.
How many snails? Write the number.
7 snails and 1 snail _6_ snails
The number of snails = 6.
Explanation:
In the above-given question,
given that,
there are 7 snails.
out of 7 snails 1 runs away.
7 – 1 = 6.
The number of snails = 6.
Question 3.
Circle the number sentences that show the same addends in a different order.
The number sentences that show the same addends = 6 + 2 = 8 and 2 + 6 = 8.
Explanation:
In the above-given question,
given that,
The number sentences are
7 + 1 = 8, 6 + 2 = 8 and 2 + 6 = 8.
The number sentences that show the same addends = 6 + 2 = 8 and 2 + 6 = 8.
### Lesson 2.3 Model Taking Apart
Essential Question
How do you model taking apart?
Listen and Draw
Use to model the problem. Draw and color to show your model. Write the numbers and a subtraction sentence.
_5_ red apples _2_ yellow apples
Jeff has _2_ yellow apples.
Jeff has 2 yellow apples.
Explanation:
In the above-given question,
given that
There are 5 red apples.
3 yellow apples.
5 – 3 = 2.
Jeff has 2 yellow apples.
Math Talk
MATHEMATICAL PRACTICES
Use Tools How does using the counters help you write the subtraction sentence?
Model and Draw
Subtract to find how many small cups there are.
Maria has 6 cups. 2 cups are big. The rest are small. How many cups are small?
_ small cups
The number of small cups = 4.
Explanation:
In the above-given question,
given that,
Maria ha 6 cups.
2 cups are big.
the rest are small.
6 – 2 = 4.
The number of small cups = 4.
Share and Show MATH BOARD
Use to solve. Draw to show your work. Write the number sentence and how many.
Question 1.
There are 7 folders. 6 folders are red. The rest are yellow. How many folders are yellow?
_1_ yellow folder
The number of yellow folders = 1.
Explanation:
In the above-given question,
given that,
There are 7 folders.
6 folders are red.
the rest are yellow.
7 – 6 = 1.
The number of yellow folders = 1.
Question 2.
There are 8 pencils. 3 pencils are short. The rest are long. How many pencils are long?
_5_ long pencils
The number of long pencils = 5.
Explanation:
In the above-given question,
given that,
There are 8 pencils.
3 pencils are short.
the rest are long.
8 – 5 = 3.
The number of long pencils = 5.
MATHEMATICAL PRACTICE
Write an Equation
Use to solve. Draw to show your work. Write the number sentence and how many.
Question 3.
There are 9 fish. 5 fish are red. The rest are yellow. How many fish are yellow?
_4_ yellow fish
The number of yellow fish = 4.
Explanation:
In the above-given question,
given that,
There are 9 fish.
5 fish are red.
the rest are yellow.
9 – 5 = 4.
The number of yellow fish = 4.
Question 4.
There are 7 ants. 4 ants are big. The rest are small. How many ants are small?
_3_ small ants
The number of small ants = 3.
Explanation:
In the above-given question,
given that,
There are 7 ants.
4 ants are big.
the rest are small.
7 – 4 = 3.
The number of small ants = 3.
Question 5.
There are 5 trees. 1 tree is short. The rest are tall. How many trees are tall?
_4_ tall trees
The number of tall trees = 4.
Explanation:
In the above-given question,
given that,
There are 5 trees.
1 tree is short.
the rest are tall.
5 – 1 = 4.
The number of tall trees = 4.
Question 6.
GO DEEPER
There are 8 birds. 1 bird flies away. Then 2 more birds fly away. How many birds are there now?
_5_ birds
The number of birds= 5.
Explanation:
In the above-given question,
given that,
there are 8 birds.
out of 8 birds, 3 fly away.
8 – 3 = 5.
The number of birds = 5.
Problem Solving • Applications
Solve. Draw a model to explain.
Question 7.
There are 6 bears. 4 are big. The rest are small. How many bears are small?
_2_ small bears
The number of small bears = 2.
Explanation:
In the above-given question,
given that,
There are 6 bears.
4 bears are big.
the rest are small.
6 – 4 = 2.
The number of small bears = 2.
Question 8.
THINK SMARTER
There are 7 bears. 1 bear walks away. Then 4 more bears walk away. How many bears are there now?
_2_ bears
The number of bears= 2.
Explanation:
In the above-given question,
given that,
there are 7 bears.
out of 7 bears, 5 walk away.
7 – 5 = 2.
The number of bears = 2.
Question 9.
GO DEEPER
There are 4 bears. Some are black and some are brown. There are fewer than 2 black bears. How many bears are brown?
__ brown bears
The number of brown bears = 2.
Explanation:
In the above-given question,
given that,
There are 4 bears.
2 bears are black.
the rest are brown.
4 – 2 = 2.
The number of brown bears = 2.
Question 10.
THINK SMARTER
Draw to solve. Write the number sentence. There are 8 flowers. 4 flowers are red. The rest are yellow. How many flowers are yellow?
The number of yellow flowers = 4.
Explanation:
In the above-given question,
given that,
There are 8 flowers.
4 flowers are red.
the rest are yellow.
8 – 4 = 4.
The number of yellow flowers = 4.
TAKE HOME ACTIVITY
• Have your child collect a group of up to 10 small objects of the same kind and use them to make up subtraction stories.
### Model Taking Apart Homework & Practice 2.3
Use to solve. Draw to show your work. Write the number sentence and how many.
Question 1.
There are 7 bags. 2 bags are big. The rest are small. How many bags are small?
__5 small bags
The number of small bags = 5.
Explanation:
In the above-given question,
given that,
There are 7 bags.
2 bags are big.
the rest are small.
7 – 2 = 5.
The number of small bags = 5.
Problem Solving
Solve. Draw a model to explain.
Question 2.
There are 8 cats. 6 cats are white. The rest are black. How many cats are black?
_2_ black cats
The number of black cats = 2.
Explanation:
In the above-given question,
given that,
There are 8 cats.
6 cats are white.
the rest are black.
8 – 6 = 2.
The number of black cats = 2.
Question 3.
WRITE Math
Use pictures and numbers to model 8 – 3.
The number of fish = 5.
Explanation:
In the above-given question,
given that,
there are 8 fish.
out of 8 fish, 3 fish swim away.
8 – 3 = 5.
The number of fish = 5.
Lesson Check
Question 1.
Solve. Draw a model to explain. There are 8 blocks. 3 blocks are white. The rest are blue. How many blocks are blue?
_5_ blue blocks
The number of blue blocks = 5.
Explanation:
In the above-given question,
given that,
There are 8 block.
3 blocks are white.
the rest are blue.
8 – 3 = 5.
The number of blue blocks = 5.
Spiral Review
Question 2.
Draw to solve. Write the number sentence and how many. There are 4 green grapes and 5 red grapes. How many grapes are there?
_9_ grapes
The number of grapes = 9.
Explanation:
In the above-given question,
given that,
There are 9 grapes.
4 grapes are green.
the rest are red grapes.
5 + 4 = 9
The number of grapes = 9.
Question 3.
Solve. Complete the model and the number sentence. 3 ducks swim in the pond. 2 more join them. How many ducks are in the pond now?
3 + 2 = _5_ ducks
The number of ducks in the pond = 5.
Explanation:
In the above-given question,
given that,
3 ducck swim in the pond.
2 more join them.
3 + 2 = 5.
The number of ducks in the pond = 5.
Question 4.
Draw the . Write the sum. What is the sum of 1 and 4?
1 + 4 = _5_
Anwer:
The sum of 1 and 4 is 5.
Explanation:
In the above-given question,
given that,
The sum of 1 and 4 is 5.
1 + 4 = 5.
### Lesson 2.4 Problem Solving • Model Subtraction
Essential Question
How do you solve subtraction problems by making a model?
Tom has 6 crayons in a box. He takes 2 crayons out of the box. How many crayons are in the box now? How can you use a model to find out?
Unlock the Problem
What do I need to find?
What information do I need to use?
The number of crayons in the box = 4.
Explanation:
In the above-given question,
given that,
Tom has 6 crayons in a box.
He takes 2 crayons out of the box.
6 – 2 = 4.
The number of crayons in the box = 4.
Show how to solve the problem.
6 – 2 = _4_
The number of crayons in the box = 4.
Explanation:
In the above-given question,
given that,
Tom has 6 crayons in a box.
He takes 2 crayons out of the box.
6 – 2 = 4.
The number of crayons in the box = 4.
HOME CONNECTION
• Your child used a bar model to help him or her understand and solve the subtraction problem.
Try Another Problem
Read the problem. Use the bar model to solve. Complete the model and the number sentence.
Question 1.
There are 10 stickers. 7 stickers are orange. The rest are brown. How many stickers are brown?
10 – 7 = __
The number of brown stickers = 3.
Explanation:
In the above-given question,
given that,
There are 10 stickers.
7 stickers are orange.
the rest are brown.
10 – 7 = 3.
The number of brown stickers = 3.
Question 2.
Eight birds were in the tree. 2 birds flew away. How many birds were left in the tree?
8 – 2 = _6_
The number of birds in the tree= 6.
Explanation:
In the above-given question,
given that,
There are 8 birds.
2 birds flew away.
8 – 2 = 6.
The number of birds in the tree = 6.
Question 3.
There were 5 cars. Some cars drove away. Then there was 1 car. How many cars drove away?
5 – __ = 1
The number of cars drove away = 4.
Explanation:
In the above-given question,
given that,
There are 5 cars.
some cars drove away.
there was 1 car.
5 – 4 = 1.
The number of cars drove away = 4.
Math Talk
MATHEMATICAL PRACTICES
Model What does each part of the model show?
Share and Show MATH BOARD
MATHEMATICAL PRACTICE
Use Models Read the problem. Use the bar model to solve. Complete the model and the number sentence.
Question 4.
Some goats were in the field. 3 goats ran away. Then there were 4 goats. How many goats were in the field before?
_7_ – 3 = 4
The number of goats in the field before = 7.
Explanation:
In the above-given question,
given that,
some goats were in the field.
3 goats ran away.
then there were 4 goats.
7 – 3 = 4.
The number of goats in the field before = 7.
Question 5.
There are 8 sleds. Some sleds slide down the hill. Then there are 4 sleds. How many sleds slide down the hill?
8 – __ = 4
The number of sleds slides down the hill = 4.
Explanation:
In the above-given question,
given that,
There are 8 sleds.
some sleds slide down the hill.
there are 4 sleds.
8 – 4 = 4.
The number of sleds slide down the hill = 4.
Question 6.
There are 10 buttons. 3 buttons are small. The rest are big. How many buttons are big?
10 – 3 = __
The number of big buttons = 7.
Explanation:
In the above-given question,
given that,
There are 10 buttons.
3 buttons are small.
the rest are big.
10 – 3 = 7.
The number of big buttons = 7.
Solve.
Question 7.
GO DEEPER
There were 8 spiders in the grass. Some spiders crawled away. Then there were 3 spiders. How many spiders crawled away? Complete the number sentence.
_5_ spiders
The number of spiders crawled away = 5.
Explanation:
In the above-given question,
given that,
there are 8 spiders.
out of 8 spiders, 5 crawled away.
there were 3 spiders.
8 – 5 = 3.
The number of spiders crawled away = 5.
Question 8.
THINK SMARTER
Write your own story problem using the model.
The number of big buttons = 7.
Explanation:
In the above-given question,
given that,
There are 9 buttons.
2 buttons are small.
the rest are big.
9 – 2 = 7.
The number of big buttons = 7.
Question 9.
THINK SMARTER
Complete the model and number sentence. There are 7 toy rockets. 4 toy rockets are red. The rest are black. How many toy rockets are black?
7 – 4 = _3_
The number of black rockets = 3.
Explanation:
In the above-given question,
given that,
There are 7 rockets.
4 rockets are red.
the rest are black.
7 – 4 = 3.
The number of black rockets = 3.
TAKE HOME ACTIVITY
• Ask your child to describe what the bottom part of a bar model means when subtracting.
### Problem Solving • Model Subtraction Homework & Practice 2.4
Read the problem. Use the model to solve. Complete the model and the number sentence.
Question 1.
There were 7 ducks in the pond. Some ducks swam away. Then there were 4 ducks. How many ducks swam away?
The number of ducks swan away = 3.
Explanation:
In the above-given question,
given that,
There are 7 ducks.
some ducks swan away.
there were 4 ducks.
7 – 3 = 4.
The number of ducks swan away = 3.
Question 2.
The number of gifts did he give away = 3.
Explanation:
In the above-given question,
given that,
9 – 3 = 6.
The number of gifts did he give away = 3.
Question 3.
Some ponies were in a barn. 3 ponies walked out. Then there were 2 ponies. How many ponies were in the barn before?
The number of ponies was in the barn = 5.
Explanation:
In the above-given question,
given that,
some ponies were in a barn.
3 ponies walked out.
then there were 2 ponies.
5 – 3 = 2.
The number of ponies was in the barn = 5.
Question 4.
WRITE Math
Choose a model from a problem you solved. Write a new subtraction problem to match.
__________
__________
__________
The number of gifts did he give away = 6.
Explanation:
In the above-given question,
given that,
9 – 6 = 3.
The number of gifts did he give away = 6.
Lesson Check
Question 1.
Complete the model and the number sentence. There are 8 shells. 6 shells are white. The rest are pink. How many shells are pink?
86 = __2_
The number of pink shells = 2.
Explanation:
In the above-given question,
given that,
There are 8 shells.
6 shells are white.
The rest are pink.
The number of pink shells = 2.
Spiral Review
Question 2.
Circle the number sentences that show the same addends in a different order.
The number sentences that show the same addends = 2 + 4 = 6 and 4 + 2 = 6.
Explanation:
In the above-given question,
given that,
The number sentences are
6 – 2 = 4, 2 + 4 = 6 and 4 + 2 = 6.
The number sentences that show the same addends = 2 + 4 = 6 and 4 + 2 = 6.
Question 3.
What is the sum? Write the sum.
The sum for 4 + 3 = 7.
Explanation:
In the above-given question,
given that,
the sum of numbers 4 and 3.
4 + 3 = 7.
the sum for 4 + 3 = 7.
### Lesson 2.5 Use Pictures and Subtraction to Compare
Essential Question
How can you use pictures to compare and subtract?
Listen and Draw
Draw bowls to show the problems. Draw lines to match.
The number of brown bowls = 9.
Explanation:
In the above-given question,
given that,
There are 16 bowls.
brown bowls = 9.
white bowls = 7.
16 – 7 = 9.
The number of brown bowls = 9.
Math Talk
MATHEMATICAL PRACTICES
Compare How do the drawings show the answers?
Model and Draw
Compare the groups. Subtract to find how many fewer or how many more.
The more nests = 2.
Explanation:
In the above-given question,
given that,
There are 6 nests.
4 birds.
6 – 4 = 2.
the more nests = 2.
Share and Show MATH BOARD
Draw lines to match. Subtract to compare.
Question 1.
The fewer flowers = 3.
Explanation:
In the above-given question,
given that,
There are 8 honey bees.
There are 5 flowers.
8 – 5 = 3.
the fewer flowers = 3.
MATHEMATICAL PRACTICE
Reason Quantitatively
Draw lines to match. Subtract to compare.
Question 2.
The more bees = 6.
Explanation:
In the above-given question,
given that,
There are 9 bees.
there are 3 leaves.
9 – 3 = 6.
the more bees = 6.
Question 3.
The fewer frogs = 4.
Explanation:
In the above-given question,
given that,
There are 10 leaves.
There are 6 frogs.
10 – 6 = 4.
the fewer frogs = 4.
Question 4.
GO DEEPER
The fewer wood = 5.
Explanation:
In the above-given question,
given that,
There are 7 squirrles.
There are 2 wood.
7 – 2 = 5.
the fewer wood = 5.
Question 5.
GO DEEPER
Evie has 3 red marbles and 4 blue marbles. Kai has 6 marbles. How many more marbles does Evie have than Kai?
_1 more
Evie have more marbles = 1 marbles.
Explanation:
In the above-given question,
given that,
Evie has 3 red marbles and 4 blue marbles.
Kai has 6 marbles.
Evie have more marbles = 1.
Problem Solving • Applications WRITE Math
Draw a picture to show the problem. Write a subtraction sentence that your picture shows.
Question 6.
Aki has 5 baseball bats and 3 baseballs. How many fewer baseballs does Aki have?
The fewer baseball = 2.
Explanation:
In the above-given question,
given that,
There are 5 baseball bats.
There are 3 baseballs.
5 – 3 = 2.
the fewer baseball = 2.
Question 7.
THINK SMARTER
Jill has 2 more cats than she has dogs. How many fewer dogs does Jill have?
_0_ fewer dogs
The fewer dogs = 0.
Explanation:
In the above-given question,
given that,
jill has 2 more cats than she has dogs.
There are 0 fewer dogs.
2 – 0 = 0
the fewer dogs = 0.
Question 8.
THINK SMARTER
Look at the picture. How many fewer leaves are there than ladybugs? Choose the answer.
The fewer leaves = 3.
Explanation:
In the above-given question,
given that,
There are 3 fewer leaves.
There are 8 leaves.
8 – 3 = 5.
the fewer leaves = 3.
### Use Pictures and Subtraction to Compare Homework & Practice 2.5
Draw lines to match. Subtract to compare.
Question 1.
The more dogs = 3.
Explanation:
In the above-given question,
given that,
There are 8 dogs.
there are 5 bones.
8 – 5 = 3.
the more dogs = 3.
Problem Solving
Draw a picture to show the problem. Write a subtraction sentence to match your picture.
Question 2.
Jo has 4 golf clubs and 2 golf balls. How many fewer golf balls does Jo have?
_4_ – _2_ = _2_
_2_ fewer
The fewer balls = 2.
Explanation:
In the above-given question,
given that,
There are 2 fewer balls.
There are 4 golf clubs.
4 – 2 = 2.
the fewer balls = 2.
Question 3.
WRITE Math
Draw pictures to compare to find 8 – 2.
The more dogs = 6.
Explanation:
In the above-given question,
given that,
There are 8 dogs.
there are 2 bones.
8 – 2 = 6.
the more dogs = 6.
Lesson Check
Question 1.
Draw lines to match. Subtract to compare. How many fewer are there?
The fewer teddys = 1.
Explanation:
In the above-given question,
given that,
There is 1 fewer teddys.
There are 4 caps.
4 – 3 = 1.
the fewer teddys = 1.
Spiral Review
Question 2.
Draw the . Write the sum. What is the sum of 5 and 1?
5 + 1 = _6_
The sum for 5 + 1 = 6.
Explanation:
In the above-given question,
given that,
the sum of numbers 5 and 1.
5 + 1 = 6.
the sum for 5 + 1 = 6.
Question 3.
Write the sum. What is the sum?
The sum for 4 + 5 = 9.
Explanation:
In the above-given question,
given that,
the sum of numbers 4 and 5.
4 + 5 = 9.
the sum for 4 + 5 = 9.
Question 4.
How many flowers? Write the number.
4 flowers and 3 flowers _7_ flowers
The sum for 4 + 3 = 7.
Explanation:
In the above-given question,
given that,
the sum of numbers 4 and 3.
4 + 3 = 7.
the sum for 4 + 3 = 7.
### Lesson 2.6 Subtract to Compare
Essential Question
How can you use models to compare and subtract?
Listen and Draw
Use to show the problem. Draw the . Model the problem using the bar model.
Math Talk
MATHEMATICAL PRACTICES
Explain how counters and a bar model can be used to find how many more puzzle pieces Mindy has than David.
Model and Draw
James has 4 stones. Heather has 7 stones. How many fewer stones does James have than Heather?
__ fewer stones
The number of fewer stones = 3.
Explanation:
In the above-given question,
given that,
james has 4 stones.
heather has 7 stones.
the fewer tones = 3
7 – 4 = 3.
The number of fewer stones = 3.
Share and Show MATH BOARD
Read the problem. Use the bar model to solve. Write the number sentence. Then write how many.
Question 1.
Abby has 8 stamps. Ben has 6 stamps. How many more stamps does Abby have than Ben?
__ more stamps
The number of more stamps Abby has = 2.
Explanation:
In the above-given question,
given that,
Abby has 8 stamps.
ben has 6 stamps.
Abby has more stamps = 2.
8 – 6 = 2.
The number of more stamps Abby has = 2.
Question 2.
Daniel has 3 books. Vicky has 6 books. How many fewer books does Daniel have than Vicky?
_3_ fewer books
The number of fewer books daniel have = 3.
Explanation:
In the above-given question,
given that,
Daniel has 3 books.
Vicky has 6 books.
6 – 3 = 3.
The number of fewer books daniel have = 3.
MATHEMATICAL PRACTICE
Use Models Read the problem. Use the bar model to solve. Write the number sentence. Then write how many
Question 3.
THINK SMARTER
Pam has 4 marbles. Rick has 10 marbles. How many fewer marbles does Pam have than Rick?
_6_ fewer marbles
The number of fewer marbles pam have = 6.
Explanation:
In the above-given question,
given that,
pam has 4 marbles.
Rick has 10 marbles.
10 – 4 = 6.
The number of fewer marbles pam have = 6.
Question 4.
Sally has 5 feathers. James has 2 feathers. How many more feathers does Sally have than James?
_3_ more feathers
The number of more feathers does sally have = 3.
Explanation:
In the above-given question,
given that,
sally has 5 feathers.
james has 2 feathers.
5 – 2 = 3.
The number of more feathers does sally have = 3.
Question 5.
GO DEEPER
Kyle has 6 keys. Kyle has 4 more keys than Lee. How many keys does Lee have?
_2_ keys
The number of keys lee have = 2.
Explanation:
In the above-given question,
given that,
kyle has 6 keys.
kyle has 4 more keys than lee.
6 – 4 = 2.
The number of keys lee have = 2.
TAKE HOME ACTIVITY
• Have your child explain how he or she solved exercise 3 using the bar model.
### Subtract to Compare Homework & practice 2.6
Read the problem. Use the bar model to solve. Write the number sentence. Then write how many.
Question 1.
Ben has 7 flowers. Tim has 5 flowers. How many fewer flowers does Tim have than Ben?
_2_ fewer flowers
The number of fewer flowers Tim have = 2.
Explanation:
In the above-given question,
given that,
ben has 7 flowers.
Tim has 5 flowers.
7 – 5 = 2.
The number of fewer flowers Tim have = 2.
Problem Solving
Complete the number sentence to solve.
Question 2.
Maya has 7 pens. Sam has 1 pen. How many more pens does Maya have than Sam?
_7_ – _1_ = __6
_6_ more pens
The number of more pens maya have = 6.
Explanation:
In the above-given question,
given that,
maya has 7 pens.
sam has 1 pen.
7 – 1 = 6.
The number of more pens maya have = 6.
Question 3.
WRITE Math
Write a subtract to compare problem and draw a bar model to solve it.
____________
____________
Lesson Check
Question 1.
Use the bar model to solve. Write the number sentence. Jesse has 2 stickers. Sara has 8 stickers. How many fewer stickers does Jesse have than Sara?
_8_ – _2 = _6_
_6_ fewer stickers
The number of fewer stickers jesse have = 6.
Explanation:
In the above-given question,
given that,
jesse has 2 stickers.
sara has 8 stickers.
8 – 2 = 6.
The number of fewer tickers jesse have = 6.
Spiral Review
Question 2.
Solve. There are 6 sheep. 5 sheep walk away. How many sheep are there now?
__ sheep
The number of sheeps = 1.
Explanation:
In the above-given question,
given that,
there are 6 sheeps.
out of 6 sheeps, 5 sheep walk away.
6 – 5 = 1.
The number of sheeps = 1.
Question 3.
Complete the bar model and the number sentence. 5 cows stand in a field. 2 more cows join them. How many cows are in the field now?
5 + 2 = __
The number of cows in the field = 7.
Explanation:
In the above-given question,
given that,
5 cows stand in a field.
2 more coins join them.
5 + 2 = 7.
The number of cows in the field = 7.
### Subtraction Concepts Mid-Chapter Checkpoint
Concepts and Skills
Circle the part you are taking from the group. Then cross it out. Write how many there are now.
Question 1.
7 birds 3 birds fly away. __ birds now
Use to solve. Draw to show your work. Write the number sentence and how many.
Question 2.
There are 4 cans. 1 can is red. The rest are yellow. How many cans are yellow?
_3_ yellow cans
The number of yellow cans = 3.
Explanation:
In the above-given question,
given that,
there are 5 cans.
out of 5 cans, 1 can is red.
the rest are yellow.
4 – 1 = 3.
The number of yellow cans= 3.
Question 3.
THINK SMARTER
Jennifer has 3 crayons. Brad has 9 crayons. How many fewer crayons does Jennifer have than Brad?
9 – 3 = __
The number of fewer crayons does Jennifer have = 6.
Explanation:
In the above-given question,
given that,
Jennifer has 3 crayons.
9 – 3 = 6.
The number of fewer crayons does Jennifer have = 6.
### Lesson 2.7 Subtract All or Zero
Essential Question
What happens when you subtract 0 from a number?
Listen and Draw
Use to show the problem. Draw the . Write the numbers.
The number of yellow cans = 3.
Explanation:
In the above-given question,
given that,
there are 3 cans.
out of 3 cans, 0 can is red.
the rest are yellow.
3 – 0 = 3.
The number of yellow cans = 3.
Math Talk
MATHEMATICAL PRACTICES
Generalize What happens when you subtract zero from a number?
Share and Show MATH BOARD
Use the picture to complete the subtraction sentence.
Question 1.
_2_ – 0 = _2_
The number of frogs = 2.
Explanation:
In the above-given question,
given that,
there are 2 frogs.
out of 2 frogs, 2 is sitting.
2 – 0 = 2.
the number of frogs = 2.
Question 2.
_2_ – _2_ = 0
The number of frogs = 0.
Explanation:
In the above-given question,
given that,
there are 2 frogs.
out of 2 frogs, 2 are jumping.
1 – 1 = 0.
The number of frogs = 0.
Question 3.
_3_ – _3_ = 0
The number of frogs = 0.
Explanation:
In the above-given question,
given that,
there are 6 frogs.
out of 6 frogs, 6 are jumping.
3 – 3 = 0.
The number of frogs = 0.
Question 4.
_6_ – 0 = _6_
The number of frogs = 6.
Explanation:
In the above-given question,
given that,
there are 6 frogs.
out of 6 frogs, 6 are sitting.
6 – 0 = 6.
The number of frogs = 6.
Question 5.
_4_ – 0 = _4_
The number of frogs = 4.
Explanation:
In the above-given question,
given that,
there are 4 frogs.
out of 4 frogs, 4 are sitting.
4 – 0 = 4.
The number of frogs = 4.
Question 6.
_2_ – _2_ = 0
The number of frogs = 0.
Explanation:
In the above-given question,
given that,
there are 4 frogs.
out of 4 frogs, 4 are jumping.
2 – 2 = 0.
The number of frogs = 0.
MATHEMATICAL PRACTICE
Use Repeated Reasoning
Complete the subtraction sentence.
Question 7.
1 – 0 = _1_
The number of squirrels = 1.
Explanation:
In the above-given question,
given that,
there is 1 squirrel.
out of 1 squirrel,
1 – 0 = 1.
The number of squirrels = 0.
Question 8.
_0_ = 6 – 6
The number of squirrels = 0.
Explanation:
In the above-given question,
given that,
there is 6 squirrels.
out of 6 squirrels, 6 are jumping.
6 – 6 = 0.
The number of squirrels = 0.
Question 9.
0 = _3_ – 3
The number of squirrels = 3.
Explanation:
In the above-given question,
given that,
there is 3 squirrels.
out of 3 squirrels, 3 are jumping.
3 – 3 = 0.
The number of squirrels = 3.
Question 10.
1 – 1 = _0_
The number of squirrels = 0.
Explanation:
In the above-given question,
given that,
there is 1 squirrel.
out of 1 squirrel, 1 is jumping.
1 – 1 = 0.
The number of squirrels = 0.
Question 11.
0 = _3_ – 3
3 – 0 = 3.
0 = 3 – 3.
Explanation:
In the above-given question,
Given that,
3 – 0 = 3.
0 = 3 – 3.
Question 12.
1 – 1 = _0_
8 = 8 – 0.
1 – 1 = 0
Explanation:
In the above-given question,
Given that,
8 = 8 – 0.
1 – 1 = 0.
Question 13.
7 – _0 = 7
7 – 0 = 7.
Explanation:
In the above-given question,
Given that,
7 – 0 = 7.
Question 14.
8 – 8 = _0__
8 – 8 = 0.
Explanation:
In the above-given question,
Given that,
8 – 8 = 0.
Question 15.
5 – 5 = _0_
5 – 5 = 0.
Explanation:
In the above-given question,
Given that,
5 – 5 = 0.
Question 16.
_0_ = 0 – 0
0 = 0 – 0
Explanation:
In the above-given question,
Given that,
0 = 0 – 0.
GO DEEPER
Choose numbers to complete the subtraction sentence.
Question 17.
_1_ – _1_ = 0
1 – 1 = 0.
Explanation:
In the above-given question,
Given that,
1 – 1 = 0.
Question 18.
_2_ – _2_ = 0
2 – 2 = 0.
Explanation:
In the above-given question,
Given that,
2 – 2 = 0.
Problem Solving • Applications
Write the number sentence and tell how many.
Question 19.
There are 6 bookmarks on the table. 0 are blue and the rest are yellow. How many bookmarks are yellow?
__ yellow bookmarks
The number of yellow bookmarks = 6.
Explanation:
In the above-given question,
given that,
There are 6 bookmarks on the table.
0 are blue and the rest are yellow.
6 – 0 = 6.
The number of yellow bookmarks = 6.
Question 20.
Jared has 8 pictures. He gave some to Wendy. Jared has 0 pictures now. How many pictures did Jared give to Wendy?
The number of pictures jared give to wendy = 8.
Explanation:
In the above-given question,
given that,
jared has 8 pictures.
he gave some to wendy.
jared has 0 pictures.
8 – 0 = 8.
The number of pictures jared give to wendy = 8.
Question 21.
THINK SMARTER
Kevin has 3 fewer leaves than Sandy. Sandy has 3 leaves. How many leaves does Kevin have?
The number of leaves kevin have = 0.
Explanation:
In the above-given question,
given that,
kevin has 3 fewer leaves than sandy.
sandy has 3 leaves.
3 – 3 = 0.
The number of leaves kevin have = 0.
Question 22.
THINK SMARTER
Is the answer correct? Choose Yes or No.
5 – 0 = 5.
5 – 0 = 5.
5 – 5 = 0.
Explanation:
In the above-given question,
given that,
The numbers are
5 – 0 = 5.
5 – 0 = 5.
5 – 5 = 0.
TAKE HOME ACTIVITY
• Have your child explain how 4 − 4 and 4 − 0 are different.
### Subtract All or Zero Homework & Practice 2.7
Complete the subtraction sentence.
Question 1.
3 – 0 = _3_
The number of parrots = 3.
Explanation:
In the above-given question,
given that,
there are 3 parrots.
out of 3 parrots, 3 are sitting.
3 – 0 = 3.
The number of parrots = 3.
Question 2.
2 – 2 = _0_
The number of parrots = 0.
Explanation:
In the above-given question,
given that,
there are 2 parrots.
out of 2 parrots, 2 are jumping.
2 – 2 = 0.
The number of parrots = 0.
Question 3.
5 – 0 = _5_
The number of parrots = 5.
Explanation:
In the above-given question,
given that,
there are 5 parrots.
out of 5 parrots, 5 are sitting.
5 – 0 = 5.
The number of parrots = 5.
Question 4.
_1_ = 1 – 0
The number of parrots = 1.
Explanation:
In the above-given question,
given that,
there is 1 parrot.
out of 1 parrot, 1 is sitting.
1 – 0 = 1.
The number of parrots = 1.
Question 5.
6 – 6 = _0_
The number of parrots = 0.
Explanation:
In the above-given question,
given that,
there are 6 parrots.
out of 6 parrots, 6 are jumping.
6 – 6 = 0.
The number of parrots = 0.
Question 6.
0 = _8_ – 8
The number of parrots = 3.
Explanation:
In the above-given question,
given that,
there are 3 parrots.
out of 3 parrots, 3 are sitting.
3 – 0 = 3.
The number of parrots = 3.
Problem Solving
Write the number sentence and tell how many.
Question 7.
There are 9 books on the shelf. 0 are blue and the rest are green. How many books are green?
_9_ green books
The number of green books = 9
Explanation:
In the above-given question,
given that,
There are 9 books on the table.
0 are blue and the rest are green.
9 – 0 = 9.
The number of green books = 9.
Question 8.
WRITE Math
Use pictures and numbers to show 2 − 0.
The number of parrots = 2.
Explanation:
In the above-given question,
given that,
there are 2 parrots.
out of 2 parrots, 2 are sitting.
2 – 0 = 2.
The number of parrots = 2.
Lesson Check
Question 1.
Complete the subtraction sentence. What is the difference for 4 – 0?
_4_ – _0_ = _4_
4 – 0 = 4.
Explanation:
In the above-given question,
Given that,
4 – 0 = 4.
Question 2.
Complete the subtraction sentence. What is the difference for 6 – 6?
_6_ – _6_ = _0_
6 – 6 = 0.
Explanation:
In the above-given question,
Given that,
6 – 6 = 0.
Spiral Review
Question 3.
Write the number. How many bunnies?
3 bunnies and 3 bunnies _6_ bunnies
The number of bunnies = 6.
Explanation:
In the above-given question,
Given that,
3 bunnies and 3 bunnies.
3 + 3 = 6.
Question 4.
Complete an addition sentence for each model.
9 + 1 = 10.
2 + 7 = 9.
3 + 4 = 7.
4 + 4 = 8.
Explanation:
In the above-given question,
given that,
the cubes
9 + 1 = 10.
2 + 7 = 9.
3 + 4 = 7.
4 + 4 = 8.
### Lesson 2.8 Algebra • Take Apart Numbers
Essential Question
How can you show all the ways to take apart a number?
Listen and Draw
Use to show all the ways to take apart 5. Color and draw to show your work.
Math Talk
MATHEMATICAL PRACTICES
Apply How do you know you showed all the ways?
Share and Show MATH BOARD
Use Color and draw to show how to take apart 9. Complete the subtraction sentence.
Question 3.
9 – _2_ = _7_
9 – 2 = 7.
Explanation:
In the above-given question,
given that,
there are 9 boxes, 2 cross out.
9 – 2 = 7.
Question 4.
9 – _3_ = _6_
9 – 3 = 6.
Explanation:
In the above-given question,
given that,
there are 9 boxes, 3 cross out.
9 – 3 = 6
Question 5.
9 – _0_ = _0_
9 – 0 = 0.
Explanation:
In the above-given question,
given that,
there are 9 boxes, 0 cross out.
9 – 0 = 9.
Question 6.
9 – _2_ = _7_
9 – 2 = 7.
Explanation:
In the above-given question,
given that,
there are 9 boxes, 2 cross out.
9 – 2 = 7.
Question 7.
9 – _3_ = _6_
9 – 3 = 6.
Explanation:
In the above-given question,
given that,
there are 9 boxes, 3 cross out.
9 – 3 = 6.
Question 8.
9 – _4_ = _5_
9 – 4 = 5.
Explanation:
In the above-given question,
given that,
there are 9 boxes, 4 cross out.
9 – 4 = 5.
Question 9.
9 – _5_ = _4_
9 – 5 = 4.
Explanation:
In the above-given question,
given that,
there are 9 boxes, 5 cross out.
9 – 5 = 4.
Question 10.
9 – _6_ = _3_
9 – 6 = 3.
Explanation:
In the above-given question,
given that,
there are 9 boxes, 6 cross out.
9 – 6 = 3.
MATHEMATICAL PRACTICE
Look for a Pattern Use . Color and draw to show how to take apart 10. Complete the subtraction sentence.
Question 11.
10 – _1_ = _9_
10 – 1 = 9.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 1 cross out.
10 – 1 = 9.
Question 12.
10 – _2_ = _8
10 – 2 = 8.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 2 cross out.
10 – 2 = 8.
Question 13.
10 – _3_ = _6
10 – 3 = 6.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 3 cross out.
10 – 3 = 6.
Question 14.
10 – _4_ = _6_
10 – 4 = 6.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 4 cross out.
10 – 4 = 6.
Question 15.
10 – _5_ = _5_
10 – 5 = 5.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 5 cross out.
10 – 5 = 5.
Question 16.
10 – __ = __
10 – 6 = 9.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 6 cross out.
10 – 6 = 4.
Question 17.
10 – __ = __
10 – 7 = 3.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 7 cross out.
10 – 7 = 3.
Question 18.
10 – _8_ = _2_
10 – 8 = 2.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 8 cross out.
10 – 8 = 2.
Question 19.
10 – _9_ = _1_
10 – 9 = 1.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 9 cross out.
10 – 9 = 1.
+Question 20.
10 – __ = __
10 – 10 = 0.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 10 cross out.
10 – 10 = 0.
Question 21.
10 – __ = __
10 – 0 = 10.
Explanation:
In the above-given question,
given that,
there are 10 boxes, 0 cross out.
10 – 0 = 10.
Problem Solving • Applications WRITE Math
Question 22.
James has 10 books. He shares them with his sister. Draw one way he can share the books.
10 – 1 = 9.
Explanation:
In the above-given question,
given that,
james has 10 books, 1 cross out.
10 – 1 = 9.
Question 23.
THINK SMARTER
Hannah has 7 shells. She shares them with Emily. Draw two ways she can share the shells.
7 – 2 = 5.
Explanation:
In the above-given question,
given that,
there are 7 shells, 1 cross out.
7 – 2 = 5.
Question 24.
GO DEEPER
I use 6 marbles to play a game. I lose 1 marble. Then I lose 1 more marble. How many marbles do I have left?
__4_ marbles
The number of marbles i have left = 4.
Explanation:
In the above-given question,
given that,
I use 6 marbles to play a game.
I lose 1 marble.
Then i lose 1 more marble.
6 – 2 = 4.
The number of marbles i have left = 4.
Question 25.
THINK SMARTER
Circle all the models that show a way to take apart 8.
8 – 2 = 6.
8 – 5 = 3
Explanation:
In the above-given question,
given that,
there are 8 boxes, 2 cross out.
8 – 2 = 6.
there are 8 boxes, 5 cross out.
8 – 5 = 3.
TAKE HOME ACTIVITY
• Write 5 − 0 = 5 and 5 − 1 = 4. Ask your child to subtract from 5 another way. Take turns to show all the ways to subtract from 5.
5 – 2 = 3.
5 – 3 = 2.
5 – 4 = 1.
5 – 5 = 0.
Explanation:
In the above-given question,
given that,
5 – 2 = 3.
5 – 3 = 2.
5 – 4 = 1.
5 – 5 = 0.
Algebra • Take Apart Numbers Homework & Practice 2.8
Use Color and draw to show how to take apart 5. Complete the subtraction sentence.
Question 1.
5 – _0_ = _5_
5 – 0 = 5.
Explanation:
In the above-given question,
given that,
there are 5 boxes, 0 cross out.
5 – 0 = 5.
Question 2.
5 – _1_ = _4_
5 – 1 = 5.
Explanation:
In the above-given question,
given that,
there are 5 boxes, 1 cross out.
5 – 1 = 5.
Question 3.
5 – _2_ = _3_
5 – 2 = 3.
Explanation:
In the above-given question,
given that,
there are 5 boxes, 2 cross out.
5 – 2 = 3.
Question 4.
5 – _3_ = _2_
5 – 3 = 2.
Explanation:
In the above-given question,
given that,
there are 5 boxes, 2 cross out.
5 – 3 = 2.
Question 5.
5 – _4_ = _1_
5 – 4 = 1.
Explanation:
In the above-given question,
given that,
there are 5 boxes, 1 cross out.
5 – 4 = 1.
Question 6.
5 – _5_ = _0_
5 – 5 = 0.
Explanation:
In the above-given question,
given that,
there are 5 boxes, 5 cross out.
5 – 5 = 0.
Problem Solving
Solve.
Question 7.
Joe has 9 marbles. He gives them all to his sister. How many marbles does he have now?
_0_ marbles
The number of marbles joe has left = 0.
Explanation:
In the above-given question,
given that,
Joe has 9 marbles.
He gives them all to his sister.
9 – 9 = 0
The number of marbles joe has left = 0.
Question 8.
WRITE Math
Use pictures and numbers to show all the ways to take apart 8.
Lesson Check
Question 1.
Draw the . Show a way to take apart 8. Complete a number sentence to match your model.
8 – __ = __
Spiral Review
Question 2.
What is the sum? Write the sum.
The sum for 6 + 4 = 10.
Explanation:
In the above-given question,
given that,
the sum of numbers 4 and 6.
4 + 6 = 10.
the sum for 4 + 6 = 10.
Question 3.
Solve. There are 7 fish. 3 fish swim away. How many fish are there now?
_4_ fish
The number of fish = 4.
Explanation:
In the above-given question,
given that,
there are 7 fish.
out of 7 fish, 3 fish swim away.
7 – 3 = 4.
The number of fish = 4.
Question 4.
Solve. There are 10 bugs. 8 hop away. How many bugs are there now?
__ bugs
The number of bugs = 2.
Explanation:
In the above-given question,
given that,
there are 10 bugs.
out of 10 bugs, 8 hop away.
10 – 8 = 2.
The number of bugs = 2.
### Lesson 2.9 Subtraction from 10 or Less
Essential Question
Why are some subtraction facts easy to subtract?
Listen and Draw
Draw a picture to show the problem. Then write the subtraction problem two different ways.
The number of fish = 4.
Explanation:
In the above-given question,
given that,
there are 8 fish.
out of 8 fish, 4 fish swim away.
8 – 4 = 4.
The number of fish = 4.
The number of fish = 2.
Explanation:
In the above-given question,
given that,
there are 8 fish.
out of 8 fish, 6 fish swim away.
8 – 6 = 2.
The number of fish = 2.
Math Talk
MATHEMATICAL PRACTICES
Look at the top problem. Explain why the difference is the same.
Model and Draw
Write the subtraction problem.
The number of capsicums = 2.
Explanation:
In the above-given question,
given that,
there are 8 capsicums.
out of 8 capsicums, 6 take away.
8 – 6 = 2.
The number of capsicums = 2.
The number of birds = 3.
Explanation:
In the above-given question,
given that,
there are 7 birds.
out of 7 birds, 4 fly away.
7 – 4 = 3.
The number of birds = 3.
Share and Show MATH BOARD
Write the subtraction problem.
Question 1.
The number of bugs = 6.
Explanation:
In the above-given question,
given that,
there are 10 bugs.
out of 10 bugs, 4 fly away.
10 – 4 = 6.
The number of bugs = 6.
Question 2.
The number of tomatoes = 7.
Explanation:
In the above-given question,
given that,
there are 7 tomatoes.
out of 7 tomatoes, 2 takes away.
7 – 2 = 5.
The number of tomatoes = 7
Question 3.
The number of leaves = 2.
Explanation:
In the above-given question,
given that,
there are 4 leaves.
out of 4 leaves, 2 take away.
4 – 2 = 2.
The number of leaves = 2.
Question 4.
The number of rabbits = 1.
Explanation:
In the above-given question,
given that,
there are 3 rabbits.
out of 3 rabbits, 2 takes away.
3 – 2 = 1.
The number of rabbits = 1.
MATHEMATICAL PRACTICE
Attend to Precision
Write the difference.
Question 5.
The difference for 2 – 1 = 1.
Explanation:
In the above-given question,
given that,
the difference of numbers 2 and 1.
2 – 1 = 1.
the difference for 2 – 1 = 1.
Question 6.
The difference for 3 – 3 = 0.
Explanation:
In the above-given question,
given that,
the difference of numbers 3 and 3.
3 – 3 = 0.
the difference for 3 – 3 = 0.
Question 7.
The difference for 5 – 4 = 1.
Explanation:
In the above-given question,
given that,
the difference of numbers 5 and 4.
5 – 4 = 1.
the difference for 5 – 4 = 1.
Question 8.
The difference for 7 – 3 = 4.
Explanation:
In the above-given question,
given that,
the difference of numbers 7 and 3.
7 – 3 = 4.
the difference for 7 – 3 = 4.
Question 9.
The difference for 6 – 2 = 4.
Explanation:
In the above-given question,
given that,
the difference of numbers 2 and 6.
6 – 2 = 4.
the difference for 6 – 2 = 4.
Question 10.
The difference for 10 – 7 = 3.
Explanation:
In the above-given question,
given that,
the difference of numbers 10 and 7.
10 – 7 = 3.
the difference for 10 – 7 = 3.
Question 11.
The difference for 9 – 9 = 0.
Explanation:
In the above-given question,
given that,
the difference of numbers 9 and 9.
9 – 9 = 0.
the difference for 9 – 9 = 0.
Question 12.
The difference for 8 – 2 = 6.
Explanation:
In the above-given question,
given that,
the difference of numbers 8 and 2.
8 – 2 = 6.
the difference for 8 – 2 = 6.
Question 13.
The difference for 7 – 4 = 3.
Explanation:
In the above-given question,
given that,
the difference of numbers 7 and 4.
7 – 4 = 3.
the difference for 7 – 4 = 3.
Question 14.
The difference for 6 – 3 = 3.
Explanation:
In the above-given question,
given that,
the difference of numbers 6 and 3.
6 – 3 = 3.
the difference for 6 – 3 = 3.
Question 15.
The difference for 8 – 0 = 8.
Explanation:
In the above-given question,
given that,
the difference of numbers 8 and 0.
8 – 0 = 8.
the difference for 8 – 0 = 8.
Question 16.
The difference for 9 – 4 = 5.
Explanation:
In the above-given question,
given that,
the difference of numbers 9 and 4.
9 – 4 = 5.
the difference for 9 – 4 = 5.
Question 17.
THINK SMARTER
Write the number sentence. 9 ants are on a log. 3 ants walk away. How many ants are still on the log?
_9_ – _3_ = _6_
The number of ants = 6.
Explanation:
In the above-given question,
given that,
there are 9 ants.
out of 9 ants, 3 ants walk away.
9 – 3 = 6.
The number of ants = 6.
Question 18.
THINK SMARTER
Explain how the picture shows subtraction.
The number of cherries = 4.
Explanation:
In the above-given question,
given that,
there are 9 cherries.
out of 9 cherries, 5 take away.
9 – 5 = 4.
The number of cherries = 4.
Problem Solving • Applications
Question 19.
GO DEEPER
Draw a picture to show subtraction. Write the subtraction problem to match the picture.
The number of bugs = 4.
Explanation:
In the above-given question,
given that,
there are 7 bugs.
out of 7 bugs, 3 hop away.
7 – 3 = 4.
The number of bugs = 4.
Question 20.
Write the number sentence. 10 ducks are at the pond. All the ducks fly away. How many ducks are still at the pond?
_10_ – _10_ = _0_
The number of ducks = 0.
Explanation:
In the above-given question,
given that,
there are 10 ducks.
out of 10 ducks, 10 fly away.
10 – 10 = 0.
The number of ducks = 0.
Question 21.
THINK SMARTER
Write all of the subtraction sentences that can tell the story. Tell how you know. Max has 6 carrots. He eats more than 1 carrot. He has more than 1 carrot left. How many carrots are left?
___________________
___________________
___________________
The number of carrots = 2.
Explanation:
In the above-given question,
given that,
Max has 6 carrots.
He eats more than 1 carrot.
6 – 4 = 2.
The number of carrots = 2.
TAKE HOME ACTIVITY
• Tell your child a subtraction problem. Have your child write the problem to subtract two different ways. Then have your child find the difference.
### Subtraction from 10 or Less Homework & Practice 2.9
Write the difference.
Question 1.
The difference for 5 – 1 = 4.
Explanation:
In the above-given question,
given that,
the difference of numbers 5 and 1.
5 – 1 = 4.
the difference for 5 – 1 = 4.
Question 2.
The difference for 3 – 2 = 1.
Explanation:
In the above-given question,
given that,
the difference of numbers 3 and 2.
3 – 2 = 1.
the difference for 3 – 2 = 1.
Question 3.
The difference for 8 – 3 = 5.
Explanation:
In the above-given question,
given that,
the difference of numbers 8 and 3.
8 – 3 = 5.
the difference for 8 – 3 = 5.
Question 4.
The difference for 6 – 4 = 2.
Explanation:
In the above-given question,
given that,
the difference of numbers 6 and 4.
6 – 4 = 2.
the difference for 6 – 4 = 2.
Question 5.
The difference for 7 – 0 = 7.
Explanation:
In the above-given question,
given that,
the difference of numbers 7 and 0.
7 – 0 = 7.
the difference for 7 – 0 = 7.
Question 6.
The difference for 5 – 3 = 2.
Explanation:
In the above-given question,
given that,
the difference of numbers 5 and 3.
5 – 3 = 2.
the difference for 5 – 3 = 2.
Question 7.
The difference for 4 – 4 = 0.
Explanation:
In the above-given question,
given that,
the difference of numbers 4 and 4.
4 – 4 = 0.
the difference for 4 – 4 = 0.
Question 8.
The difference for 8 – 1 = 7.
Explanation:
In the above-given question,
given that,
the difference of numbers 8 and 1.
8 – 1 = 7.
the difference for 8 – 1 = 7.
Problem Solving
Solve.
Question 9.
6 birds are in the tree. None of the birds fly away. How many birds are left?
_6_ – _0_ = _6_
The number of birds = 6.
Explanation:
In the above-given question,
given that,
there are 6 birds.
out of 6 birds, 0 fly away.
6 – 0 = 6.
The number of birds = 6.
Question 10.
WRITE Math Find 10 – 3. Write the subtraction fact two ways.
___________________
___________________
___________________
The difference for 10 – 3 = 6.
Explanation:
In the above-given question,
given that,
the difference of numbers 10 and 3.
10 – 3 = 6.
the difference for 10 – 3 = 6.
Lesson Check
Question 1.
Write the difference.
The difference for 4 – 0 = 4.
Explanation:
In the above-given question,
given that,
the difference of numbers 4 and 0.
4 – 0 = 4.
the difference for 4 – 0 = 4.
Spiral Review
Question 2.
Solve. Write a number sentence. There are 8 pens. 3 pens are blue. The rest are red. How many pens are red?
_8_ – _3_ = 5
The number of red pens = 5.
Explanation:
In the above-given question,
given that,
there are 8 pens.
out of 8 pens, 3 pens are blue.
8 – 3 = 5.
The number of red pens = 5.
Question 3.
Circle the number sentences that show the same addends in a different order.
The number sentences that show the same addends = 5 + 4 = 9 and 4 + 5 = 9.
Explanation:
In the above-given question,
given that,
The number sentences are
4 – 9 = 5, 5 + 4 = 9 and 4 + 5 = 9.
The number sentences that show the same addends = 5 + 4 = 9 and 4 + 5 = 9.
### Subtraction Concepts Chapter 2 Review/Test
Question 1.
Circle the part you are taking from the group. Then cross it out. Write how many there are now.
5 zebras 3 zebras walk away. __ zebras now
The number of zebras = 2.
Explanation:
In the above-given question,
given that,
there are 5 zebras.
out of 5 zebras, 3 walk away.
5 – 3 = 2.
The number of zebras = 2.
GO DEEPER
Circle the part you take away from the group. Then cross it out. Write the difference.
Question 2.
There are 6 cats. 5 cats run away.
6 – 5 = _1_
The number of cats = 1.
Explanation:
In the above-given question,
given that,
there are 6 cats.
out of 6 cats, 5 walk away.
6 – 5 = 1.
The number of cats = 1.
Question 3.
There are 4 dogs. 1 dog runs away.
4 – 1 = _3_
The number of dogs = 3.
Explanation:
In the above-given question,
given that,
there are 4 dogs.
out of 4 dogs, 1 walk away.
4 – 1 = 3.
The number of dogs = 3.
Question 4.
Is the subtraction sentence true? Choose Yes or No.
5 – 5 = 0.
2 – 2 = 0.
4 – 0 = 4
Explanation:
In the above-given question,
given that,
the numbers are
5 – 5 = 0.
2 – 2 = 0.
4 – 0 = 4
Question 5.
Color to solve. Write the number sentence and how many.
There are 9 pencils. 5 pencils are red. The rest are yellow. How many pencils are yellow?
__ yellow pencils
Question 6.
Read the problem. Use the model to solve. Complete the model and the number sentence.
There are 6 frogs on a log. 1 frog is big. The others are small. How many frogs are small?
6 – 1 = _5_
The small frogs = 5.
Explanation:
In the above-given question,
given that,
There are 6 frogs.
1 frog is big.
6 – 1 = 5.
the small frogs = 5.
Question 7.
Look at the picture. How many fewer bats are there than balls? Choose the number.
The fewer bats = 2.
Explanation:
In the above-given question,
given that,
There are 5 balls.
There are 3 bats.
5 – 3 = 2.
the fewer bats = 2.
Question 8.
Read the problem. Use the bar model to solve. Maria has 2 rocks. Peter has 8 rocks. How many more rocks does Peter have than Maria?
_6_ rocks
The rocks does peter have = 6.
Explanation:
In the above-given question,
given that,
Peter has 8 rocks.
Maria has 2 rocks.
8 – 2 = 6.
the rocks peter have = 6.
Question 9.
The models show two ways to take apart 6. Complete the subtraction sentences. Use these numbers.
The difference for 6 – 1 = 5.
Explanation:
In the above-given question,
given that,
the difference of numbers 6 and 1.
6 – 1 = 5.
the difference for 6 – 1 = 5.
The difference for 6 – 2 = 4.
Explanation:
In the above-given question,
given that,
the difference of numbers 6 and 2.
6 – 2 = 4.
the difference for 6 – 2 = 4.
Question 10.
Write the subtraction sentence in the box that shows the difference.
The difference for 10 – 5 = 5.
Explanation:
In the above-given question,
given that,
the difference of numbers 10 and 5.
10 – 5 = 5.
the difference for 10 – 5 = 5.
The difference for 5 – 1 = 4.
Explanation:
In the above-given question,
given that,
the difference of numbers 5 and 1.
5 – 1 = 4.
the difference for 5 – 1 = 4.
The difference for 7 – 4 = 3.
Explanation:
In the above-given question,
given that,
the difference of numbers 7 and 4.
7 – 4 = 3.
the difference for 7 – 4 = 3.
Question 11.
THINK SMARTER
Read the problem. Draw a model to solve. Complete the number sentence.
Mr. Bear catches 8 fish. He takes 3 fish home. He throws the other fish back in the water. How many fish does he throw back in the water?
_8_ – _3_= _5_ fish
The number of fishes = 5.
Explanation:
In the above-given question,
given that,
there are 8 fishes.
out of 8 fish, he takes 3 fish home.
8 – 3 = 5.
The number of fish = 5.
Question 12.
Write the subtraction sentence the picture shows.
_8_ – _3_ = _5_
Explain.
____________
____________
____________ |
# 0.2 Practice tests (1-4) and final exams (Page 17/36)
Page 22 / 36
8 . The 99% confidence interval, because it includes all but one percent of the distribution. The 95% confidence interval will be narrower, because it excludes five percent of the distribution.
## 8.2: confidence interval, single population mean, standard deviation unknown, student’s t
9 . The t -distribution will have more probability in its tails (“thicker tails”) and less probability near the mean of the distribution (“shorter in the center”).
10 . Both distributions are symmetrical and centered at zero.
11 . df = n – 1 = 20 – 1 = 19
12 . You can get the t -value from a probability table or a calculator. In this case, for a t -distribution with 19 degrees of freedom, and a 95% two-sided confidence interval, the value is 2.093, i.e.,
The calculator function is invT(0.975, 19).
13 .
98.4 ± 0.14 = (98.26, 98.54).
The calculator function Tinterval answer is (98.26, 98.54).
14 . ${t}_{\frac{\alpha }{2}}=2.861.$ The calculator function is invT(0.995, 19).
$EBM={t}_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=\left(2.861\right)\left(\frac{0.3}{\sqrt{20}}\right)=0.192$
98.4 ± 0.19 = (98.21, 98.59). The calculator function Tinterval answer is (98.21, 98.59).
15 . df = n – 1 = 30 – 1 = 29.
98.4 ± 0.11 = (98.29, 98.51). The calculator function Tinterval answer is (98.29, 98.51).
## 8.3: confidence interval for a population proportion
16 . ${p}^{\prime }=\frac{280}{500}=0.56$
${q}^{\prime }=1-{p}^{\prime }=1-0.56=0.44$
$s=\sqrt{\frac{pq}{n}}=\sqrt{\frac{0.56\left(0.44\right)}{500}}=0.0222$
17 . Because you are using the normal approximation to the binomial, ${z}_{\frac{\alpha }{2}}=1.96$ .
Calculate the error bound for the population ( EBP ):
Calculate the 95% confidence interval:
0.56 ± 0.0435 = (0.5165, 0.6035).
The calculator function 1-PropZint answer is (0.5165, 0.6035).
18 . ${z}_{\frac{\alpha }{2}}=1.64$
0.56 ± 0.03 = (0.5236, 0.5964). The calculator function 1-PropZint answer is (0.5235, 0.5965)
19 . ${z}_{\frac{\alpha }{2}}=2.58$
0.56 ± 0.05 = (0.5127, 0.6173).
The calculator function 1-PropZint answer is (0.5028, 0.6172).
20 . EBP = 0.04 (because 4% = 0.04)
${z}_{\frac{\alpha }{2}}=1.96$ for a 95% confidence interval
You need 601 subjects (rounding upward from 600.25).
21 .
You need 577 subjects (rounding upward from 576.24).
22 .
You need 1,068 subjects (rounding upward from 1,067.11).
## 9.1: null and alternate hypotheses
23 . H 0 : p = 0.58
H a : p ≠ 0.58
24 . H 0 : p ≥ 0.58
H a : p <0.58
25 . H 0 : μ ≥ $268,000 H a : μ <$268,000
26 . H a : μ ≠ 107
27 . H a : p ≥ 0.25
## 9.2: outcomes and the type i and type ii errors
28 . a Type I error
29 . a Type II error
30 . Power = 1 – β = 1 – P (Type II error).
31 . The null hypothesis is that the patient does not have cancer. A Type I error would be detecting cancer when it is not present. A Type II error would be not detecting cancer when it is present. A Type II error is more serious, because failure to detect cancer could keep a patient from receiving appropriate treatment.
32 . The screening test has a ten percent probability of a Type I error, meaning that ten percent of the time, it will detect TB when it is not present.
Frequency find questions
What is nominal variable
Write short notes on, nominal variable, ordinal variable, internal variable, ratio variable.
olusola
P( /x-50/ less than or equal to 5 ) where mean =52 and Variance =25
how I get the mcq
the exploration and analysis of large data to discover meaningful patterns and rules
Hussein
how do we calculate the median
f(x)=cx(1-x)^4 as x range 4rm 0<=x<=1. Can someone pls help me find d constant C. By integration only..
uses of statistics in Local Government
Hi
Tamuno
hello
Saleema
Atul
District statistical officer
Atul
statistical services
Atul
Please is this part of the IMT program
Tamuno
testing of drugs
Shambhavi
hii 2
Qamar-ul-
Tamuno
Hello every one
Okoi
sample survey is done by local government in each and every field.
syeda
statistics is used in almost every government organisations such as health department, economic department, census, weather forecasting fields
raghavendra
that's true
syeda
statistics is one of the tool that represents the falling and rising of any cases in one sheet either that is in population census whether forecast as well as economic growth
statistic is a technique, and statistics is a subject
syeda
Probability tells you the likelihood of an event happening. ... The higher the probability, the more likely it is to happen. Probability is a number or fraction between 0 and 1. A probability of 1 means something will always happen, and a probability of 0 means something will never happen...
Saying it's a number between zero and one means it is a fraction so you could remove "or fraction" from you definition.
Carlos
wouldn't be correct to remove fractions, saying a number is justified as probabilities can also be decimals between 0 and 1.
Denzel
Saying "a number" will include it being a decimal which are themselves fractions in another form.
Carlos
I will simply say a probability is a number in the range zero to one, inclusive.
Carlos
f#\$
Carlos
How to delete an entry? This last one was a pocket print.
Carlos
what is probability
chance of occurrence
Sikander
what is data
raw facts and figures
Sikander
information of any kind
Tahir
What is Statistic
what statistical analysis can i run on growth and yield of spinach.
guillio
format of the frequency distribution table
henry
what is pearson correlation coefficient indicates?
Eticha
Statistic is the mean of the sample.
Raman
can anyone determine the value of c and the covariance and correlation for the joint probability density function Fxy(x,y)=c over the range 0<x<5,0<y,and x-1<y<x-1.
Nuhu
what actually is the definition of range
I need social statistics materials
Chinedu
the range of a set of data is the difference between the largest and smallest values
La
I need more explanation about cluster sampling
Hafsat
write the set of old number that are greater than or equal to minutes 7 butl less than 5 in both of the set notation |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# Sequences
While the idea of a sequence of numbers, $$a_1,a_2,a_3,\ldots$$ is straightforward, it is useful to think of a sequence as a function. We have up until now dealt with functions whose domains are the real numbers, or a subset of the real numbers, like $$f(x)=\sin x$$. A sequence is a function with domain the natural numbers $$N=\{1,2,3,\ldots\}$$ or the non-negative integers, $$Z^{\ge0}=\{0,1,2,3,\ldots\}$$. The range of the function is still allowed to be the real numbers; in symbols, we say that a sequence is a function $$f\colon N \to R$$. Sequences are written in a few different ways, all equivalent; these all mean the same thing:
$\displaylines{a_1,a_2,a_3,\ldots\cr \left\{a_n\right\}_{n=1}^{\infty}\cr \left\{f(n)\right\}_{n=1}^{\infty}\cr}$
As with functions on the real numbers, we will most often encounter sequences that can be expressed by a formula. We have already seen the sequence $$a_i=f(i)=1-1/2^i$$, and others are easy to come by:
\eqalign{ f(i)&={i\over i+1}\cr f(n)&={1\over2^n}\cr f(n)&=\sin(n\pi/6)\cr f(i)&={(i-1)(i+2)\over2^i}\cr }
Frequently these formulas will make sense if thought of either as functions with domain $$\mathcal{R}$$ or $$\mathcal{N}$$, though occasionally one will make sense only for integer values.
Faced with a sequence we are interested in the limit $$\lim_{i\to \infty} f(i) = \lim_{i\to\infty} a_i.$$ We already understand $$\lim_{x\to\infty} f(x)$$ when $$x$$ is a real valued variable; now we simply want to restrict the "input'' values to be integers. No real difference is required in the definition of limit, except that we specify, perhaps implicitly, that the variable is an integer. Compare this definition to definition 4.10.2.
Definition 11.1.1
Suppose that $$\left\{a_n\right\}_{n=1}^{\infty}$$ is a sequence. We say that $$\lim_{n\to \infty}a_n=L$$ if for every $$\epsilon>0$$ there is an $$N > 0$$ so that whenever $$n>N$$, $$|a_n-L| < \epsilon$$. If $$\lim_{n\to\infty}a_n=L$$ we say that the sequence converges, otherwise it diverges.
If $$f(i)$$ defines a sequence, and $$f(x)$$ makes sense, and $$\lim_{x\to\infty}f(x)=L$$, then it is clear that $$\lim_{i\to\infty}f(i)=L$$ as well, but it is important to note that the converse of this statement is not true. For example, since $$\lim_{x\to\infty}(1/x)=0$$, it is clear that also $$\lim_{i\to\infty}(1/i)=0$$, that is, the numbers
${1\over1},{1\over2},{1\over3},{1\over4},{1\over5},{1\over6},\ldots$
get closer and closer to 0. Consider this, however: Let $$f(n)=\sin(n\pi)$$.
This is the sequence
$\sin(0\pi), \sin(1\pi),\sin(2\pi),\sin(3\pi),\ldots=0,0,0,0,\ldots$
since
$\sin(n\pi)=0$
when $$n$$ is an integer. Thus $$\lim_{n\to\infty}f(n)=0$$. But $$\lim_{x\to\infty}f(x)$$, when $$x$$ is real, does not exist: as $$x$$ gets bigger and bigger, the values $$\sin(x\pi)$$ do not get closer and closer to a single value, but take on all values between $$-1$$ and $$1$$ over and over. In general, whenever you want to know $$\lim_{n\to\infty}f(n)$$ you should first attempt to compute $$\lim_{x\to\infty}f(x)$$, since if the latter exists it is also equal to the first limit. But if for some reason $$\lim_{x\to\infty}f(x)$$ does not exist, it may still be true that $$\lim_{n\to\infty}f(n)$$ exists, but you'll have to figure out another way to compute it.
It is occasionally useful to think of the graph of a sequence. Since the function is defined only for integer values, the graph is just a sequence of dots. In figure 11.1.1 we see the graphs of two sequences and the graphs of the corresponding real functions.
{C}{C}
Figure 11.1.1. Graphs of sequences and their corresponding real functions.
Not surprisingly, the properties of limits of real functions translate into properties of sequences quite easily. Theorem 2.3.6 about limits becomes
Theorem 11.1.2
Suppose that $$\lim_{n\to\infty}a_n=L$$ and $$\lim_{n\to\infty}b_n=M$$ and $$k$$ is some constant. Then
\eqalign{ &\lim_{n\to\infty} ka_n = k\lim_{n\to\infty}a_n=kL\cr &\lim_{n\to\infty} (a_n+b_n) = \lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n=L+M\cr &\lim_{n\to\infty} (a_n-b_n) = \lim_{n\to\infty}a_n-\lim_{n\to\infty}b_n=L-M\cr &\lim_{n\to\infty} (a_nb_n) = \lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}b_n=LM\cr &\lim_{n\to\infty} {a_n\over b_n} = {\lim_{n\to\infty}a_n\over \lim_{n\to\infty}b_n}={L\over M},\hbox{ if $$M$$ is not 0}\cr }
{C}
Likewise the Squeeze Theorem (4.3.1) becomes
Theorem 11.1.3
Suppose that
$a_n \le b_n \le c_n$
for all $$n>N$$, for some $$N$$. If
$\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n=L,$
then
$\lim_{n\to\infty}b_n=L.$
{C}
And a final useful fact:
Theorem 11.1.4
$\lim_{n\to\infty}|a_n|=0$
if and only if
$\lim_{n\to\infty}a_n=0.$
This theorem says simply that the size of $$a_n$$ gets close to zero if and only if $$a_n$$ gets close to zero.
Example 11.1.5
Determine whether $$\left\{{n\over n+1}\right\}_{n=0}^{\infty}$$ converges or diverges. If it converges, compute the limit.
SOLUTION
Since this makes sense for real numbers we consider
$\lim_{x\to\infty}{x\over x+1}=\lim_{x\to\infty}1-{1\over x+1}=1-0=1.$
Thus the sequence converges to 1.
{C}
Example 11.1.6
Determine whether $$\bigg\{{\ln n\over n}\bigg\}_{n=1}^{\infty}$$ converges or diverges. If it converges, compute the limit.
SOLUTION
We compute $$\lim_{x\to\infty}{\ln x\over x}=\lim_{x\to\infty}{1/x\over 1}= 0,$$ using L'Hôpital's Rule. Thus the sequence converges to 0.
Example 11.1.7
Determine whether $$\{(-1)^n\}_{n=0}^{\infty}$$ converges or diverges. If it converges, compute the limit.
SOLUTION
This does not make sense for all real exponents, but the sequence is easy to understand: it is $$1,-1,1,-1,1\ldots$$ and clearly diverges.
{C}
Example 11.1.8
Determine whether $$\{(-1/2)^n\}_{n=0}^{\infty}$$ converges or diverges. If it converges, compute the limit.
SOLUTION
We consider the sequence
$\{|(-1/2)^n|\}_{n=0}^{\infty}=\{(1/2)^n\}_{n=0}^{\infty}$. Then $\lim_{x\to\infty}\left({1\over2}\right)^x=\lim_{x\to\infty}{1\over2^x}=0,$
so by theorem 11.1.4 the sequence converges to 0.
Example 11.1.9
Determine whether $$\{(\sin n)/\sqrt{n}\}_{n=1}^{\infty}$$ converges or diverges. If it converges, compute the limit.
SOLUTION
Since $$|\sin n|\le 1$$, $$0\le|\sin n/\sqrt{n}|\le 1/\sqrt{n}$$ and we can use theorem 11.1.3 with $$a_n=0$$ and $$c_n=1/\sqrt{n}$$. Since $$\lim_{n\to\infty} a_n=\lim_{n\to\infty} c_n=0$$, $$\lim_{n\to\infty}\sin n/\sqrt{n}=0$$ and the sequence converges to 0.
Example 11.1.10
A particularly common and useful sequence is $$\{r^n\}_{n=0}^{\infty}$$, for various values of $$r$$. Some are quite easy to understand: If $$r=1$$ the sequence converges to 1 since every term is 1, and likewise if $$r=0$$ the sequence converges to 0. If $$r=-1$$ this is the sequence of example 11.1.7 and diverges. If $$r>1$$ or $$r < -1$$ the terms $$r^n$$ get large without limit, so the sequence diverges. If $$0 < r < 1$$ then the sequence converges to 0. If $$-1 < r < 0$$ then $$|r^n|=|r|^n$$ and $$0 < |r| < 1$$, so the sequence $$\{|r|^n\}_{n=0}^{\infty}$$ converges to 0, so also $$\{r^n\}_{n=0}^{\infty}$$ converges to 0. converges. In summary, $$\{r^n\}$$ converges precisely when $$-1 < r\le1$$ in which case $$\lim_{n\to\infty} r^n=\cases{ 0& if \(-1 < r < 1$$\cr 1& if $$r=1$$\cr} \)
Sometimes we will not be able to determine the limit of a sequence, but we still would like to know whether it converges. In some cases we can determine this even without being able to compute the limit.
A sequence is called increasing or sometimes strictly increasing if $$a_i < a_{i+1}$$ for all $$i$$. It is called non-decreasing or sometimes (unfortunately) increasing if $$a_i\le a_{i+1}$$ for all $$i$$. Similarly a sequence is decreasing if $$a_i>a_{i+1}$$ for all $$i$$ and non-increasing if $$a_i\ge a_{i+1}$$ for all $$i$$. If a sequence has any of these properties it is called monotonic.
Example 11.1.11
The sequence
$\left\{{2^i-1\over2^i}\right\}_{i=1}^{\infty}= {1\over2},{3\over4},{7\over8},{15\over16},\ldots, \) is increasing, and \[ \left\{{n+1\over n}\right\}_{i=1}^{\infty}= {2\over1},{3\over2},{4\over3},{5\over4},\ldots$
is decreasing.
A sequence is bounded above if there is some number $$N$$ such that $$a_n\le N$$ for every $$n$$, and bounded below if there is some number $$N$$ such that $$a_n\ge N$$ for every $$n$$. If a sequence is bounded above and bounded below it is bounded. If a sequence $$\{a_n\}_{n=0}^{\infty}$$ is increasing or non-decreasing it is bounded below (by $$a_0$$), and if it is decreasing or non-increasing it is bounded above (by $$a_0$$). Finally, with all this new terminology we can state an important theorem.
Theorem 11.1.12
If a sequence is bounded and monotonic, then it converges.
We will not prove this; the proof appears in many calculus books. It is not hard to believe: suppose that a sequence is increasing and bounded, so each term is larger than the one before, yet never larger than some fixed value $$N$$. The terms must then get closer and closer to some value between $$a_0$$ and $$N$$. It need not be $$N$$, since $$N$$ may be a "too-generous'' upper bound; the limit will be the smallest number that is above all of the terms $$a_i$$.
Example 11.1.13
All of the terms $$(2^i-1)/2^i$$ are less than 2, and the sequence is increasing. As we have seen, the limit of the sequence is 1---1 is the smallest number that is bigger than all the terms in the sequence. Similarly, all of the terms $$(n+1)/n$$ are bigger than $$1/2$$, and the limit is 1---1 is the largest number that is smaller than the terms of the sequence.
We do not actually need to know that a sequence is monotonic to apply this theorem---it is enough to know that the sequence is "eventually'' monotonic, that is, that at some point it becomes increasing or decreasing. For example, the sequence $$10$$, $$9$$, $$8$$, $$15$$, $$3$$, $$21$$, $$4$$, $$3/4$$, $$7/8$$, $$15/16$$, $$31/32,\ldots$$ is not increasing, because among the first few terms it is not. But starting with the term $$3/4$$ it is increasing, so the theorem tells us that the sequence $$3/4, 7/8, 15/16, 31/32,\ldots$$ converges. Since convergence depends only on what happens as $$n$$ gets large, adding a few terms at the beginning can't turn a convergent sequence into a divergent one.
Example 11.1.14
Show that $$\{n^{1/n}\}$$ converges.
SOLUTION
We first show that this sequence is decreasing, that is, that $$n^{1/n}> (n+1)^{1/(n+1)}$$. Consider the real function $$f(x)=x^{1/x}$$ when $$x\ge1$$. We can compute the derivative, $$f'(x)=x^{1/x}(1-\ln x)/x^2$$, and note that when $$x\ge 3$$ this is negative. Since the function has negative slope, $$n^{1/n}> (n+1)^{1/(n+1)}$$ when $$n\ge 3$$. Since all terms of the sequence are positive, the sequence is decreasing and bounded when $$n\ge3$$, and so the sequence converges. (As it happens, we can compute the limit in this case, but we know it converges even without knowing the limit; see exercise 1.)
Example 11.1.15
Show that $$\{n!/n^n\}$$ converges.
SOLUTION
Again we show that the sequence is decreasing, and since each term is positive the sequence converges. We can't take the derivative this time, as $$x!$$ doesn't make sense for $$x$$ real. But we note that if $$a_{n+1}/a_n < 1$$ then $$a_{n+1} < a_n$$, which is what we want to know. So we look at
$a_{n+1}/a_n: {a_{n+1}\over a_n} = {(n+1)!\over (n+1)^{n+1}}{n^n\over n!}= {(n+1)!\over n!}{n^n\over (n+1)^{n+1}}= {n+1\over n+1}\left({n\over n+1}\right)^n= \left({n\over n+1}\right)^n < 1.$
(Again it is possible to compute the limit; see exercise 2.)
### Exercises 11.1
Ex 11.1.1 Compute $$\lim_{x\to\infty} x^{1/x}$$. (answer)
Ex 11.1.2 Use the squeeze theorem to show that $$\lim_{n\to\infty} {n!\over n^n}=0$$.
Ex 11.1.3 Determine whether $$\{\sqrt{n+47}-\sqrt{n}\}_{n=0}^{\infty}$$ converges or diverges. If it converges, compute the limit. (answer)
Ex 11.1.4 Determine whether $$\left\{{n^2+1\over (n+1)^2}\right\}_{n=0}^{\infty}$$ converges or diverges. If it converges, compute the limit. (answer)
Ex 11.1.5 Determine whether $$\left\{{n+47\over\sqrt{n^2+3n}}\right\}_{n=1}^{\infty}$$ converges or diverges. If it converges, compute the limit. (answer)
Ex 11.1.6 Determine whether $$\left\{{2^n\over n!}\right\}_{n=0}^{\infty}$$ converges or diverges. (answer) |
# Lesson 9-5 Dilations.
## Presentation on theme: "Lesson 9-5 Dilations."— Presentation transcript:
Lesson 9-5 Dilations
5-Minute Check on Lesson 9-5
Transparency 9-6 5-Minute Check on Lesson 9-5 Determine whether each regular polygon tessellates the plane. Explain quadrilateral octagon gon Determine whether a semi-regular tessellation can be created from each figure. Assume each figure has a side length of 1 unit. 4. triangle and square pentagon and square Which regular polygon will not tessellate the plane? Yes, Square’s interior angle = 90 = 4 No, Octagon’s interior angle = 135 ≠ integer No, 15-gon’s interior angle = 156 ≠ integer Yes, 360 + 290 = 360 No, 108n + 90m ≠ 360 Standardized Test Practice: A triangle B quadrilateral C C pentagon D hexagon Click the mouse button or press the Space Bar to display the answers.
Objectives Determine whether a dilation is an enlargement, a reduction, or a congruence transformation enlargement – gotten larger |r| > 1 reduction – gotten smaller |r| < 1 congruence transformation – stayed the same |r| = 1 where r is the scaling factor (similar triangles)
Vocabulary Dilation – a transformation that may change the size of a figure
Dilations y x Five hexagons that are each 50% reduction (1/2) of the size of the one before it Scale factor is ½, and the center point is the origin (center of the figures)
Dilations y x A B A’ B’ CP A* B* Small dashed lines are the rays along which the dilations occur and the center point is CP . AB = 16, A’B’ = 8 so r = ½ (a reduction |r|<1) A*B* = 8 but since its on the opposite side of CP, r = - ½ (negative = opposite, but still a reduction)
Dilations |r| > 1 Enlargement 1 < r |r| < 1 Reduction A
y x |r| > 1 Enlargement |r| < 1 Reduction |r| = 1 Congruence Transformation 1 < r A B 0 < r < 1 C CP – Center Point CP These are the lines that the figure’s vertices “travel” on as the figure is dilated C’ -1 < r < 0 A’B’C’ is with r = -1. B’ A’ -1 > r
b. Find the measure of the dilation image or the preimage
a. Find the measure of the dilation image or the preimage of using the given scale factor of CD = 15, r = 3. Dilation Theorem Multiply. Answer: 45 b. Find the measure of the dilation image or the preimage of using the given scale factor C’D’ = 7, r = - ⅔. Dilation Theorem Multiply each side by Answer: 10.5 Example 5-1a
Find the measure of the dilation image or the preimage of
Find the measure of the dilation image or the preimage of using the given scale factor. a. b. Answer: 32 Answer: 36 Example 5-1b
Draw the dilation image of trapezoid PQRS with center C and r = - 3
Since |r| > 1 the dilation is an enlargement of trapezoid PQRS. R' S' P' Q' Draw Since r is negative, P', Q', R', and S' will lie on respectively. Locate P', Q', R', and S' so that Example 5-2a
Draw the dilation image of trapezoid PQRS with center C and
Since the dilation is an enlargement of trapezoid PQRS. R' S' P' Q' Answer: Draw trapezoid P'Q'R'S'. Example 5-2a
Answer: A'(–2, 2), B'(4, –4), C' (–2, –4)
COORDINATE GEOMETRY Triangle ABC has vertices A(–1, 1), B(2, –2), and C(–1, –2). Find the image of ABC after a dilation centered at the origin with a scale factor of 2. Sketch the preimage and the image. Answer: A'(–2, 2), B'(4, –4), C' (–2, –4) Example 5-3b
image length preimage length
Determine the scale factor used for the dilation with center C. Determine whether the dilation is an enlargement, reduction, or congruence transformation. image length preimage length Simplify. Since the image is on the opposite side of C from the preimage, the scale factor is negative. So the scale factor is –1. The absolute value of the scale factor equals 1, so the dilation is a congruence transformation. Answer: –1; congruence transformation Example 5-4a
Determine the scale factor used for each dilation with center C
Determine the scale factor used for each dilation with center C. Determine whether the dilation is an enlargement, reduction, or congruence transformation. a b. Answer: reduction Answer: 2; enlargement Example 5-4b
MULTIPLE-CHOICE TEST ITEM Sharetta built a frame for a photograph that is 20 centimeters by 25 centimeters. The frame measures 400 millimeters by 500 millimeters. Which scale factor did she use? A 2 B 3 C D Read the Test Item The photograph’s dimensions are given in centimeters, and the frame’s dimensions are in millimeters. You need to convert from millimeters to centimeters in the problem. Example 5-5a
Step 1 Convert from millimeters to centimeters.
Solve the Test Item Step 1 Convert from millimeters to centimeters. or 40 centimeters or 20 centimeters Step 2 Find the scale factor. frame length photo length Simplify. Step 3 Sharetta used a scale factor of 2 to build the frame. Choice A is the correct answer. Answer: A Example 5-5a
factor of What size poster board will he need to
MULTIPLE-CHOICE TEST ITEM Ruben is making a scale drawing of the front of his house. His house is 48 feet wide and 30 feet high at its highest point. Ruben decides on a dilation reduction factor of What size poster board will he need to make a complete drawing? A 19 in. by 26 in. B 22 in. by 30 in. C in. by 28 in. D 16 in. by 29 in. Answer: B Example 5-5b
Summary & Homework Summary: Homework: Dilations can be
enlargements, (|scaling factor| > 1) reductions, or (|scaling factor| < 1 congruence transformations (|scaling factor|=1) Homework: pg ; 14, 15, 20-23, 30-35
Similar presentations |
# SOCR EduMaterials Activities Distributions
(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)
Jump to: navigation, search
## SOCR Educational Materials - Activities - SOCR Computing Probabilities Activity
Goal: Using SOCR to compute probability for different distribution.
Example: Here we use normal distribution (mean=5, sigma=4) as an example.
Step 1: Set up parameters. Select the Normal Distribution from the drop-down list on the top-left. Set Mean = 5 Standard Deviation = 4
Now the density function of normal (mean=5, sigma=4) will show up in the top-right with red color. See following figure.
Step 2: Compute P(X<7) Move cursor from 21, the most right side, to the right until 7. Now the red area means X<7. And the probability of red area is always represented by the value of “Between”. Hence P(X<7)=0.69. See attached figure.
To get a snapshot of this result, click on snapshot. Then save the file with a filename and extension .jpeg , for example, hwk1-parta.jpeg. When done, you can open that new .jpeg file and go to edit-> copy.. Open separately a Word file and paste into that file. You can reduce the image, but make sure that it is big enough for us to see the numbers
Step 3: Compute P(X>3) Move cursor from -11, the most right side, to the left until 3. Red area means X>3. And the probability of red area is always recorded in the value of “Between”. Therefore, P(X>3)=0.69. See attached figure.
Step 4: Compute P(2<X<6)
First, move cursor from -11, the most left side, to the right until 2. Then move cursor from 21, the most right side, to the left until 6. Find “Between” in the bottom-right widow, which is just the probability of red area. Then P(2<X<6)=0.375. See attached figure.
Similarly, we can use SOCR to compute the probability in other known distribution, such as uniform distribution, exponential distribution. |
# How do you find the limit of (2x) / (x+7sqrt(x)) as x approaches 0?
Oct 9, 2016
${\lim}_{x \rightarrow 0} \frac{2 x}{x + 7 \sqrt{x}} = 0$
#### Explanation:
${\lim}_{x \rightarrow 0} \frac{2 x}{x + 7 \sqrt{x}}$
Factor $\sqrt{x}$ from the denominator:
$= {\lim}_{x \rightarrow 0} \frac{2 x}{\sqrt{x} \left(\sqrt{x} + 7\right)}$
Dividing $\frac{\sqrt{x}}{\sqrt{x}}$:
$= {\lim}_{x \rightarrow 0} \frac{2 \sqrt{x}}{\sqrt{x} + 7}$
Now we can evaluate the limit without having a denominator of $0$:
$= \frac{2 \sqrt{0}}{\sqrt{0} + 7} = \frac{0}{7} = 0$ |
# How do you solve the system of equations -4x + 6y = 4 and - x + 8y = 1?
Nov 28, 2016
$\left\{\begin{matrix}x = - 1 \\ y = 0\end{matrix}\right.$
#### Explanation:
$\left\{\begin{matrix}- 4 x + 6 y = 4 \\ - x + 8 y = 1\end{matrix}\right.$
$\left\{\begin{matrix}- 4 x + 6 y = 4 \\ \left(- x + 8 y = 1\right) \cdot \left(- 4\right)\end{matrix}\right.$
$\left\{\begin{matrix}- 4 x + 6 y = 4 \\ 4 x - 32 y = - 4\end{matrix}\right.$
$- 26 y = 0$
Therefore
$y = 0$
Plug $0$ in for $y$ in one of the equations like so:
$- x + \left(8 \cdot 0\right) = 1$
$- x = 1$
$x = - 1$
Check your answer by putting $0$ in for $y$ and $- 1$ in for $x$ in the other equation like so:
$\left(- 4 x - 1\right) + \left(6 \cdot 0\right) = 4$
$4 + 0 = 4$
$4 = 4$
:) |
# Multiplication
Photo by: Dario Sabljak
Multiplication is often described as repeated addition. For example, the product 3 × 4 is equal to the sum of three 4s: 4 + 4 + 4.
## Terminology
In talking about multiplication, several terms are used. In the expression 3 × 4, the entire expression, whether it is written as 3 × 4 or as 12, is called the product. In other words, the answer to a multiplication problem is the product. In the original expression, the numbers 3 and 4 are each called multipliers, factors, or terms. At one time, the words multiplicand and multiplier were used to indicate which number got multiplied (the multiplicand) and which number did the multiplying (the multiplier). That terminology has now fallen into disuse. Now the term multiplier applies to either number.
Multiplication is symbolized in three ways: with an ×, as in 3 × 4; with a centered dot, as in 3 · 4; and by writing the numbers next to each other, as in 3(4), (3)(4), 5x, or (x + y)(x − y).
## Rules of multiplication for numbers other than whole—or natural—numbers
Common fractions. The numerator of the product is the product of the numerators; the denominator of the product is the product of the denominators. For example, .
Decimals. Multiply the decimal fractions as if they were natural numbers. Place the decimal point in the product so that the number of places in the product is the sum of the number of places in the multipliers. For example, 3.07 × 5.2 = 15.964.
Signed numbers. Multiply the numbers as if they had no signs. If the two factors both have the same sign, give the product a positive sign or omit the sign entirely. If the two factors have different signs, give the product a negative sign. For example, (3x)(−2y) = −6xy; (−5)(−4) = +20.
## Words to Know
Factor: A number used as a multiplier in a product.
Multiplier: One of two or more numbers combined by multiplication to form a product.
Product: The result of multiplying two or more numbers.
Powers of the same base. To multiply two powers of the same base, add the exponents. For example 10 2 × 10 3 = 10 5 and x 5 × x −2 = x 3 .
Monomials. To multiply two monomials, find the product of the numerical and literal parts of the factors separately. For example, (3x 2 y)(5xyz) = 15x 3 y 2 z.
Polynomials. To multiply two polynomials, multiply each term of one by each term of the other, combining like terms. For example, (x + y)(x − y) = x 2 − xy + xy − y 2 = x 2 − y 2 .
## Applications
Multiplication is used in almost every aspect of our daily lives. Suppose you want to buy three cartons of eggs, each containing a dozen eggs, at 79 cents per carton. You can find the total number of eggs purchased (3 cartons times 12 eggs per carton = 36 eggs) and the cost of the purchase (3 cartons at 79 cents per carton = \$2.37).
Specialized professions use multiplication in an endless variety of ways. For example, calculating the speed with which the Space Shuttle will lift off its launch pad involves untold numbers of multiplication calculations. |
Assignment Help Online
# Maths assignment help online
In the previous article we explored multiple area of maths assignment help online. There is always something about math assignment that makes students feel fidgety. There are many students who are good at math,but there will always be some assignments involving concepts that would require an expert’s help. This is where AAH- your academic hand comes into picture to provide you maths assignment help online.
Algebra assignment help online:
Lets us discuss what is algebra about and how important is the algebra assignment help is. Sometimes simple equality problems become headache. If i take an example and try to solve it using the basic rules of inequality we will see how easy and interesting math assignment help online can be. Lets take an example and start with the very basic concept of inequality.
Find the value of x that satisfies the following inequality : (x+1) > 0
Most of the students will be able to find the value of x for which the equation holds true. i.e if we substitute any value of x > -1 left hand side of the equation will always be greater than zero. Now, lets make this inequality a bit more interesting.
Find the value of x that satisfies the following inequality : ( x+1)/(x -1 )^2 > 0
Here we have an additional value in the denominator, which is (x-1)^2 . If we use our mathematical aptitude, we will see that this equation is same as the last one. Let us see how it is same as the last one.
We know a perfect square cannot be negative and if we multiply two sides of an inequality with a positive number the sign of inequality doesn’t change. For example , 4 > 3 and if we multiple both sides by 2 , new inequality will be 8 >6 ,so the inequality holds. On the other hand if we multiply both sides by -1 the sign will get reversed. For example, -4 < -3 ( after multiplying it by -1 ).
Let us take our equation again. As I said the sign of the inequality will hold on multiplication with a positive quantity, hence on multiplication with (x-1)^2 we can safely say that (x+1) > 0 * (x-1)^2 or (x+1) >0.
So students must be convinced here. However, we have missed one of the basic concept of maths assignment help online. If we go by the last solution we can say that for x> -1 this equation holds. Let me disapprove this solution. What if we take x = 1 ?
This will lead to (x+1)/0 , and any number divided by zero becomes not defined, hence we have to eliminate those points for which the denominator becomes zero. Therefore the solution to this equation would be x > -1 and x != 1.
If we keep these basic concepts of maths assignment help online in mind we can easily deal with mathematics homework. If you have problems with the basic fundamental of maths assignment help online our writers can help you. We provide detailed solution to your problems without missing out any important information. Apart from maths assignment help online, our writers actively provide help with the assignments of the following subjects:
Apart from the offline online assignment help, we have started providing online assignment services as well. We understand the money constraint students have therefore we are very much considerate and provide affordable assignment services online. Apart from the affordable assignment service we also have started the free assignment service online. You can post your query on our forum where our tutors are active. The turnaround time is minimal. Thank you for your time ! |
# DIVIDING MIXED NUMBERS WORKSHEET
## About "Dividing mixed numbers worksheet"
Dividing mixed numbers worksheet :
Worksheet on dividing mixed numbers is much useful to the students who would like to practice problems on mixed numbers.
## Dividing mixed numbers worksheet
1. Divide 3 2/5 by 6/7
2. Divide 2 3/5 by 3
3. Divide 1 1/2 by 2 3/5
4. If 1/4 cup of rice is used to make each sushi roll, how many sushi rolls can be made using 2 1/2 cups of rice ?
5. We have 1 1/4 pounds of cheese. And we are going to pack this 1 1/4 pounds of cheese in two cans where each can has the capacity of 5/8 pound. In how many cans can we fill 1 1/4 pounds of cheese ?
6. One pizza can be made in 1/2 hour. How many pizzas can be made in 2 1/2 hours ?
7. David eats 1 1/8 pizzas and divides into two equal parts for his two kids. What is the part of the pizza will each kid receive ?
## Dividing mixed numbers worksheet - Solution
Problem 1 :
Divide 3 2/5 by 6/7
Solution :
Using the method explained above, we have
3 2/5 ÷ 6/7 = 17/5 ÷ 7/6
3 2/5 ÷ 6/7 = (17/5) x (6/7)
3 2/5 ÷ 6/7 = (17x6) / (5x7)
3 2/5 ÷ 6/7 = 102/35
3 2/5 ÷ 6/7 = 2 32/35
Problem 2 :
Divide 2 3/5 by 3
Solution :
Using the method explained above, we have
2 3/5 ÷ 3 = 13/5 ÷ 3
2 3/5 ÷ 3 = (13/5) x (1/3)
2 3/5 ÷ 3 = (13x1) x (5x3)
2 3/5 ÷ 3 = 13 / 15
Problem 3 :
Divide 1 1/2 by 2 3/5
Solution :
Using the method explained above, we have
1 1/2 ÷ 2 3/5 = 3/2 ÷ 13/5
1 1/2 ÷ 2 3/5 = 3/2 x 5/13
1 1/2 ÷ 2 3/5 = (3x5) / (2x13)
1 1/2 ÷ 2 3/5 = 15/26
Problem 4 :
If 1/4 cup of rice is used to make each sushi roll, how many sushi rolls can be made using 2 1/2 cups of rice ?
Solution :
To find the number of sushi rolls that can be made, we need to determine how many fourths are in 2 1/2 .
Since 1/4 cup of rice is used to make each sushi roll, in the diagram given below, 2 1/2 is divided into fourths.
In 1 cup, we have four 1/4 cups of rice.
But, we have 2 1/2 cups of rice.
So, we have to count the number of 1/4 cups in 2 1/2 cups in the above diagram.
And there are ten 1/4 cups in 1 cup of rice.
Hence, we can make 10 sushi rolls from 2 1/2 cups of rice.
Problem 5 :
We have 1 1/4 pounds of cheese. And we are going to pack this 1 1/4 pounds of cheese in two cans where each can has the capacity of 5/8 pound. In how many cans can we fill 1 1/4 pounds of cheese ?
Solution :
To find the number of cans required, let us use the diagram given below.
The diagram below represents 2 pounds and it is divided into quarters and into eighths.
We have to fill 1 1/4 pounds of cheese in two cans where each can has the capacity of 5/8 pound.
So, we have to count the number of 5/8 pounds in 1 1/4 pounds.
And there are two 5/8 pounds in 1 1/4 pounds of cheese.
Hence, we can fill 1 1/4 pounds of cheese in two cans.
Problem 6 :
One pizza can be made in 1/2 hour. How many pizzas can be made in 2 1/2 hours ?
Solution :
Time taken to make one pizza = 1/2 hour
No. of pizzas made in 2 1/2 hours = 2 1/2 ÷ 1/2
No. of pizzas made in 2 1/2 hours = 5/2 ÷ 1/2
No. of pizzas made in 2 1/2 hours = 5/2 x 2/1
No. of pizzas made in 2 1/2 hours = (5x2) / (2x1)
No. of pizzas made in 2 1/2 hours = 5
Problem 7 :
David eats 1 1/8 pizzas and divides into two equal parts for his two kids. What is the part of the pizza will each kid receive ?
Solution :
1 1/8 pizzas is divided in to two equal parts for his two kids.
So, amount of part pizza received by each kid is
= 1 1/8 ÷ 2
= 9/8 ÷ 2
= 9/8 x 1/2
= (9x1) / (8x2)
= 9/16
Hence, the amount of pizza received by each kid is 9/16.
After having gone through the stuff given above, we hope that the students would have understood "Dividing mixed numbers worksheet".
Apart from the stuff given above, if you want to know more about "Dividing mixed numbers worksheet", please click here
Apart from "Dividing mixed numbers worksheet", if you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# ACT Math : Polynomials
## Example Questions
### Example Question #1 : Binomials And Foil
Which of the following expressions is equivalent to: 6x (m2 +yx2 3)?
6xm2 + 6yx2 -18x
6xm2 + 6yx3 -18x
6xm2 + 7x3 -18
6xm2 + 6yx3 -18
xm2 + 7x3 -18
6xm2 + 6yx3 -18x
Explanation:
6x (m2 +yx2 3)= 6x∙m2 + 6xyx2 – 6x∙3= 6xm2 + 6yx3 -18x (Use Distributive Property)
### Example Question #1 : How To Multiply Binomials With The Distributive Property
Which of the following expressions is equivalent to: ?
Explanation:
Use the distributive property to multiply by all of the terms in :
### Example Question #2 : How To Multiply Binomials With The Distributive Property
If and are constants and is equivalent to , what is the value of ?
Cannot be determined from the given information.
Explanation:
The question gives us a quadratic expression and its factored form. From this, we know
At this point, solve for t.
Now, we can plug in to get
.
Now, use FOIL to get s.
### Example Question #3 : How To Multiply Binomials With The Distributive Property
Which expression is equal to ?
Explanation:
In this problem, we are to multiply the two binomials using the FOIL method. This method stands for the order in which you multiply the variables. F stands for the first term of each binomial. O stands for the outside terms, meaning the first term of the first binomial and the last term of the second. I stands for the inside terms, meaning the second term of the first binomial and the first term of the second. L stands for the last term of each binomial. Once you do this, you simply add each part together to recieve your polynomial answer. Here is how our problem is solved:
### Example Question #4 : How To Multiply Binomials With The Distributive Property
Which expression is equal to ?
Explanation:
In this problem, we are to multiply the two binomials using the FOIL method. This method stands for the order in which you multiply the variables. F stands for the first term of each binomial. O stands for the outside terms, meaning the first term of the first binomial and the last term of the second. I stands for the inside terms, meaning the second term of the first binomial and the first term of the second. L stands for the last term of each binomial. Once you do this, you simply add each part together to recieve your polynomial answer. Here is how our problem is solved:
### Example Question #5 : How To Multiply Binomials With The Distributive Property
Which expression is equal to ?
Explanation:
In this problem, we are to multiply the two binomials using the FOIL method. This method stands for the order in which you multiply the variables. F stands for the first term of each binomial. O stands for the outside terms, meaning the first term of the first binomial and the last term of the second. I stands for the inside terms, meaning the second term of the first binomial and the first term of the second. L stands for the last term of each binomial. Once you do this, you simply add each part together to recieve your polynomial answer. Here is how our problem is solved:
### Example Question #1 : Binomials And Foil
Which expression is equal to ?
Explanation:
In this problem, we are to multiply the two binomials using the FOIL method. This method stands for the order in which you multiply the variables. F stands for the first term of each binomial. O stands for the outside terms, meaning the first term of the first binomial and the last term of the second. I stands for the inside terms, meaning the second term of the first binomial and the first term of the second. L stands for the last term of each binomial. Once you do this, you simply add each part together to recieve your polynomial answer. Here is how our problem is solved:
### Example Question #7 : How To Multiply Binomials With The Distributive Property
Which expression is equal to ?
Explanation:
In this problem, we are to multiply the two binomials using the FOIL method. This method stands for the order in which you multiply the variables. F stands for the first term of each binomial. O stands for the outside terms, meaning the first term of the first binomial and the last term of the second. I stands for the inside terms, meaning the second term of the first binomial and the first term of the second. L stands for the last term of each binomial. Once you do this, you simply add each part together to recieve your polynomial answer. Here is how our problem is solved:
### Example Question #1 : Trinomials
Find the -intercepts:
and
and
only
and
and
Explanation:
-intercepts occur when .
1. Set the expression equal to and rearrange:
2. Factor the expression:
3. Solve for :
and...
4. Rewrite the answers as coordinates:
becomes and becomes .
### Example Question #2 : How To Factor A Trinomial
Solve for when . |
# A surveyor stands 150ft from the base of a tree and measures the angle of elevation to be 46.2. His eye level is 6feet above the ground. What is the height of the tree to the nearest foot?
## I'd be interested in seeing your work. I get
(h-6)/150 = tan 46.2°
(h-6)/150 = 1.04
h-6 = 156.42
h = 162.42
## Well, well, well, we have a height-hunting surveyor here! Let's embrace our inner math clown and solve this riddle!
So, the surveyor is standing 150 feet away from the tree, and the angle of elevation to the top of the tree is 46.2 degrees. We also know that the surveyor's eye level is 6 feet above the ground. Quite the precise measurements!
Now, to find the height of the tree, we need to break out our mathematical magic. We can use a little bit of trigonometry here. Specifically, the tangent function.
The tangent of an angle is equal to the opposite side divided by the adjacent side. In our case, the opposite side is the height of the tree, and the adjacent side is the distance from the surveyor to the tree. So, we can set up the following equation:
tan(46.2) = (height of the tree) / (150 + 6)
Now, let's do a little calculation dance. We can solve for the height of the tree:
(height of the tree) = tan(46.2) * (150 + 6)
Grab your calculator and let's crunch those numbers! Once you've calculated it, you'll know the height of the tree to the nearest foot. Good luck, height-hunting surveyor!
## To find the height of the tree, we can use trigonometry. We have the length of the adjacent side (150ft) and the measure of the angle of elevation (46.2 degrees).
Let's start by drawing a diagram:
* T (top of the tree)
/
/|
/ |
/ |
/ |
/θ |
/ |
/ |
S /________| G (surveyor's position)
In this diagram, G represents the surveyor's position, T represents the top of the tree, and S represents the base of the tree. The angle θ is the angle of elevation.
We can consider the right triangle GST, where GS represents the horizontal distance from the surveyor to the base of the tree, and ST represents the height of the tree.
Now, we have the adjacent side (GS) and the angle θ. We can use the trigonometric function tangent (tan) to find the height of the tree.
The tangent of an angle in a right triangle is defined as the ratio of the opposite side to the adjacent side. In this case, the opposite side is ST and the adjacent side is GS.
So, we can use the formula:
tan(θ) = ST / GS
In our case, tan(46.2 degrees) = ST / 150ft
To find ST, rearrange the formula:
ST = tan(46.2 degrees) * 150ft
Now, calculate ST:
ST ≈ tan(46.2 degrees) * 150ft
Using a scientific calculator, we find:
ST ≈ 149.9ft
However, we need to consider that the surveyor's eye level is 6ft above the ground. So, we add that height to the calculated value of ST to get the final height of the tree:
Height of the tree = ST + eye level height
Height of the tree = 149.9ft + 6ft
Height of the tree ≈ 155.9ft
Therefore, the height of the tree to the nearest foot is approximately 156 feet.
## If the height is h, then
(h-6)/150 = tan 46.2° |
# Factors of 3
Definition of factors of 3: If a number completely divides 3 with the remainder zero, then that number is called a factor of 3.
WhatsApp Group Join Now
Telegram Group Join Now
By the above definition, we can say that the factors of 3 are the divisors of 3. In this section, we will learn about the factors of 3 and the prime factors of 3.
Also Read: Basic concepts of factors
Table of Contents
## Highlights of Factors of 3
• 3=1×3 is the prime factorization of 3.
• The factors of 3 are 1 and 3.
• The only prime factor of 3 is 3.
• Negative factors of 3 are –1 and –3.
## What are the Factors of 3?
To get the factors of 3 we need to first write the number 3 multiplicatively in all possible ways. Note that we have:
3 = 1×3
This is the only way we can express 3 multiplicatively. Therefore the factors of 3 in pairs are given as follows:
Thus the only pair factor of 3 is (1, 3). We know that all the numbers appearing in pair factors are the factors of 3. So we conclude that
## Number of Factors of 3
From above we have calculated the factors of which are 1 and 3. Thus the total number of factors of 3 is two.
## Prime Factors of 3
Note that the factors of 3 are 1 and 3. Among those factors, we observe that only 3 is a prime number as it does not have any proper divisors.
∴ the only prime factor of 3 is the number 3 itself.
Question: What are the factors of 3?
Video Solution:
## Is 3 a Prime Number?
As the factors of 3 are 1 and the number 3 itself, by the definition of prime numbers, we can say that 3 is a prime number.
Conclusion: 3 is a prime number.
## How to Find Factors of 3?
Now we will determine the factors of 3 by division method. In this method, we will find the numbers that can divide 3 with no remainder. See that
Note that no numbers other than the numbers in violet color can divide 3. So the numbers in violet color, that is, 1 and 3 are the complete list of factors of 3.
Share via:
WhatsApp Group Join Now
Telegram Group Join Now |
Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### High School Mathematics5.3 Revision of Algebraic Expressions - III
Examples: Expand, (x + 2y)2. Given that, (x + 2y)2, this is in the form of (a + b)2 = a2 + 2ab + b2. Using above formula, (x + 2y)2 = x2 + 4xy + y2 Expand, (2x - 3y)2. Given that, (2x - 3y)2, this is in the form of (a - b)2 = a2 - 2ab + b2. Using above formula, (2x - 3y)2 = 4x2 - 12xy + 9y2 Expand,(3x - 4y) * (3x + 4y). Given that, (3x - 4y) * (3x + 4y), this is in the form of (a + b)(a - b) = a2 - b2. Using above formula, (3x - 4y) * (3x + 4y) = (3x)2 - (4y)2 = 9x2 - 16y2 Directions: Solve the following problems. Also write at least 10 examples of your own.
Q 1: Expand, (4x - y)2.16x2 - 16xy + y216x2 + 8xy + y216x2 - 8xy + y2 Q 2: Expand, (3x - 2y) (3x + 2y).9x2 - 4y29x2 - 2y23x2 - 4y2 Q 3: Expand, (5x - 2y)2.25x2 + 20xy + 4y225x2 - 20xy + 4y225x2 - 24xy + 4y2 Q 4: Expand, (2x + 3y)2 - (2x - 3y)2.16xy12xy24xy Q 5: Expand, (3x + y)2.9x2 - 6xy + y29x2 + 12xy + y29x2 + 6xy + y2 Q 6: Expand, (4x + 3y)2 + (4x - 3y)2.16x2 + 18y232x2 + 9y232x2 + 18y2 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! |
# How to Find the Number of Solutions to a Linear Equation?
A linear equation is an equation in which the largest power of the variable is equal to $$1$$. In this article, in a few short steps, we will explain the method of identifying the number of solutions of a linear equation.
## A step-by-step guide to Finding the Number of Solutions to a Linear Equation
An algebraic expression in which there is equality is called an equation. An equation in which there is only one variable is called a single unknown equation. If the highest variable exponent in the equation is equal to $$1$$, we call this equation a linear equation.
To solve a linear equation, we first move the variables to one side and the known numbers to the other side of the equation. After that, we simplify the expression. According to the statement obtained after simplification, there are $$3$$ different modes for solving the linear equation:
If the statement obtained is false, there is no solution to the linear equation.
If the obtained statement is true for only one value, then the linear equation has a solution.
If the obtained statement is always true, then there are infinite solutions for the linear equation.
### Finding the Number of Solutions to a Linear Equation-Example 1
How many solutions does the following equation have?
$$-12x+25=-12x$$
Solution:
Solve for $$x: -12x+25=-12x$$$$25=+12x-12x$$$$25=0$$.
$$25=0$$ is a false statement. The linear equation has no solution because by solving the linear equation you get a false statement as an answer.
### Finding the Number of Solutions to a Linear Equation-Example 2
How many solutions does the following equation have?
$$10x=32x-22x$$
Solution:
Simplify the right side of the equation: $$10x=32x-22x$$$$10x=10x$$.
$$10x=10x$$ is a statement that is always true. So, the equation has infinitely many solutions because by solving a linear equation you get a statement that is always true.
## Exercises forFinding the Number of Solutions to a Linear Equation
### How many solutions does the following equation have?
• $$\color{blue}{11x+12=17x}$$
• $$\color{blue}{no\:solution}$$
### What people say about "How to Find the Number of Solutions to a Linear Equation? - Effortless Math: We Help Students Learn to LOVE Mathematics"?
No one replied yet.
X
45% OFF
Limited time only!
Save Over 45%
SAVE $40 It was$89.99 now it is \$49.99 |
# Taking socks out of drawers, conditional probability
• I
• CGandC
In summary: D_i ## is a subset of a sample space which is the product of the drawers and the socks.In summary, the conversation discussed the odds of randomly selecting a white sock from a dresser with 3 drawers, each containing a different number and combination of black and white socks. The first attempt at solving the problem used a sample space of drawers and socks, but the calculation of probabilities was incorrect. The correct solution involves considering the probabilities of choosing a particular drawer and a white sock from that drawer, and using the law of total probability to find the overall probability of selecting a white sock.
CGandC
TL;DR Summary
x
Problem: In a dresser there are 3 drawers. In one drawer there are two black socks and one white sock, in the second drawer there are two white socks, and in the third drawer there is a black and white sock. Suppose I chose a drawer randomly ( meaning, in a uniform distribution ) and I took a sock out of it randomly. What are the odds that the sock I took out is white?
Attempt:
Denote as ## D_i ## the event in which we chose drawer ## i ##. Notice that for every drawer we pick, we also pick a sock. Hence the sample space is ## \Omega = \{ (d,s) : d \in \{ 1 , 2 , 3 \} , s \in \{ \text{ White Sock , Black sock}\} \} ##.
Note that ## \Omega = 3 \cdot ( \frac{3!}{2! \cdot 1! } + \frac{2!}{2!} + \frac{2!}{1!1!} ) = 3\cdot ( 3+1+2 ) = 18 ##.
Let ## D_i ## denote the event of choosing the ## i ##-th drawer, note that ## D_1 = \{ (1,\text{ white sock }), (1,\text{ black sock } \})## , ## D_3 = \{ (3,\text{ white sock } ),(3,\text{ black sock }) \}##, ## D_2 = \{ (2,\text{ white sock }) \}##.
Denote as ## W ## the event of choosing a white sock. Note that
## W = \{ (1,\text{ white sock }) , (2,\text{ white sock }) , (3,\text{ white sock }) \} ##
Denote as ## S ## the event of choosing a sock. Note that ## S = \Omega ##.
We'll Calculate ## P(W| D \cap S ) = P(W| D ) ##,
Note that ## D_1 \cup D_2 \cup D_3 = \Omega ##, hence
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{18} = \frac{1}{9} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{18} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{18} = \frac{1}{9} ##
## P(W|D_1) = \frac{ | W \cap D_1 |}{ |D_1|} = \frac{1}{2} ##
## P(W|D_2) = \frac{ | W \cap D_2 |}{ |D_2|} = \frac{1}{1} ##
## P(W|D_3) = \frac{ | W \cap D_3 |}{ |D_3|} = \frac{1}{2} ##
And from the law of total probability,
Thus, ## P(W| D ) = P(W|D_1)P(D_1) + P(W|D_2)P(D_2) + P(W|D_3)P(D_3) = \frac{1}{6} ##Solution from another student which I think is the correct one judging by other answers on the web:
W - a white sock was taken out.
We'll want to calculate ## P(W) ##.
## E_i ## - we choose the i-th drawer. ( since we are talking about uniform distribution space, ## P(E_i) = 1/3 ## for ## i \in \{ 1,2,3 \} ## )
From the law of total probability, notice that
## P(W,E_1) = 1/3 ## , ## P(W,E_2) = 1 ##, ## P(W,E_3) = 1/2 ##
Hence:
## P(W) = \sum_{i=1}^3 P(W|E_i)P(E_i) = P(W|E_1)P(E_1) + P(W|E_2)P(E_2) + P(W|E_3)P(E_3) = (1/3)\cdot (1/3) + 1 \cdot (1/3) + (1/2)\cdot (1/3) = 11/18 ##
My question:
Where did my attempt fail ( the final calculation is incorrect ) ? Is it because of how I defined the sample space and the ## D_i ## ? I can't figure out, can you please help me?
Last edited:
Notice that the sum of the ##D_i##s is not 1. That should tell you that something is wrong there.
If the choice of the drawer is uniform from three drawers, why aren't those probabilities 1/3?
You're right, but I still don't understand about the relation between the drawer choice and the socks in the sample set. If the probability of ## D_i ## were ## 1/3 ## each then I have don't see how the sock choosing is manifested in this problem since we don't refer to them in the sample set, rather we consider the sample set as ## \Omega = \{ 1,2,3 \} ## where the numbers are the different drawers .
How can one model the sample set as ## \Omega = \{ (d,s) : d \in \{ 1 , 2 , 3 \} , s \in \{ \text{ White Sock , Black sock}\} \}
##, and yet maintain the probability of ## D_i ## to be ## 1/3 ## ?
Because we're not talking only about choosing drawers but also about choosing socks.
You are only given the probability of a white sock given you have chosen a particular drawer. Look at how that is directly used in the other student's solution. I do not understand your calculation of the ##P(D_i)##s. Those probabilities are wrong.
hutchphd
I just don't understand what sample space to look at, on the one hand when calculating ## P(D_i) ## then ## D_i ## is a subset of some sample space, which I think, according to the offered probabilties in the student's answer, is ## \{ 1,2,3 \} ##.
On the otherhand, when calculating ## P(D_i \cap W ) ##, we are looking at the event ## D_i \cap W ## and clearly it isn't ## \{ i\} ## , but rather something like ## \{ (i, \text{ "White Sock" }) \} ##.
So how exactly the elements of the sample space/set are supposed to look like?
CGandC said:
On the otherhand, when calculating ## P(D_i \cap W ) ##, we are looking at the event ## D_i \cap W ## and clearly it isn't ## \{ i\} ## , but rather something like ## \{ (i, \text{ "White Sock" }) \} ##.
So how exactly the elements of the sample space/set are supposed to look like?
When you want to find ##P(D_i \cap W)## and it is not directly given, you can use either:
##P(D_i \cap W) = P(W | D_i) P(D_i)##
or
##P(D_i \cap W) = P(D_i | W) P(W)##
Chose wisely. One of these is very easy and the numbers are virtually given. The other is hard.
I can reach the answer by calculating the probabilities, but I want to understand the "behind the scenes" and that stems from me not understanding what ## \Omega ## ( the sample space ) looks like in this problem; what do its elements look like? are they ordered pairs?
There are no hard and fast rules about sample spaces, but I'd say that the sample space is the socks. Each sample is one sock. The problem is to find the probability for each sock.
Sure, you can use ordered pairs of (drawer, sock) to identify the socks. But that is just a naming convention.
Looking again at the sample space with freshened mindset, I think the natural sample space is ## \Omega = \{ (Drawer_1,White),(Drawer_1,Black), (Drawer_2,White) , (Drawer_3,White) , (Drawer_3,Black) \} ##, from here we see that
## | \Omega | = 5##, ## |D_1| =2 , |D_2| =1, |D_3| =2 ##,
Now, I indeed get that
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
And indeed ## P(D_1) +P(D_2)+P(D_3) = 1 ##.
Seems like I've calculated the sample set size wrongly in my first attempt ( and it wasen't entirely the sample set in the question since there were more elements, but that is ok since I could've attributed zero probability to those elements in our discrete probability space ).
Initially I tried to combinatorically calculate the sample space's size with the following mindset: First choose a drawer ( there are 3 options for it ) and then for the drawer, pick up a sock ( for ## D_1 ## there are 3 options, for ## D_2 ## there is 1 option, for ## D_3 ## there are 2 options, ) hence I've got ## |\Omega | = 3 \cdot ( \frac{3!}{2! \cdot 1! } + \frac{2!}{2!} + \frac{2!}{1!1!} ) = 3\cdot ( 3+1+2 ) = 18## ; but clearly this was wrong, so:
1. If ## \sum{P(E_i)} = 1 ## where ## E_i ## are some events, then does this imply that ## E_1 \cup \cdots \cup E_k = \Omega ## ?
2. How can one combinatorically correctly calculate the size of the sample set as I've tried to do ( but failed as I've got wrong result )?
CGandC said:
Looking again at the sample space with freshened mindset, I think the natural sample space is ## \Omega = \{ (Drawer_1,White),(Drawer_1,Black), (Drawer_2,White) , (Drawer_3,White) , (Drawer_3,Black) \} ##, from here we see that
## | \Omega | = 5##, ## |D_1| =2 , |D_2| =1, |D_3| =2 ##,
Now, I indeed get that
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
And indeed ## P(D_1) +P(D_2)+P(D_3) = 1 ##.
Please define the event, ##D_1##. If it is the event that the selected sock was from drawer one, then ##P(D_1)=1/3##. The problem states clearly that the drawer is selected from three drawers with equal probability:
CGandC said:
I chose a drawer randomly ( meaning, in a uniform distribution )
You are making this part more complicated than it should be.
pbuk
CGandC, sysprog and FactChecker
Hornbein said:
I would quibble that this strongly implies a sample space of socks.
Yes, in the same way as ## x = 2 + 2 ## implies a function ## f(a, b) \to c ## over a commutative group but until you can reliably solve ## x = 4 ## you are not ready for group theory.
PeroK
So the worked solution which should gain full marks at any exam in which this is an appropriate level question without any mention of sample spaces or conditional probability is:
Probability of selecting a white sock from:
- the first drawer: ## \frac 1 3 \times \frac 1 3 = \frac 1 9 ##
- the second drawer: ## \frac 1 3 \times 1 = \frac 1 3 ##
- the third drawer: ## \frac 1 3 \times \frac 1 2 = \frac 1 6 ##
Total probability of selecting a white sock: ## \frac 1 9 + \frac 1 3 + \frac 1 6 = \frac 2 {18} + \frac 6 {18} + \frac 3 {18} = \frac {11} {18} ##.
DrClaude and FactChecker
Thanks, I understand all of these but my question is about something else. I still don't understand why my combinatorial calculation of the sample space is wrong ( I know it's wrong but I don't understand why ). I explained in my comment above as follows
CGandC said:
Initially I tried to combinatorically calculate the sample space's size with the following mindset: First choose a drawer ( there are 3 options for it ) and then for the drawer, pick up a sock ( for ## D_1 ## there are 3 options, for ## D_2 ## there is 1 option, for ## D_3 ## there are 2 options, ) hence I've got ## |\Omega | = 3 \cdot ( \frac{3!}{2! \cdot 1! } + \frac{2!}{2!} + \frac{2!}{1!1!} ) = 3\cdot ( 3+1+2 ) = 18## ; but clearly this was wrong, so:
How does one correctly calculate the sample space's size using combinatorial argumentation as I have tried? ( Again, I'm not talking about calculating a probability but rather calculating size of a finite set )
( which is ##
\Omega = \{ (Drawer_1,White),(Drawer_1,Black), (Drawer_2,White) , (Drawer_3,White) , (Drawer_3,Black) \}
## )
I think a calculation using trees as offered above will be a good idea, I'll do it that way and try to solve the problem again.
Last edited:
CGandC said:
How does one correctly calculate the sample space's size using combinatorial argumentation as I have tried?
By constructing a tree: the size of the sample space is the number of leaf nodes.
pbuk said:
So the worked solution which should gain full marks at any exam in which this is an appropriate level question without any mention of sample spaces or conditional probability is:
Probability of selecting a white sock from:
- the first drawer: ## \frac 1 3 \times \frac 1 3 = \frac 1 9 ##
- the second drawer: ## \frac 1 3 \times 1 = \frac 1 3 ##
- the third drawer: ## \frac 1 3 \times \frac 1 2 = \frac 1 6 ##
Total probability of selecting a white sock: ## \frac 1 9 + \frac 1 3 + \frac 1 6 = \frac 2 {18} + \frac 6 {18} + \frac 3 {18} = \frac {11} {18} ##.
I don't know any level where these problems are done and conditional probabilities are not mentioned. Certainly, they are being used here in such an intuitive way that there is no need to name them. But a rose by any other name is still a rose. ;-)
CGandC said:
Thanks, I understand all of these but my question is about something else. I still don't understand why my combinatorial calculation of the sample space is wrong ( I know it's wrong but I don't understand why ).
To focus in on the problem, your values of the ##P(D_i)## are all wrong. I can't follow how you got your numbers, so I can not help more than that.
FactChecker said:
I don't know any level where these problems are done and conditional probabilities are not mentioned.
Oh absolutely, I just meant that I wouldn't expect them in the solution to this problem because:
FactChecker said:
they are being used here in such an intuitive way that there is no need to name them
Another example at an appropriate level (which in the UK is Year 10/11) https://www.bbc.co.uk/bitesize/guides/zsrq6yc/revision/8
FactChecker
Well, I can solve it using intuition and basic stuff, but I'm trying to solve it by first principles ( a more complicated way) for the sake of learning, not just solving.
Here's what I've tried to calculate the probability:
First, I've written the sample space ## \Omega = \{ (d_1,w),(d_1,b), (d_2,w) , (d_3,w) ,(d_3,b) \} ##. ( ## d_1,d_2,d_3 ## stand for drawers ##1,2,3 ## respectively and ##w,b## stand for White,Black )
Denote ## D_1,D_2,D_3 ## as the events of choosing drawers ##1,2,3 ## respectively. Denote ## W ## as the event of choosing a white sock.
Note that,
## D_1 = \{ (d_1,w),(d_1,b) \} ##, ## D_2 = \{ (d_2,w) \} ##, ## D_3 = \{ (d_3,w) ,(d_3,b) \} ##.
Note that
## W\cap D_1 = \{ (d_1,w) \} ##,## W\cap D_2 = \{ (d_2,w) \} ##,## W\cap D_3 = \{ (d_3,w) \} ##.
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
( Note that their sum is 1 )
## P( W| D_1) = \frac{ P(W\cap D_1) }{ P(D_1) } = \frac{ \frac{1}{5} }{ \frac{2}{5} } = \frac{1}{2} ##
## P( W| D_2) = \frac{ P(W\cap D_2) }{ P(D_2) } = \frac{ \frac{1}{5} }{ \frac{1}{5} } = 1 ##
## P( W| D_3) = \frac{ P(W\cap D_3) }{ P(D_3) } = \frac{ \frac{1}{5} }{ \frac{2}{5} } = \frac{1}{2} ##
Hence, from law of total probability we get:
## P(W ) = P(W|D_1)P(D_1) + P(W|D_2)P(D_2) + P(W|D_3)P(D_3) = 1/2 \cdot 2/5 + 1 \cdot 1/5 + 1/2 \cdot 2/5 = 3/5 ##
I still don't see what I've done wrong in my above analysis, if you have any idea then that might pinpoint my misunderstanding.
CGandC said:
Well, I can solve it using intuition and basic stuff, but I'm trying to solve it by first principles ( a more complicated way) for the sake of learning, not just solving.
Here's what I've tried to calculate the probability:
First, I've written the sample space ## \Omega = \{ (d_1,w),(d_1,b), (d_2,w) , (d_3,w) ,(d_3,b) \} ##. ( ## d_1,d_2,d_3 ## stand for drawers ##1,2,3 ## respectively and ##w,b## stand for White,Black )
Denote ## D_1,D_2,D_3 ## as the events of choosing drawers ##1,2,3 ## respectively. Denote ## W ## as the event of choosing a white sock.
Note that,
## D_1 = \{ (d_1,w),(d_1,b) \} ##, ## D_2 = \{ (d_2,w) \} ##, ## D_3 = \{ (d_3,w) ,(d_3,b) \} ##.
Note that
## W\cap D_1 = \{ (d_1,w) \} ##,## W\cap D_2 = \{ (d_2,w) \} ##,## W\cap D_3 = \{ (d_3,w) \} ##.
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
Stop right there. You keep ignoring the FACT that the drawers are equally likely with probabilities ##P(D_i)=1/3##.
This is very clearly stated in the problem statement, whereas I can not follow any logic behind your numbers.
Until you get that right, there is no reason to go further.
CGandC said:
Well, I can solve it using intuition and basic stuff, but I'm trying to solve it by first principles ( a more complicated way) for the sake of learning, not just solving.
Nobody is suggesting using intuition, we are trying to show you how to solve it by first principles. This should not be complicated, the complication is coming from your refusal to listen to what you are being told and believing that mathematics is about fancy symbols rather than understanding the concepts the symbols represent. This ends up with you writing things that are obviously incorrect like:
CGandC said:
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
where it should be obvious that ## P(D_1) = P(D_2) = P(D_3) = \frac 1 3 ##.
If you want to learn, draw a tree:
• Start with a node with three branches to nodes labelled ## D_1 ## etc.
• Write the probability ## \frac 1 3 ## on each branch.
• Add a second level of branches to nodes labelled ## W ## and ## B ## .
• Write the appropriate probabilities on each branch (e.g. the branch from ## D_1 ## to ## W ## should be labelled ## \frac 1 3 ##).
• Multiply the probabilities that lead to ## W ## leaf nodes and write them next to the nodes.
• Add up the probabilities on the ## W ## leaf nodes to get ## \frac {11}{18} ##.
Now do the same with symbols:
• Write ## P(D_1) = P(D_2) = P(D_3) = \frac 1 3 ##. Notice how these correspond to the probabilities on the first level of branches in the tree.
• Write probabilities for ## P(W|D_1) ## etc. Notice how these correspond to the conditional probabilities on the second level of branches in the tree.
• Write ## P(W \cap D_1) = P(W | D_1) P(D_1) ## etc. Notice how these correspond to the probabilities on the ## W ## leaf nodes.
• Write ## P(W) = P(W \cap D_1) + P(W \cap D_2) + P(W \cap D_3) = \frac {11}{18} ##. Notice how this is the same sum as you performed on the tree.
CGandC
The sample space could be taken to be the set of pairs (drawer chosen, colour of sock chosen). It can be taken to have 3 x 2 = 6 elements, though one of them has probability zero, so could simply be erased. After that, give these 6 points probabilities as done above. Notice that the event ‘first drawer is chosen’ ={(1,w), (1,b)}. The event “sock chosen is white” = {(1,w), (2,w), (3,w)}. To allocate probabilities to these 6 points you need to *think* about the probabilities of drawers, and about the conditional probabilities of colours given drawers.
You also could take the sample space as being a set of socks. Number the socks in each drawer, giving the black socks lower numbers than the white socks. You could even have an imaginary “number 3” sock in each drawer with only two socks. Now there are 3x3 points in your sample space. Two of them have zero probability. The probabilities of the socks in each drawer add up to 1/3. The socks with non zero probability in each drawer have equal probabilities, so either 1/3 is shared inequality parts over three socks, or over two socks.
Any rich enough sample space will do, but you will have to use the implied strong meaning of “chosen at random” in order to figure out what all those probabilities are.
What is wrong is blindly assuming that each point of your sample space has equal probability. If you thought that “first principles” means using the rule P(A) = #A / #Omega you are badly wrong. That “rule” is not a first principle but a mathematical model for the experiment “choosing a point uniformly at random from a finite set Omega”.
CGandC
The complete sample space is
{##(D_1, b_1), (D_1, b_2), (D_1, w_1),
(D_2, w_1), (D_2, w_2),
(D_3, b_1), (D_3, w_2)##}.
It can also be specified other ways, such as {##b_{1,1}, b_{1,2}, w_{1,1}, w_{2,1}, w_{2,2}, b_{3,1}, w_{3,1}##}, where the first index in the box number and the second is the sock number of the color in that box.
The next step is to assign probabilities to each event of picking that sock. That requires several applications of the probability of the box times the conditional probability of picking that sock given that box. They are like this:
##P(b_{1,1})=1/3*1/3=1/9##
Then you can add up all the probabilities of the white socks.
CGandC
I would call that *a* complete sample space. Complete in the sense that it is rich enough to be able to talk about all the probabilities of all events described in the original problem. It is also minimal, in the sense that it has the smallest possible number of elementary outcomes. But it is not the only possible solution, it is not the only sensible solution. Of course, all sensible solutions end up with the same answer to the original question.
FactChecker
pbuk said:
Nobody is suggesting using intuition, we are trying to show you how to solve it by first principles. This should not be complicated, the complication is coming from your refusal to listen to what you are being told and believing that mathematics is about fancy symbols rather than understanding the concepts the symbols represent. This ends up with you writing things that are obviously incorrect like:
How do I learn from basics if you're letting me jump straight to deep waters by thinking of probabilities? ( not the the problem's hard, I'm talking about establishing an understanding ) first I need to think of my sample space, then I would think what probability model I'd like on the elements of the sample space, in my case in is uniform model, then I'd start thinking about probabilities but only after I've combinatorically reasoned about the problem?
The tree helps for calculating the probabilities but I want to work bottom-up but first looking at my sample space. Thanks for the algorithm though, I find it helpful.
gill1109 said:
The sample space could be taken to be the set of pairs (drawer chosen, colour of sock chosen). It can be taken to have 3 x 2 = 6 elements, though one of them has probability zero, so could simply be erased. After that, give these 6 points probabilities as done above. Notice that the event ‘first drawer is chosen’ ={(1,w), (1,b)}. The event “sock chosen is white” = {(1,w), (2,w), (3,w)}. To allocate probabilities to these 6 points you need to *think* about the probabilities of drawers, and about the conditional probabilities of colours given drawers.
You also could take the sample space as being a set of socks. Number the socks in each drawer, giving the black socks lower numbers than the white socks. You could even have an imaginary “number 3” sock in each drawer with only two socks. Now there are 3x3 points in your sample space. Two of them have zero probability. The probabilities of the socks in each drawer add up to 1/3. The socks with non zero probability in each drawer have equal probabilities, so either 1/3 is shared inequality parts over three socks, or over two socks.
Any rich enough sample space will do, but you will have to use the implied strong meaning of “chosen at random” in order to figure out what all those probabilities are.
What is wrong is blindly assuming that each point of your sample space has equal probability. If you thought that “first principles” means using the rule P(A) = #A / #Omega you are badly wrong. That “rule” is not a first principle but a mathematical model for the experiment “choosing a point uniformly at random from a finite set Omega”.
I have thought of the first sample-space you proposed regarding the 6 elements ( but not about “sock chosen is white” = {(1,w), (2,w), (3,w)} ) and giving one element probability zero, but it seemed to me too artificial ( also in relation to the possible rich sample spaces ). And by first principles I meant defining my sample space then a probability space in correspondence to each result in the sample; it doesn't have to be a uniform probability space but that is what "feels" right for the problem.
FactChecker said:
The complete sample space is
{##(D_1, b_1), (D_1, b_2), (D_1, w_1),
(D_2, w_1), (D_2, w_2),
(D_3, b_1), (D_3, w_2)##}.
It can also be specified other ways, such as {##b_{1,1}, b_{1,2}, w_{1,1}, w_{2,1}, w_{2,2}, b_{3,1}, w_{3,1}##}, where the first index in the box number and the second is the sock number of the color in that box.
The next step is to assign probabilities to each event of picking that sock. That requires several applications of the probability of the box times the conditional probability of picking that sock given that box. They are like this:
##P(b_{1,1})=1/3*1/3=1/9##
Then you can add up all the probabilities of the white socks.
I have thought about this sample space but I don't see how the probability for each result will be uniform given we don't differentiate between the socks (The tricky thing about the sample space imposed in the question is that we don't differentiate between the socks, and If we don't differentiate between the socks, a better sample space would be {##b_{1}, w_{1}, w_{2}, b_{3}, w_{3}##} )
Last edited:
CGandC said:
I have thought about this sample space but I don't see how the probability for each result will be uniform given we don't differentiate between the socks
The results are not uniform over all the socks. Each one has to be calculated. (They are uniform over the socks in a given drawer.)
##P(b_{1,1})=P(b_{1,2})=P(w_{1,1})=1/3*1/3=1/9##,
##P(w_{2,1})=P(w_{2,2})=1/3*1/2=1/6##, and
##P(b_{3,1})=P(w_{3,1})=1/3*1/2=1/6##
Now add up the probabilities of the white socks.
CGandC said:
(The tricky thing about the sample space imposed in the question is that we don't differentiate between the socks, and If we don't differentiate between the socks, a better sample space would be {##b_{1}, w_{1}, w_{2}, b_{3}, w_{3}##} )
You can differentiate as much as you want to make the probability calculations simple. Then if the problem question does not differentiate, you add up the probabilities of the events that should be considered equivalent results.
FactChecker said:
The results are not uniform over all the socks. Each one has to be calculated. (They are uniform over the socks in a given drawer.)
##P(b_{1,1})=P(b_{1,2})=P(w_{1,1})=1/3*1/3=1/9##,
##P(w_{2,1})=P(w_{2,2})=1/3*1/2=1/6##, and
##P(b_{3,1})=P(w_{3,1})=1/3*1/2=1/6##
Now add up the probabilities of the white socks.
You can differentiate as much as you want to make the probability calculations simple. Then if the problem question does not differentiate, you add up the probabilities of the events that should be considered equivalent results.
Just one thing that bothers me is how do I know a-priori without the answer, If I should've differentiated between the different socks in a given drawer ( then, the sample space would be ## b_{1,1}, b_{1,2}, w_{1,1}, w_{2,1}, w_{2,2}, b_{3,1}, w_{3,1} ## ) or if I shouldn't have ( in which case, the sample space would be ## b_{1}, w_{1}, w_{2}, b_{3}, w_{3} ## )? because it seems to me like you are differentiating between them only because you know what the answer for the probability of choosing a sock is supposed to be.
CGandC said:
Just one thing that bothers me is how do I know a-priori without the answer, If I should've differentiated between the different socks in a given drawer ( then, the sample space would be ## b_{1,1}, b_{1,2}, w_{1,1}, w_{2,1}, w_{2,2}, b_{3,1}, w_{3,1} ## ) or if I shouldn't have ( in which case, the sample space would be ## b_{1}, w_{1}, w_{2}, b_{3}, w_{3} ## )? because it seems to me like you are differentiating between them only because you know what the answer for the probability of choosing a sock is supposed to be.
Not "supposed to be", I differentiated enough so that the individual probabilities were obvious. You have to differentiate at least to the extent that you can calculate individual probabilities and at least enough to answer the problem (often more). For that, you need to be able to interpret the problem presented so that your set of tools can be applied. Then you calculate probabilities at a fine enough resolution. Once that is done, you can combine and add probabilities to answer the particular question that the problem asks for.
CGandC said:
How do I learn from basics if you're letting me jump straight to deep waters by thinking of probabilities? ( not the the problem's hard, I'm talking about establishing an understanding )
You are mistaken, it is you that is "trying to jump straight into the deep waters". I have shown you the way this material is taught to more than a million 14 year olds in the UK each year; IME most of them understand it.
CGandC said:
first I need to think of my sample space, then I would think what probability model I'd like on the elements of the sample space, in my case in is uniform model,
And this is where you are going wrong: because you are jumping straight to the sample space without considering how that sample space is constructed you can't see that the distribution is not uniform. This is what the tree is for.
CGandC said:
then I'd start thinking about probabilities but only after I've combinatorically reasoned about the problem?
The tree is the best way to combinatorially reason about the problem.
FactChecker
CGandC said:
first I need to think of my sample space, then I would think what probability model I'd like on the elements of the sample space, in my case in is uniform model,
I think this is your fundamental error. The probability distribution of the sample space is not always uniform. It is up to you to calculate the correct probabilities using a valid mathematical process.
This is a very common situation, where the process presented is done in a sequence of steps (pick a drawer, then pick a sock from the drawer). The conditional probabilities of the results of the individual steps are clear. Use those to calculate the correct probabilities of the sample space. Your assumption that the sample space has a uniform distribution is just wishful thinking.
pbuk
pbuk said:
You are mistaken, it is you that is "trying to jump straight into the deep waters". I have shown you the way this material is taught to more than a million 14 year olds in the UK each year; IME most of them understand it.
I've been taught this material in a school also and I understood it. But in school it is learned in a matter which doesn't make you think about what you're doing, why you're doing it and where do these steps stem from, these are research questions one intends to find answers in a place of scholarship like the academia.
You seem to be missing the important point here but I've tried explaining it to you but you keep ignoring it.
I will give you another example that can relate to the way you're viewing the current discussion: A kid knows that if an object drops down then a force of gravity is acting on it and that its weight is ## w = m \cdot g ##. One who learns physics in the the academia will look again at the problem of the object falling down but will try to answer it in a much more research and rigorous oriented way, for example, he'll make a proper free body diagram, define coordinates, define the vectors, define the assumptions, maybe look at the energy expression of the object, etc... In short, he'll try to dive deeper into the details in order to gain a better understanding of the subject which will in turn merit him later on when he will deal with more complicated problems/situations concerning the area of his study, that is - the student is intended to expand his ways of thinking about those details that concern his field of erudition.
I hope I convinced you, If so then I hope it will make you a virtue; if you still remain unconvinced and continue to be the discerning critic of everything one has to say then there won't be any agreement between us as to the idea of how one should look at problems ( which, I say, sometimes there are isn't a "single" way to look at them and the many efforts of differentiation and attempts of understanding the convoluted paths to establish a clear understanding should also be appreciated ).
FactChecker said:
I think this is your fundamental error. The probability distribution of the sample space is not always uniform. It is up to you to calculate the correct probabilities using a valid mathematical process.
This is a very common situation, where the process presented is done in a sequence of steps (pick a drawer, then pick a sock from the drawer). The conditional probabilities of the results of the individual steps are clear. Use those to calculate the correct probabilities of the sample space. Your assumption that the sample space has a uniform distribution is just wishful thinking.
Thanks, I think that's enough for now. I'll try to unthink of this problem for the mean-time and approach to it later; and if that was not helpful then I'll discuss with my professor.
I often like to use extreme examples for "sanity checks" of my thinking.
Suppose you had 10 drawers. Nine have one black sock each and the tenth has a million white socks. You pick one drawer out of the ten with uniform probability and then pick a sock from that drawer with uniform probability. Even though the vast majority of socks are white, their probabilities add up to only 1/10.
phinds
FactChecker said:
I often like to use extreme examples for "sanity checks" of my thinking.
Yep. Often clarifies things quite nicely and usually with very little effort.
CGandC said:
How do I learn from basics if you're letting me jump straight to deep waters by thinking of probabilities? ( not the the problem's hard, I'm talking about establishing an understanding ) first I need to think of my sample space, then I would think what probability model I'd like on the elements of the sample space, in my case in is uniform model, then I'd start thinking about probabilities but only after I've combinatorically reasoned about the problem?
I don't think you can achieve what you hope to with this problem. As others have pointed out, for the sample space you constructed, the probabilities are not uniform, so you can't simply count to calculate the probabilities. You have to calculate them using different reasoning.
You might want to consider a really simple case. You toss two unfair coins. The sample space is {HH, HT, TH, TT}. How are you going to reason using combinatorics to derive the probability for each outcome? You can't. You need some other method and use more information to assign probabilities.
While I can understand your desire to solve the problem from first principles, I don't think it's particularly useful here. It would be similar to trying to find the derivative of ##\sin x^2## using the limit definition of the derivative rather than making use of the chain rule.
DrClaude, CGandC, pbuk and 1 other person
• Set Theory, Logic, Probability, Statistics
Replies
10
Views
918
• Set Theory, Logic, Probability, Statistics
Replies
10
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
276
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
2K
• Set Theory, Logic, Probability, Statistics
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
901
• Computing and Technology
Replies
4
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
10
Views
2K
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
2K |
# How do you find the first, fourth, and eighth terms of the sequence A(n) = -2*5^(n-1)?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
7
Jul 26, 2017
See a solution process below:
#### Explanation:
To find any term in the sequence, substitute the number of the term for $n$ and calculate the result:
First Term
$A \left(\textcolor{red}{n}\right) = - 2 \cdot {5}^{\textcolor{red}{n} - 1}$ becomes:
$A \left(\textcolor{red}{1}\right) = - 2 \cdot {5}^{\textcolor{red}{1} - 1} = - 2 \cdot {5}^{0} = - 2 \cdot 1 = - 2$
Fourth Term
$A \left(\textcolor{red}{n}\right) = - 2 \cdot {5}^{\textcolor{red}{n} - 1}$ becomes:
$A \left(\textcolor{red}{4}\right) = - 2 \cdot {5}^{\textcolor{red}{4} - 1} = - 2 \cdot {5}^{3} = - 2 \cdot 125 = - 250$
Eight Term
$A \left(\textcolor{red}{n}\right) = - 2 \cdot {5}^{\textcolor{red}{n} - 1}$ becomes:
$A \left(\textcolor{red}{8}\right) = - 2 \cdot {5}^{\textcolor{red}{8} - 1} = - 2 \cdot {5}^{7} = - 2 \cdot 78125 =$
$156 , 250$
• An hour ago
• An hour ago
• An hour ago
• An hour ago
• 3 minutes ago
• 12 minutes ago
• 14 minutes ago
• 38 minutes ago
• 51 minutes ago
• 58 minutes ago
• An hour ago
• An hour ago
• An hour ago
• An hour ago |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
4.1: Line Integrals
[ "article:topic", "Line Integrals", "work", "authorname:mcorral", "line integral over a scalar field", "line integral over a vector field" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
In single-variable calculus you learned how to integrate a real-valued function $$f (x)$$ over an interval $$[a,b]$$ in $$\mathbb{R}^1$$ . This integral (usually called a Riemann integral) can be thought of as an integral over a path in $$\mathbb{R}^1$$ , since an interval (or collection of intervals) is really the only kind of “path” in $$\mathbb{R}^1$$ . You may also recall that if $$f (x)$$ represented the force applied along the $$x$$-axis to an object at position $$x$$ in $$[a,b]$$, then the work $$W$$ done in moving that object from position $$x = a \text{ to }x = b$$ was defined as the integral:
$W=\int_a^b f (x)dx$
In this section, we will see how to define the integral of a function (either real-valued or vector-valued) of two variables over a general path (i.e. a curve) in $$\mathbb{R}^2$$ . This definition will be motivated by the physical notion of work. We will begin with real-valued functions of two variables.
In physics, the intuitive idea of work is that
$\text{Work = Force × Distance}$
Suppose that we want to find the total amount $$W$$ of work done in moving an object along a curve $$C$$ in $$\mathbb{R}^2$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, with a force $$f (x, y)$$ which varies with the position $$(x, y)$$ of the object and is applied in the direction of motion along $$C$$ (see Figure $$\PageIndex{1}$$ below).
Figure $$\PageIndex{1}$$ Curve $$C : x = x(t),\, y = y(t) \text{ for }t \text{ in }[a,b]$$
We will assume for now that the function $$f (x, y)$$ is continuous and real-valued, so we only consider the magnitude of the force. Partition the interval $$[a,b]$$ as follows:
$a = t_0 < t_1 < t_2 < ··· < t_{n−1} < t_n = b ,\text{ for some integer }n ≥ 2$
As we can see from Figure $$\PageIndex{1}$$, over a typical subinterval $$[t_i ,t_{i+1}]$$ the distance $$∆s_i$$ traveled along the curve is approximately $$\sqrt{∆x_i^2 +∆y_i^2}$$ , by the Pythagorean Theorem. Thus, if the subinterval is small enough then the work done in moving the object along that piece of the curve is approximately
$\text{Force × Distance} \approx f (x_{i∗}, y_{i∗}) \sqrt{ ∆x_i^2 +∆y_i^2}\label{Eq4.1}$
where $$(x_{i∗}, y_{i∗}) = (x(t_{i∗}), y(t_{i∗}))$$ for some $$t_{i∗} \text{ in }[t_i ,t_{i+1}]$$, and so
$W \approx \sum_{i=0}^{n-1} f (x_{i∗}, y_{i∗}) \sqrt{ ∆x_i^2 +∆y_i^2}\label{Eq4.2}$
is approximately the total amount of work done over the entire curve. But since
$\sqrt{ ∆x_i^2 +∆y_i^2} = \sqrt{\left ( \dfrac{∆x_i}{∆t_i} \right )^2 +\left ( \dfrac{∆y_i}{∆t_i}\right )^2}∆t_i$
where $$∆t_i = t_{i+1} − t_i$$ , then
$W \approx \sum_{i=0}^{n-1}f (x_{i∗}, y_{i∗})\sqrt{\left ( \dfrac{∆x_i}{∆t_i} \right )^2 + \left ( \dfrac{∆y_i}{∆t_i} \right )^2}∆t_i \label{Eq4.3}$
Taking the limit of that sum as the length of the largest subinterval goes to 0, the sum over all subintervals becomes the integral from $$t = a \text{ to }t = b$$, $$∆x_i ∆t_i \text{ and }∆y_i ∆t_i$$ become $$x ′ (t) \text{ and }y ′ (t)$$, respectively, and $$f (x_{i∗}, y_{i∗})$$ becomes $$f (x(t), y(t))$$, so that
$W=\int_a^b f (x(t), y(t)) \sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \label{Eq4.4}$
The integral on the right side of the above equation gives us our idea of how to define, for any real-valued function $$f (x, y)$$, the integral of $$f (x, y)$$ along the curve $$C$$, called a line integral:
Definition $$\PageIndex{1}$$: Line Integral of a scalar field
For a real-valued function $$f (x, y)$$ and a curve $$C$$ in $$\mathbb{R}^2$$, parametrized by $$x = x(t), y = y(t), a ≤ t ≤ b$$, the line integral of $$f (x, y)$$ along $$C$$ with respect to arc length $$s$$ is
$\int_C f (x, y)\,ds = \int_a^b f (x(t), y(t))\sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \label{Eq4.5}$
The symbol $$ds$$ is the differential of the arc length function
$s = s(t) = \int_a^t \sqrt{x ′ (u)^2 + y ′ (u)^2}\,du \label{Eq4.6}$
which you may recognize from Section 1.9 as the length of the curve $$C$$ over the interval $$[a,t]$$, for all $$t$$ in $$[a,b]$$. That is,
$ds = s ′ (t)\,dt = \sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt, \label{Eq4.7}$
For a general real-valued function $$f (x, y)$$, what does the line integral $$\int_C f (x, y)\,ds$$ represent? The preceding discussion of $$ds$$ gives us a clue. You can think of differentials as infinitesimal lengths. So if you think of $$f (x, y)$$ as the height of a picket fence along $$C$$, then $$f (x, y)\,ds$$ can be thought of as approximately the area of a section of that fence over some infinitesimally small section of the curve, and thus the line integral $$\int_C f (x, y)\,ds$$ is the total area of that picket fence (see Figure $$\PageIndex{2}$$).
Figure $$\PageIndex{2}$$: Area of shaded rectangle = height × width ≈ $$f (x, y)\,ds$$
Example $$\PageIndex{1}$$
Use a line integral to show that the lateral surface area $$A$$ of a right circular cylinder of radius $$r$$ and height $$h$$ is $$2\pi rh$$.
Solution:
We will use the right circular cylinder with base circle $$C$$ given by $$x^2 + y^2 = r^2$$ and with height $$h$$ in the positive $$z$$ direction (see Figure $$\PageIndex{3}$$). Parametrize $$C$$ as follows:
$x = x(t) = r \cos t , y = y(t) = r \sin t , 0 ≤ t ≤ 2π$
Figure $$\PageIndex{3}$$
\nonumber \begin{align} A&=\int_C f (x, y)\,ds = \int_a^b f (x(t), y(t))\sqrt{x ′ (t)^2 + y ′ (t)^2}\,dt \\ \nonumber &=\int_0^{2\pi} h \sqrt{(−r \sin t)^2 +(r \cos t)^2}\,dt \\ \nonumber &=h\int_0^{2\pi} r \sqrt{\sin^2 t+\cos^2 t}\,dt \\ \nonumber &=rh\int_0^{2\pi} 1\,dt = 2\pi rh \\ \end{align}
Note in Example $$\PageIndex{1}$$ that if we had traversed the circle $$C$$ twice, i.e. let t vary from $$0 \text{ to }4\pi$$ then we would have gotten an area of $$4\pi rh$$, i.e. twice the desired area, even though the curve itself is still the same (namely, a circle of radius $$r$$). Also, notice that we traversed the circle in the counter-clockwise direction. If we had gone in the clockwise direction, using the parametrization
$x = x(t) = r \cos (2π− t) , y = y(t) = r \sin (2π− t) , 0 ≤ t ≤ 2π ,\label{Eq4.8}$
then it is easy to verify (Exercise 12) that the value of the line integral is unchanged.
In general, it can be shown (Exercise 15) that reversing the direction in which a curve $$C$$ is traversed leaves $$\int_C f (x, y)\,ds$$ unchanged, for any $$f (x, y)$$. If a curve $$C$$ has a parametrization $$x = x(t), y = y(t), a ≤ t ≤ b,$$ then denote by $$−C$$ the same curve as $$C$$ but traversed in the opposite direction. Then $$−C$$ is parametrized by
$x = x(a+ b − t) , y = y(a+ b − t) , a ≤ t ≤ b ,\label{Eq4.9}$
and we have
$\int_C f (x, y)\,ds =\int_{-C}f (x, y)\,ds .\label{Eq4.10}$
Notice that our definition of the line integral was with respect to the arc length parameter $$s$$. We can also define
$\int_C f (x, y)\,dx=\int_a^b f (x(t), y(t)) x ′ (t)\,dt\label{Eq4.11}$
as the line integral of $$f (x, y)$$ along $$C$$ with respect to $$x$$, and
$\int_C f (x, y)\,d y=\int_a^b f (x(t), y(t)) y ′ (t)\,dt \label{Eq4.12}$
as the line integral of $$f (x, y)$$ along $$C$$ with respect to $$y$$.
In the derivation of the formula for a line integral, we used the idea of work as force multiplied by distance. However, we know that force is actually a vector. So it would be helpful to develop a vector form for a line integral. For this, suppose that we have a function $$f(x, y)$$ defined on $$\mathbb{R}^2$$ by
$\nonumber \textbf{f}(x, y) = P(x, y)\textbf{i} + Q(x, y)\textbf{j}$
for some continuous real-valued functions $$P(x, y)$$ and $$Q(x, y) \text{ on }\mathbb{R}^2$$ . Such a function $$f$$ is called a vector field on $$\mathbb{R}^2$$ . It is defined at points in $$\mathbb{R}^2$$ , and its values are vectors in $$\mathbb{R}^2$$ . For a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, let
$\nonumber \textbf{r}(t) = x(t)\textbf{i} + y(t)\textbf{j}$
be the position vector for a point $$(x(t), y(t))$$ on $$C$$. Then $$\textbf{r}'(t) = x'(t)\textbf{i} + y'(t)\textbf{j}$$ and so
\nonumber \begin{align} \int_C P(x, y)\,dx+ \int_C Q(x, y)\,d y &=\int_a^b P(x(t), y(t)) x ′ (t)\,dt+\int_a^b Q(x(t), y(t)) y ′ (t)\,dt \\ \nonumber &=\int_a^b (P(x(t), y(t)) x ′ (t)+Q(x(t), y(t)) y ′ (t))\,dt \\ \nonumber &=\int_a^b \textbf{f}(x(t), y(t))\cdot \textbf{r} ′ (t)dt \\ \end{align}
by definition of $$f(x, y)$$. Notice that the function $$f(x(t), y(t))\cdot r ′ (t)$$ is a real-valued function on $$[a,b]$$, so the last integral on the right looks somewhat similar to our earlier definition of a line integral. This leads us to the following definition:
Definition $$\PageIndex{2}$$: Line Integral of a vector feild
For a vector field $$\textbf{f}(x, y) = P(x, y)\textbf{i} +Q(x, y)\textbf{j}$$ and a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$, the line integral of f along $$C$$ is
\begin{align} \int_C \textbf{f}\cdot d\textbf{r} &= \int_C P(x, y)\,dx+\int_C Q(x, y)\,d y \label{Eq4.13} \\ &=\int_a^b \textbf{f}(x(t), y(t))\cdot \textbf{r} ′ (t)\,dt \label{Eq4.14} \\ \end{align}
where $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$ is the position vector for points on $$C$$.
We use the notation $$d\textbf{r} = \textbf{r} ′ (t)\,dt = dx\textbf{i}+ d y\textbf{j}$$ to denote the differential of the vector-valued function r. The line integral in Definition $$\PageIndex{2}$$ is often called a line integral of a vector field to distinguish it from the line integral in Definition $$\PageIndex{1}$$ which is called a line integral of a scalar field. For convenience we will often write
$\nonumber \int_C P(x, y)\,dx +\int_C Q(x, y)\,d y =\int_C P(x, y)\,dx+Q(x, y)\,d y ,$
where it is understood that the line integral along $$C$$ is being applied to both $$P \text{ and }Q$$. The quantity $$P(x, y)\,dx +Q(x, y)\,d y$$ is known as a differential form. For a real-valued function $$F(x, y)$$, the differential of $$F$$ is $$dF = \dfrac{∂F}{∂x}\,dx+ \dfrac{∂F}{∂y}\, d y.$$ A differential form $$P(x, y)\,dx+Q(x, y)\,d y$$ is called exact if it equals $$dF$$ for some function $$F(x, y)$$.
Recall that if the points on a curve $$C$$ have position vector $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$, then $$\textbf{r} ′ (t)$$ is a tangent vector to $$C$$ at the point $$(x(t), y(t))$$ in the direction of increasing $$t$$ (which we call the direction of $$C$$). Since $$C$$ is a smooth curve, then $$\textbf{r} ′ (t) \neq \textbf{0} \text{ on }[a,b]$$ and hence
$\nonumber \textbf{T}(t) = \dfrac{\textbf{r}'(t)}{\left \lVert \textbf{r}'(t) \right \rVert}$
is the unit tangent vector to $$C$$ at $$(x(t), y(t))$$. Putting Definitions $$\PageIndex{1}$$ and $$\PageIndex{2}$$ together we get the following theorem:
Theorem $$\PageIndex{1}$$
For a vector field $$\textbf{f}(x, y) = P(x, y)\textbf{i} + Q(x, y)\textbf{j}$$ and a curve $$C$$ with a smooth parametrization $$x = x(t), y = y(t), a ≤ t ≤ b$$ and position vector $$\textbf{r}(t) = x(t)\textbf{i}+ y(t)\textbf{j}$$,
$\int_C \textbf{f}\cdot d\textbf{r} = \int_C \textbf{f}\cdot \textbf{T}\,ds,\label{Eq4.15}$
where $$\textbf{T}(t) = \dfrac{\textbf{r} ′ (t)}{ \left \lVert \textbf{r} ′ (t)\right \rVert }$$ is the unit tangent vector to $$C$$ at $$(x(t), y(t))$$.
If the vector field $$\textbf{f}(x, y)$$ represents the force moving an object along a curve $$C$$, then the work $$W$$ done by this force is
$W = \int_C \textbf{f}\cdot \textbf{T} \, ds = \int_C \textbf{f}\cdot d\textbf{r} \label{Eq4.16}$ |
Victorian Curriculum Year 4 - 2021 Edition
5.06 Compare fractions
Lesson
Being able to plot a fraction on a number line can help us compare fractions in this lesson. Let's try this problem as a review.
Plot $\frac{1}{10}$110 on the number line.
## Learn
This video looks at comparing fractions using area models.
### Apply
##### question 1
Which fraction is smaller?
1. $\frac{2}{6}$26
A
$\frac{3}{6}$36
B
$\frac{2}{6}$26
A
$\frac{3}{6}$36
B
## Learn
This video shows how to use number lines to compare fractions.
### Apply
##### question 2
Think about the fractions $\frac{3}{4}$34 and $\frac{4}{5}$45.
1. Plot the number $\frac{3}{4}$34 on the number line.
2. Plot the number $\frac{4}{5}$45 on the number line.
3. The two numbers can be shown on the same number line like this:
Which number is bigger?
$\frac{3}{4}$34
A
$\frac{4}{5}$45
B
$\frac{3}{4}$34
A
$\frac{4}{5}$45
B
## Learn
What about mixed numbers or improper fractions? This video shows us how to compare using these.
### Apply
##### question 3
Compare the two fractions by entering the greater than ($>$>) or less than ($<$<) symbol in the box.
1. $1\frac{8}{9}\editable{}1\frac{4}{9}$189149
Remember!
• The denominator tells us how many equal parts to split the number line into.
• The numerator tells us how many of those equal parts to select.
### Outcomes
#### VCMNA158
Count by quarters, halves and thirds, including with mixed numerals. Locate and represent these fractions on a number line |
# How do you do integrals in calculus?
## How do you do integrals in calculus?
We define integrals as the function of the area bounded by the curve y = f(x), a ≤ x ≤ b, the x-axis, and the ordinates x = a and x =b, where b>a. Let x be a given point in [a,b]. Then b∫af(x)dx ∫ a b f ( x ) d x represents the area function.
### Is integral calculus hard?
Integration is generally much harder than differentiation. This little demo allows you to enter a function and then ask for the derivative or integral. Differentiation is typically quite easy, taking a fraction of a second. Integration typically takes much longer, if the process completes at all!
What is integral calculus in simple words?
In calculus, an integral is the space under a graph of an equation (sometimes said as “the area under a curve”). An integral is the reverse of a derivative, and integral calculus is the opposite of differential calculus. The word “integral” can also be used as an adjective meaning “related to integers”.
Is integral the same as antiderivative?
Antiderivatives are related to definite integrals through the fundamental theorem of calculus: the definite integral of a function over an interval is equal to the difference between the values of an antiderivative evaluated at the endpoints of the interval.
## Who uses calculus in real life?
Calculus is the language of engineers, scientists, and economists. The work of these professionals has a huge impact on our daily life – from your microwaves, cell phones, TV, and car to medicine, economy, and national defense.
### What is limit in calculus?
A limit tells us the value that a function approaches as that function’s inputs get closer and closer to some number. The idea of a limit is the basis of all calculus.
What happens if you integrate position?
speed. Velocity is rate of change in position, so its definite integral will give us the displacement of the moving object. Speed is the rate of change in total distance, so its definite integral will give us the total distance covered, regardless of position.
What is beginner calculus?
Calculus is a study of rates of change of functions and accumulation of infinitesimally small quantities. It can be broadly divided into two branches: Differential Calculus. This concerns rates of changes of quantities and slopes of curves or surfaces in 2D or multidimensional space.
## Why do I like calculus?
Finding Real World Applications. Recognize that calculus is the study of change. It is easier to enjoy something if you understand it, and it is easier to understand something if you can see a purpose for it. Calculus has many real world applications every day.
### What is the definition of integral in calculus?
Integral calculus. The definite integral inputs a function and outputs a number, which gives the algebraic sum of areas between the graph of the input and the x-axis. The technical definition of the definite integral involves the limit of a sum of areas of rectangles, called a Riemann sum .
What is an indefinite integral in calculus?
the integral is called an indefinite integral, which represents a class of functions (the antiderivative) whose derivative is the integrand. The fundamental theorem of calculus relates the evaluation of definite integrals to indefinite integrals.
What does integration mean?
Integration is the calculation of an integral. Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc. When we speak about integrals, it is related to usually definite integrals. The indefinite integrals are used for antiderivatives. |
Max/Min Problems Resulting in Quadratic Functions
Max/Min Problems Resulting in Quadratic Functions
This lesson offers a variety of word/story problems that are solved by finding the maximum or minimum value of a quadratic function.
Suggested review:
General Strategy for Optimization (Max/Min) Problems
1. What are you maximizing/minimizing?
Read the problem carefully.
Answer the question: ‘What am I being asked to maximize/minimize?’
Look for ‘extreme’ wordsfor example: maximum, minimum, greatest, least, biggest, smallest, largest, most.
2. Write Expression
Write an expression for the thing you need to maximize/minimize (from step 1). You may want to make a sketch. You may need to give names to unknown quantities.
3. One Variable Only
Does your expression from step 2 involve more than one variable? If so, look for other known information that allows you to eliminate variable(s). You must end up with an expression that involves only one variable.
Give a meaningful function name to your expression, so function notation can be used to write your work concisely.
4. Find and Report
Use appropriate techniques to find the desired maximum/minimum value, and where it occurs.
Use a complete sentence to report your result.
Example: Max/Min Problem
Among all rectangles that have a perimeter of twenty feet, find the dimensions of the one with the largest area.
Pre-Solution: Stop and think! Do the dimensions actually affect the area?
• $1\text{ ft} \times 9\text{ ft}$ rectangle: perimeter is $\,20\text{ ft}\,,$ area is $\,9\text{ sq ft}$
• $2\text{ ft} \times 8\text{ ft}$ rectangle: perimeter is $\,20\text{ ft}\,,$ area is $\,16\text{ sq ft}$
Yes, the dimensions do affect the area.
(You've probably already guessed the answer at this point! But, this is a nice simple question to clearly illustrate the general procedure.)
Solution
1. What are you maximizing/minimizing?
Key word: ‘largest’.
Largest what? Largest area.
You need to maximize the area of a rectangle.
2. Write Expression
You need an expression for the area of a rectangle: area equals length times width. Choose meaningful names: let $\,\ell\,$ be the length, $\,w\,$ be the width, and $\,A\,$ be the area.
Let $\,w\,$ and $\,\ell\,$ have units of feet; $\,A\,$ therefore has units of square feet.
With these names: $A = \ell w$
At this moment, the expression for $\,A\,$ involves two variables, not one.
3. One Variable Only
Re-read the problem. Can we write down anything that is true that involves our named variables?
Yes—the rectangle must have a perimeter of $\,20\,$ feet. Thus, $\,2\ell + 2w = 20\,.$
This allows us to solve for one variable in terms of the other. You can solve for $\,\ell\,$ in terms of $\,w\,,$ or vice versa—your choice.
Sometimes it's easier to solve for one variable than another, but here there's no difference. I choose to solve for $\,\ell\,$:
$$\begin{gather} \cssId{s63}{2\ell + 2w = 20}\cr \cssId{s64}{\ell + w = 10}\cr \cssId{s65}{\ell = 10-w}\tag{*} \end{gather}$$
So: $$A = \ell w = (10-w)w$$
At this point, $\,A\,$ is a function of only one variable, $\,w\,.$
Switch to function notation:
\begin{align} A(w) \ &\cssId{s69}{= (10-w)w}\cr &\cssId{s70}{= 10w - w^2}\cr &\cssId{s71}{= -w^2 + 10w} \end{align}
Note that $\,A\,$ is a quadratic function that is concave down (sheds water), so it has a maximum value.
4. Find and Report
The vertex formula says that the vertex of $\,f(x) = ax^2 + bx + c\,$ is:
$$\cssId{s77}{\bigl(-\frac{b}{2a},f(-\frac{b}{2a})\bigr)}$$
For $\,A(w) = -w^2 + 10w\,,$ we have $\,a = -1\,,$ $\,b = 10\,.$
Thus: $$-\frac{b}{2a} = -\frac{10}{2(-1)} = 5$$
This is the $x$-value of the vertex of the quadratic area function.
That is, when the width is $\,5\,$ ($\,w = 5\,$) the maximum area occurs.
Note from (*) that when $\,w = 5\,$ we have:
$$\cssId{s84}{\ell = 10-w = 10-5 = 5}$$
The rectangle that achieves the maximum area is actually a $\,5\times 5\,$ square.
What is the maximum area?
\begin{align} A(5)\ &\cssId{s87}{= -5^2 + 10\cdot 5}\cr &\cssId{s88}{= -25 + 50}\cr &\cssId{s89}{= 25} \end{align}
This is the $y$-value of the vertex of the quadratic area function.
Among all rectangles with a perimeter of $\,20\,$ feet, the one with dimensions $\,5\text{ ft}\times 5\text{ ft}\,$ has the largest area. This largest area is $\,25\,$ square feet.
Example: Max/Min Problem
Find two positive numbers $\,x\,$ and $\,y\,$ that add to $\,30\,,$ and for which $\,2x^2 + 5y^2\,$ is a minimum.
Pre-Solution: Stop and think! Get a feeling for the problem.
• $x = 1\,,$ $\,y = 29\,$:
$2x^2 + 5y^2 = 2(1^2) + 5(29^2) = 4207$
• $x = 2\,,$ $\,y = 28\,$:
$2x^2 + 5y^2 = 2(2^2) + 5(28^2) = 3928$
Yes, the choices for $\,x\,$ and $\,y\,$ do affect the quantity $\,2x^2 + 5y^2\,.$
For this question, you're probably not able to guess the answer.
There's no restriction that $\,x\,$ and $\,y\,$ must be whole numbers, so $\,x = 1.5\,$ and $\,y = 28.5\,$ is allowable.
Solution
1. What are you maximizing/minimizing?
Key word: ‘minimum’.
You need to minimize the quantity $\,2x^2 + 5y^2\,.$
2. Write Expression
We already have the expression; let's call it $\,S\,$ for sum. So, $\,S = 2x^2 + 5y^2\,.$
At this moment, $\,S\,$ is a function of two variables.
3. One Variable Only
Re-read the problem. Can we write down anything that is true that involves $\,x\,$ and $\,y\,$?
Yes—they must add to $\,30\,.$ Thus, $\,x + y = 30\,.$
This allows us to solve for one variable in terms of the other. Choosing to solve for $\,y\,$ gives:
$$\cssId{s124}{y = 30 - x}\tag{\cssId{s125}{*}}$$
So:
\begin{align} S\ &\cssId{s127}{= 2x^2 + 5y^2}\cr &\cssId{s128}{= 2x^2 + 5(30-x)^2}\cr &\cssId{s129}{= \text{(you check these steps)}}\cr &\cssId{s130}{= 7x^2 - 300x + 4500} \end{align}
At this point, $\,S\,$ is a function of only one variable, $\,x\,.$
Switch to function notation:
$$\cssId{s133}{S(x) = 7x^2 - 300x + 4500}$$
Note that $\,S\,$ is a quadratic function that is concave up (holds water), so it has a minimum value.
4. Find and Report
The vertex formula says that the vertex of $\,f(x) = ax^2 + bx + c\,$ is:
$$\cssId{s138}{\bigl(-\frac{b}{2a},f(-\frac{b}{2a})\bigr)}$$
For $\,S(x) = 7x^2 - 300x + 4500\,,$ we have $\,a = 7\,,$ $\,b = -300\,.$
Thus:
$$\cssId{s142}{-\frac{b}{2a} = -\frac{(-300)}{2(7)}} \cssId{s143}{= \frac{150}{7}}$$
This is the $x$-value of the vertex of the quadratic function $\,S\,.$
That is, when $\,x = \frac{150}{7}\,,$ the minimum value occurs.
Note from (*) that when $\,x = \frac{150}{7}\,,$ we have:
\begin{align} y\ &\cssId{s150}{= 30 - \frac{150}{7}}\cr\cr &\cssId{s151}{= \frac{210}{7} - \frac{150}{7}}\cr\cr &\cssId{s152}{= \frac{60}{7}} \end{align}
What is the minimum value of $\,2x^2 + 5y^2\,$?
\begin{align} S(\frac{150}{7})\ &\cssId{s154}{= 2(\frac{150}{7})^2 + 5(\frac{60}{7})^2}\cr\cr &\cssId{s155}{= \text{(after a bit of work)}}\cr\cr &\cssId{s156}{= \frac{9000}{7}} \end{align}
This is the $y$-value of the vertex of the quadratic function $\,S\,.$
The two positive numbers $\,x\,$ and $\,y\,$ that add to $\,30\,$ and make $2x^2 + 5y^2\,$ as small as possible are $\,x = \frac{150}{7}\,$ and $\,y = \frac{60}{7}\,.$ This smallest value of $\,2x^2 + 5y^2\,$ is $\,\frac{9000}{7}\,.$
To make your life easier, zip up to WolframAlpha and type in:
vertex of 2x^2 + 5(30-x)^2
How easy is that?! |
Lesson 4
Equal Groups of Non-Unit Fractions
Warm-up: Notice and Wonder: Thirds (10 minutes)
Narrative
This warm-up prompts students to examine a diagram representing equal groups of non-unit fractions. The understandings elicited here allow students to discuss the relationship between the product of a whole number and a unit fraction and that of a whole number and a non-unit fraction with the same denominator.
Launch
• Groups of 2
• Display the image.
Activity
• “What do you notice? What do you wonder?”
• 1 minute: quiet think time
• 1 minute: partner discussion
• Share and record responses.
Student Facing
What do you notice? What do you wonder?
Activity Synthesis
• If no students notice or wonder about equal groups, ask, “What groups do you see and how do you see them?” (4 wholes, each whole has $$\frac{2}{3}$$ shaded)
• “How many thirds do you see?” (8 thirds)
• “How are these diagrams different than those we've seen so far in this unit?” (Previously, each whole has only one shaded part. These have two shaded parts each.)
• “Today, we will think about situations that involve equal groups but now each group has non-unit fractions.”
Activity 1: Jars of Jam (15 minutes)
Narrative
In this 5 Practices activity, students reason about a situation that involves finding the product of a whole number and a non-unit fraction. They may rely on what they previously learned about multiplying a whole number and a unit fraction, but can reason in any way that makes sense to them. The goal is to elicit different strategies and help students see the connections between strategies and with their earlier work. Students reason abstractly and quantitively as they solve the problem (MP2) and construct arguments (MP3) as they share their reasoning during the synthesis.
Monitor for the students who:
• draw a drawing or a diagram to show 5 groups with three $$\frac{1}{4}$$s in each group and count the total number of fourths
• reason additively, by finding the value of $$\frac{3}{4} + \frac{3}{4} + \frac{3}{4} + \frac{3}{4} + \frac{3}{4}$$, or by adding smaller groups of $$\frac{3}{4}$$ at a time, for instance, 2 groups of $$\frac{3}{4}$$, another 2 groups, and 1 more group
• reason multiplicatively, for instance, by thinking of $$\frac{3}{4}$$ as $$3 \times \frac{1}{4}$$ and then finding $$5 \times 3 \times \frac{1}{4}$$, or by reasoning about $$5 \times \frac{3}{4}$$
Students who see the situation as $$5 \times \frac{3}{4}$$ may, based on their earlier work, generalize that the value is $$\frac{5 \times 3}{4}$$. Encourage them to clarify how they know this is the case.
During the synthesis, sequence student presentations in the order listed.
Reading: MLR6 Three Reads. “We are going to read this 3 times.” After the 1st Read: “Tell your partner what this situation is about.” After the 2nd Read: “List the quantities. What can be counted or measured?” (number of jars, number of friends, number of cups of jam). After the 3rd Read: “What strategies can we use to solve this problem?”
Representation: Internalize Comprehension. Synthesis: Invite students to identify details they want to remember. Display the sentence frame: “The next time I need to represent the product of a whole number and a fraction, I will . . . .”
Supports accessibility for: Conceptual Processing, Organization, Memory
Launch
• Groups of 2
• Read the first problem as a class.
• Invite students to share what they know about homemade jams or any experience in making them.
• If needed, remind students that measuring cups come in different fractional amounts, such as $$\frac{1}{4}$$, $$\frac{1}{2}$$, and $$\frac{3}{4}$$.
Activity
• “Work independently on the problem. Explain or show your reasoning so that it can be followed by others. Afterwards, share your thinking with your partner.”
• 5 minutes: independent work time
• 2–3 minutes: partner discussion
• Monitor for the strategies listed in the activity narrative.
Student Facing
Elena fills 5 small jars with homemade jams to share with her friends. Each jar can fit $$\frac{3}{4}$$ cup of jam. How many cups of jam are in the jars? Explain or show your reasoning.
If you have time: Elena still has some jam left. She takes 2 large jars and puts $$\frac{5}{4}$$ cups of jam in each jar. How many cups of jam are in the jars?
Student Response
If students are not sure how to represent or reason about 5 groups of $$\frac{3}{4}$$, consider asking them: “How would you represent (or think about) 5 groups of $$\frac{1}{4}$$?” and “How can you build on that representation (or strategy) to find how much is in 5 groups of $$\frac{3}{4}$$?”
Activity Synthesis
• Select previously identified students to share their responses. Display or record their work for all to see.
• “What multiplication expression can represent the amount of jam in the jars? How do you know?” ($$5 \times \frac{3}{4}$$ or $$5 \times 3 \times \frac{1}{4}$$, because there are 5 equal groups of $$\frac{3}{4}$$.)
• “Where do you see the 5 groups in each strategy presented? Where do we see the $$\frac{3}{4}$$?”
• “How is finding the value of $$5 \times \frac{3}{4}$$ like finding the value of $$5 \times \frac{1}{4}$$?” (They're both about finding the total amount in equal groups. They both involve a whole number of groups and a fraction in each group.)
• “How is it different?” (The amount in each group is a non-unit fraction instead of a unit fraction.)
Activity 2: How Do We Multiply? (20 minutes)
Narrative
The purpose of this activity is for students to use diagrams to reason about products of a whole number and a non-unit fraction with diagrams, building on their work with diagrams that represent products of a whole number an a unit fraction. They begin to generalize that the number of shaded parts in a diagram that represents $$n \times \frac{a}{b}$$ is $$n \times a$$ and to explain that generalization (MP8).
Launch
• Groups of 2
• “Let's represent some other products of a whole number and a fraction and find their values.”
Activity
• “Take a few quiet minutes to work on the activity. Afterwards, share your responses with your partner.”
• 5–7 minutes: independent work
• 2–3 minutes: partner discussion
• Monitor for the strategies students use to reason about the last two problems.
• Identify students who reason visually (using diagrams), additively, and multiplicatively to share in the synthesis.
Student Facing
1. This diagram represents $$\frac{2}{5}$$.
1. Show how you would use or adjust the diagram to represent $$4 \times \frac{2}{5}$$.
2. What is the value of the shaded parts in your diagram?
2. This diagram represents $$\frac{5}{8}$$.
1. Show how you would use or adjust the diagram to represent $$3 \times \frac{5}{8}$$.
2. What is the value of the shaded parts in your diagram?
3. Find the value of each expression. Draw a diagram if you find it helpful. Be prepared to explain your reasoning.
1. $$2 \times \frac {1}{6}$$
2. $$2 \times \frac{4}{6}$$
3. $$2 \times \frac {5}{6}$$
4. $$4 \times \frac{5}{6}$$
4. Mai said that to multiply any fraction by a whole number, she would multiply the whole number and the numerator of the fraction and keep the same denominator. Do you agree with Mai? Explain your reasoning.
Activity Synthesis
• Discuss the four multiplication expressions in the third problem.
• Select 1–2 students who might have drawn diagrams for all expressions.
• “How does your diagram show the value of $$2 \times \frac{4}{6}$$?” (There are 2 groups of $$\frac{4}{6}$$ , so there are 8 sixths shaded, which is $$\frac{8}{6}$$.)
• Select 1–2 students who drew a diagram for some expressions and reason numerically for others.
• “Why did you choose to draw a diagram for some expressions and to do something else for others?” (After drawing the first two diagrams, I realized that I'd have to draw a lot of groups or parts, so I thought about the numbers instead.)
• Select 1–2 students who reasoned about all expressions numerically.
• “How did you find the value of the expressions without drawing diagrams at all?” (I saw a pattern in earlier problems, that we can multiply the whole number and the numerator of the fraction and keep the denominator.)
• Discuss the last problem in the lesson synthesis.
Lesson Synthesis
Lesson Synthesis
“Mai said she can multiply any fraction by a whole number by multiplying the whole number by the numerator and keeping the denominator.”
Invite students to share whether they agree or disagree with Mai's statement and to explain their reasoning.
“Let’s discuss Mai’s reasoning using the expression $$4 \times \frac{2}{3}$$ and the diagram from today’s warm-up.”
“Why can we multiply $$4 \times 2$$ to get the numerator of the product?” (We can think in terms of thirds. The diagram shows 4 groups of 2 thirds, or 8 thirds total.)
“Why is the denominator of the product the same as the fraction in the expression?” (The denominator represents the size of the equal parts in each group. The size of the part doesn’t change when the number of groups increases.) |
12.5 - Logarithmic graphs
In section 6.1 we learned about the cartesian plane, rectangular coordinates, graphs and how to identify points on graphs. Before reading this section you may want to review that section.
Henceforth we will call the graphs described in section 6.1 linear graphs. In this section we will learn how to plot quantities on a new type of graph called a logarithmic graph. We will see how this can make it easier to determine the functional relationship between certain quantities.
We will motivate logarithmic graphs by giving two examples.
Example 1: Let t represent time and let i represent the electric current flowing in some electric circuit. The data shown in the table was collected for t and i. (For later use, the natural logarithm of the entries in column 2 was taken and entered in column 3.)
Find an equation relating i and t.
Solution: First we try plotting i versus t. But this only produces a curve as shown to the right and from this curve it is difficult to find the equation.
Next we try plotting ln (i) versus t. This does produce a straight line as shown to the right and we know how to find the equation of a straight line.
To find the equation of this line we will pretend for a moment that the axes are labelled y and x. Then this line must have the equation y = m x + b, where m and b are yet to be determined. Eventually we’ll put back the proper label ln (i) for y and t for x.
To determine m and b we will use the elimination method discussed in section 6.2. Let’s use these two points which lie on the line:
(x = 2.00 , y = 0.94) and (x = 8.00, y = −1.05).
Substituting the first point into the equation of the straight line, y = m x + b, gives the equation
0.94 = 2 m + b.
Substituting the second point into the equation of the straight line, y = m x + b, gives the equation
−1.05 = 8 m + b.
This pair of equations,
0.94 = 2 m + b
−1.05 = 8 m + b
is a system of two equations for the two unknowns m and b. We can solve them by elimination. Subtracting the first equation from the second equation eliminates b and gives:
−1.99 = 6 m m = −0.33
Back-substituting this value of m into, say the first equation, 0.94 = 2 m + b, then gives 0.94 = (2)(−0.33) + b which gives b = 1.6. So the equation of the straight line is
y = −0.33 x + 1.6.
Now replace back y by ln(i) and x by t:
ln (i) = −0.33 t + 1.6.
Antilogging this equation gives the equation
i = e −0.33 t + 1.6,
or simplifying,
i = 5 e −0.33 t.
The key lesson learned from this example is that the graph of any exponential function y = a e b x is a straight line when we plot ln (y) versus x.
Example 2: The curve to the right could be the graph of y = x 2, y = x 3 or y = x 4. From a plot of y versus x it is difficult to tell which it is.
But if we plot ln (y) versus ln (x) then these three functions become straight lines. These functions are called power functions because the variable x is raised to some power.
The key lesson learned from this example is that the graph of any power function y = a x b is a straight line when we plot ln (y) versus ln (x).
Semilog And log-log graph paper
The figure above shows a logarithmic scale of the kind used on logarithmic graph paper. It is not a linear scale because larger numbers are allotted less space than smaller numbers. The lower part of the picture shows the logarithmic scale in more detail. The numbers along the axis are located where their logarithms would be placed on linear graph paper. This means that we can plot x itself on logarithmic graph paper rather than plot log (x) on linear graph paper. We avoid the job of taking logarithms on the calculator. Notice that moving to the left along a logarithmic axis only makes the numbers smaller and smaller. Zero is located an infinite distance to the left and negative numbers do not exist. The range from one power of 10 to the next is called a cycle.
Two types of logarithmic graphs are useful:
the semi-log graph, which has a logarithmic vertical scale and a linear horizontal scale, as shown below. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below.
The equations of straight lines on logarithmic graph paper
One purpose of logarithmic graph paper is simply to put wide ranges of data on one graph. Another purpose is to quickly check if a function follows an exponential law or a power law. We saw one example showing that plotting ln (y) versus ln (x) caused power law functions of the form y = a x b to become straight lines. And we saw another example showing that plotting ln (y) versus x caused an exponential function of the form y = a e b x to become a straight line. From these two examples we conclude the following:
A straight line on a semilog graph of y versus x represents an exponential function of the form y = a e b x. A straight line on a log-log graph of y versus x represents a power law function of the form y = a x b. To find the constants a and b, we can substitute two widely-spaced points which lie on the line into the appropriate equation. This gives two equations for the two unknowns a and b which can be solved by elimination.
Example: Find the equation of the straight line in the graph to the right.
Solution: A straight line on a log-log graph of Q versus T represents the power law function
Q = a T b.
We must find the values of a and b. To do this we will use a variation of the method described in section 6.2.
Take the two points shown in the graph and substitute them, one at a time, into the equation Q = a T b. This gives two equations in the two unknown constants a and b:
Here is the variation: first take natural logarithms of both equations
and then subtract the two equations. This eliminates ln (a). Then solve for b:
Note that b is the power of the power function and is often called the “slope of the log-log graph” and this equation is often used as a shortcut to compute it.
Back-substituting b into either of the previous equations gives ln (a) = 0.7120, and anti-logging gives a = 2.04. The constant a is often called the “intercept” although it occurs at T = 1, not at T = 0. (Remember that 0 does not exist in a logarithmic scale.)
So to 3 sig. figs. the equation describing the line is
Q = 2.04 t 0.558.
Example: Find the equation of the straight line in the graph to the right.
Solution: A straight line on a semi-log graph of P versus t represents the exponential function
P = a e b t.
We must find the values of a and b. To do this we will use a variation of the method described in section 6.2.
Take the two points shown in the graph and substitute them, one at a time, into the equation P = a e b t. This gives two equations in the two unknown constants a and b:
Here is the variation: first take natural logarithms of both equations
and then subtract the two equations. This eliminates ln (a). Then solve for b:
Note that b is the rate of the exponential function and is often called the “slope of the semilog graph” and this equation is often used as a shortcut to compute it.
Back-substituting b into either of the previous equations gives ln (a) = 4.377, and anti-logging gives a = 79.6. Note that the constant a is the “y intercept”. So to 3 sig. figs. the equation describing the line is
P = 79.6 e −0.461 t.
Algebra Coach Exercises
If you found this page in a web search you won’t see the |
# Interactive Charts Examining California's Prison System
Save Article
## Failed to save article
Who does California place behind bars? According to the CDCR 2011 data, most inmates originate from Southern California and are non-white. Other data also shows that around 20 percent of inmates are age 50 and up and 95 percent of all inmates are male. Explore KQED’S The Lowdown’s interactive charts that take a closer look at the demographics of inmates in California.
Find hundreds more engaging math-focused media and integrated activities, all aligned with CCSS at PBS LearningMedia.
Suggested Activity:
Learning Outcomes
Students will be able to
• solve problems involving finding a part, given the whole and the percent
• use data to make an argument
Common Core State Standards: 6.RP.A.3c, 6.SP.B.5
Vocabulary: Percentages, ratios, infographic, incarcerated
Materials: Calculators; Infographic handout
Preparation: Make a pie chart with some actual data about students. Suggestions include birthday season, eye color, or number of siblings. Data should be represented as percentages.
Procedure
1. Optional Introduction (5–10 minutes, whole group)
Begin by showing the pie chart you prepared. Ask students for their observations: What information does the chart show? Explain that the pie chart shows percentages, but that you can figure out the actual numbers that the percentages represent if you know the total number of people who were surveyed. (If necessary, remind students that percentages are ratios, and that they express quantities as part of 100.) Show students this equation:
[Total # surveyed] * [% Responding] = [# of Respondents]
Now enter data into the equation and solve for the number of respondents. Repeat for each category in the pie chart so that students see that the total number of respondents equals the total number surveyed.
After students have figured out the number of respondents for each category, ask them whether the data help them come to any conclusions about the distribution of birthdays, eye color, or number of siblings. What observations can they make based on the data?
2. Prison Population Activity (10 minutes, small groups)
Hand out the infographic. Explain to students that an infographic, or information graphic, is a visual image designed to present complex information quickly and clearly. Give students a few minutes to look at the infographic on their own and then ask if they have any questions about how the data are presented. (Note: In the “California Prison Population by Race” graph, you may wish to explain that the second and fourth bars are the distribution of races in the general, non-incarcerated population.)
Have students choose to review either the age and gender infographics or the race infographic. They should use the data in the infographic and the information about the total number of people in California’s prisons to calculate how many people of each age, gender, or race are imprisoned. Students may use calculators during this part of the activity.
3. Data Analysis Activity (10 minutes, small groups)
Once students have finished the calculations, have them look critically at the data with other groups who analyzed different graphs. Based on what they have learned, ask students to identify which populations (in terms of age, gender, and race) are overrepresented. Do prisons have the same demographics as the non-incarcerated population? What does this say about crime and punishment in California? If students were in charge of crime-prevention efforts for California, toward which groups and subgroups should they target their programming?
4. Conclusion (5 minutes, whole group)
Quickly review student calculations of the total number of people in prison, broken down by age, gender, and race. Then have students do some analysis. Ask them
• Were you surprised by any of the data?
• What conclusions can you draw from the data?
• What recommendations did you make about crime-prevention efforts? |
# Polynomials
#### Defining Polynomials
• A polynomial function is a function comprising of several terms. These terms are either constants, or a constant multiple of an integer power of a variable.
• Each term in a polynomial function is called a monomial, and the constant or integer is known as the coefficient.
• In the polynomial equation, the highest power of the variable identifies the degree of the polynomial.
• The term with the highest degree is known as the leading term, while the coefficient of the leading term is the leading coefficient.
#### Understanding Polynomial Graphs
• The graphs of polynomial functions are continuous, and without any sharp bends or gaps.
• Understanding how to plot polynomial graphs involves knowing the degree and the leading coefficient. This helps identify the basic shape of the graph.
• A polynomial of odd degree will have ends that point in opposite directions.
• A polynomial of even degree will have ends that point in the same direction.
#### Division of Polynomials
• The Factor theorem provides a quick means of checking whether (x-a) is a factor of a polynomial, P(x). If P(a) = 0, then x = a is a root of the polynomial and (x-a) is a factor.
• Long division of polynomials can be utilised when dividing a polynomial by a factor of a higher degree.
• If a polynomial f(x) is divided by (x - a) and the remainder is zero, then ‘a’ is said to be a root of the equation f(x) = 0.
#### Polynomial Inequalities
• To solve polynomial inequalities, first solve the equation obtained by setting the polynomial greater than or equal to zero, and then determine which of the intervals defined by these solutions satisfy the original inequality.
#### Rational Root Theorem
• The Rational Root Theorem is an important theorem in polynomial equations. It states that if a polynomial has rational roots or zeros, then they are a fraction derived from the ratio of the factors of the constant term to the factors of the leading coefficient.
#### Fundamental Theorem of Algebra
• The Fundamental Theorem of Algebra asserts that each polynomial equation of degree n has precisely n complex roots or zeros, including repeated roots. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Determinants
## Number calculated from the entries in a matrix primarily through multiplication and subtraction.
Estimated25 minsto complete
%
Progress
Practice Determinants
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated25 minsto complete
%
Finding Determinants of Matrices
A mathematical theorem states that a matrix is singular if and only if its determinant is zero. Is the following matrix singular?
\begin{align*}\begin{bmatrix} 2 & 1 & 3\\ 0 & 2 & 1\\ -1 & 3 & 0\end{bmatrix}\end{align*}
### Determinant of a Matrix
Each square matrix has a real number value associated with it called its determinant. This value is denoted by \begin{align*}det \ A\end{align*} or \begin{align*}|A|\end{align*}.
Finding the Determinant of a \begin{align*}2 \times 2\end{align*} matrix:
\begin{align*}det\begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{vmatrix} {\color{red}a} & {\color{blue}b}\\ {\color{blue}c} & {\color{red}d}\end{vmatrix} = {\color{red}ad}- {\color{blue}bc}\end{align*}
Finding the Determinant of a \begin{align*}3 \times 3\end{align*} matrix: To begin, we will repeat the first two columns after matrix. Next, calculate the products and sums as shown below and find the difference. The result is the determinant of the \begin{align*}3 \times 3\end{align*} matrix.
Using the determinant to find the Area of a Triangle in the coordinate plane:
We can find the area of a triangle with vertices \begin{align*}(x_1,y_1), (x_2,y_2)\end{align*} and \begin{align*}(x_3,y_3)\end{align*} using the formula below
\begin{align*}A = \pm\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix},\end{align*} where the \begin{align*}\pm\end{align*} accounts for the possibility that the determinant could be negative but area should always be positive.
Using the calculator to find the determinant of a matrix: If you are using a TI-83 or TI-84, access the Matrix menu by either pressing MATRIX or (\begin{align*}2^{nd} \ x^{-1}\end{align*} MATRIX). Now you can choose to EDIT matrix \begin{align*}A\end{align*}. Change the dimensions as needed and enter the data values. Now return to the home screen (\begin{align*}2^{nd}\end{align*} MODE QUIT) and return to the MATRIX menu. Arrow over to MATH and select 1:det( by pressing ENTER. Go into the MATRIX menu once more to select 1:\begin{align*}[A]\end{align*} under the NAMES column. Press ENTER. Your screen should show \begin{align*}det([A]\end{align*} at this time. Press ENTER once more and the result will be your determinant. These directions work for square matrices of any size.
Let's solve the following problems relating to determinants of matrices.
1. Find the \begin{align*}det\begin{bmatrix} 3 & -4\\ 1 & 5\end{bmatrix}.\end{align*}
Using the rule above for a \begin{align*}2 \times 2\end{align*} matrix, the determinant can be found as shown:
\begin{align*}det\begin{bmatrix} 3 & -4\\ 1 & 5\end{bmatrix} = \begin{vmatrix} {\color{red}3} & {\color{blue}-4}\\ {\color{blue}1} & {\color{red}5}\end{vmatrix} = {\color{red}(3)(5)}-{\color{blue}(-4)(1)} = {\color{red}15}-{\color{blue}(-4)} = 19\end{align*}
1. Find the \begin{align*}det\begin{bmatrix}2 & -3 & 5\\ -4 & 7 & 1\\ 3 & 8 & 6\end{bmatrix}.\end{align*}
First, we need to repeat the first two columns. Then we can find the diagonal products as shown:
\begin{align*}\begin{bmatrix} 2 & -3 & 5\\ -4 & 7 & 1\\ 3 & 8 & 6\end{bmatrix}\begin{matrix} 2 & -3\\ -4 & 7\\ 3 & 8\end{matrix}\\ {\color{red}(2 \cdot 7 \cdot 6)+(-3 \cdot 1 \cdot 3)+(5\cdot-4\cdot8)} & \ {\color{red}= 84+-9+-160 = -85}\\ = \ {\color{blue}(3\cdot7\cdot5)+(8\cdot 1\cdot 2)+(6\cdot-4\cdot-3)} & \ {\color{blue}= 105+16+72 = 193}\\ {\color{red}-85}-{\color{blue}193} &= -278\end{align*}
1. Find the area of the triangle with vertices (2, -1), (4, 5) and (8, 1)
The first step is to set up our matrix and find the determinant as shown:
\begin{align*}\begin{bmatrix} 2 & -1 & 1\\ 4 & 5 & 1\\ 8 & 1 & 1\end{bmatrix}\begin{matrix} 2 & -1\\ 4 & 5\\ 8 & 1\end{matrix}\\ {\color{red}(2\cdot5\cdot1)+(-1\cdot1\cdot8)+(1\cdot4\cdot1)} & \ {\color{red}=10+-8+4=6}\\ = \ {\color{blue}(8\cdot5\cdot1)+(1\cdot1\cdot2)+(1\cdot4\cdot-1)} & \ {\color{blue}= 40+2-4=38}\\ {\color{red}6}-{\color{blue}38} &= -32\end{align*}
Now we can multiply this determinant, -32, by \begin{align*}-\frac{1}{2}\end{align*} (we will multiply by the negative in order to have a positive result) to get 16. So the area of the triangle is 16 \begin{align*}u^{2}\end{align*}.
### Examples
#### Example 1
Earlier, you were asked to determine whether the following matrix is singular.
\begin{align*}\begin{bmatrix} 2 & 1 & 3\\ 0 & 2 & 1\\ -1 & 3 & 0\end{bmatrix}\end{align*}
To find the determinant, we first need to repeat the first two columns. Then we can find the diagonal products as shown:
\begin{align*}\begin{bmatrix} 2 & 1 & 3\\ 0 & 2 & 1\\ -1 & 3 & 0\end{bmatrix}\begin{matrix} 2 & 1\\ 0 & 2\\ -1 & 3\end{matrix}\\ {\color{red}(2 \cdot 2 \cdot 0)+(1 \cdot 1 \cdot -1)+(3 \cdot 0 \cdot 3)} & \ {\color{red}= 0+ -1 + 0 = -1}\\ = \ {\color{blue}(3 \cdot 2 \cdot -1)+(2 \cdot 1 \cdot 3)+(1 \cdot 0 \cdot 0)} & \ {\color{blue}= -6 + 6 + 0 = 0}\\ {\color{red}-1}-{\color{blue}0} &= -1\end{align*}
The determinant is not zero and therefore the matrix is not singular.
#### Example 2
Find the determinant of the matrix below:
\begin{align*}\begin{bmatrix} -1 & 8\\ 2 & -9\end{bmatrix}\end{align*}
\begin{align*}\begin{vmatrix} {\color{red}-1} & {\color{blue}8}\\ {\color{blue}2} & {\color{red}-9}\end{vmatrix} = {\color{red}(-1)(-9)}-{\color{blue}(8)(2)} = {\color{red}9}-{\color{blue}16} = -7\end{align*}
#### Example 3
Find the determinant of the matrix below:
\begin{align*}\begin{bmatrix} -2 & 4 & -3\\ 5 & -6 & 1\\ -4 & 1 & -2\end{bmatrix}\end{align*}
\begin{align*}\begin{vmatrix} -2 & 4 & -3\\ 5 & -6 & 1\\ -4 & 1 & -2\end{vmatrix}\begin{matrix}-2 & 4\\ 5 & -6\\ -4 & 1\end{matrix}\\ {\color{red}(-2\cdot-6\cdot-2)+(4\cdot-1\cdot-4)+(-3\cdot5\cdot1)} & \ {\color{red}= -24+16-15=-23}\\ = \ {\color{blue}(-4\cdot-6\cdot-3)+(1\cdot1\cdot-2)+(-2\cdot5\cdot4)} & \ {\color{blue}= -72-2-40=-114}\\ {\color{red}-23}-{\color{blue}(-114)} &= 91\end{align*}
#### Example 4
Find the area of the triangle with vertices (-5, 2), (8, -1) and (3, 9).
\begin{align*}\begin{vmatrix} -5 & 2 & 1\\ 8 & -1 & 1\\ 3 & 9 & 1\end{vmatrix}\begin{matrix} -5 & 2\\ 8 & -1\\ 3 & 9\end{matrix}\\= \ {\color{red}(5+6+72)}-{\color{blue}(-3 + -45 + 16)} = {\color{red}83}-{\color{blue}(-32)} = 115\end{align*}
So the area is \begin{align*}\frac{1}{2}(115) = 57.5 \ u^{2}\end{align*}.
### Review
1. .
\begin{align*}\begin{bmatrix} 2 & -1\\ 3 & 5\end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} -3 & -2\\ 6 & 4\end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 5 & 10\\ -3 & -7\end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} -4 & 8\\ 3 & 5\end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 11 & 3\\ 7 & 2\end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 9 & 3\\ 2 & -1\end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 1 & -1 & 3\\ 5 & 0 & 6\\ -4 & 8 & 2\end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 5 & -2 & 1\\ 6 & 1 & 0\\ -3 & 2 & 4\end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 4 & -1 & 2\\ 3 & 0 & 1\\ -2 & 5 & 6\end{bmatrix}\end{align*}
Find the area of each triangle with vertices given below.
1. (2, -1), (-5, 2) and (0, 6)
2. (-8, 12), (10, 5) and (1, -4)
3. (-7, 2), (8, 0) and (3, -4)
Find the value of \begin{align*}a\end{align*} in the matrices below.
1. .
\begin{align*}\begin{vmatrix} a & 3\\ 8 & 2\end{vmatrix} = -10\end{align*}
1. .
\begin{align*}\begin{vmatrix} 4 & a\\ 3 & 5\end{vmatrix} = -1\end{align*}
1. .
\begin{align*}\begin{vmatrix} 2 & -1 & 3\\ 4 & 5 & 2\\ -3 & 0 & a\end{vmatrix} = 23\end{align*}
To see the Review answers, open this PDF file and look for section 4.7.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
determinant
The determinant is a single number descriptor of a square matrix. The determinant is computed from the entries of the matrix, and has many properties and interpretations explored in linear algebra.
Sarrus’ rule
Sarrus’ rule is a memorization technique that enables you to compute the determinant of matrices efficiently.
Square matrix
A square matrix is a matrix in which the number of rows equals the number of columns. |
RD Sharma Solutions for Class 12 Maths Chapter 2 Function - GMS - Learning Simply
Students' favourite free learning app with LIVE online classes, instant doubt resolution, unlimited practice for classes 6-12, personalized study app for Maths, Science, Social Studies, video e-learning, online tutorial, and more. Join Telegram
# RD Sharma Solutions for Class 12 Maths Chapter 2 Function
Scroll Down and click on Go to Link for destination
# RD Sharma Solutions for Class 12 Maths Chapter 2 Function
RD Sharma Solutions, Maths Chapter 2, for Class 12, help students who aspire to obtain a good academic score in the exam. The solutions are designed by experts to boost confidence among students in understanding the concepts covered in this chapter and methods to solve problems in a shorter period. These materials are prepared based on Class 12 CBSE syllabus by our experts, keeping into consideration, the types of questions asked in the RD Sharma Solutions. Chapter 2 Function explains function and domain and codomain of functions. It has four exercises. Students can easily access answers to the problems present in RD Sharma Solutions for Class 12. Let us have a look at some of the important concepts that are discussed in this chapter.
• Classification of functions
• Types of functions
• Constant function
• Identity function
• Modulus function
• Integer function
• Exponential function
• Logarithmic function
• Reciprocal function
• Square root function
• Operations on real functions
• Kinds of functions
• One-one function
• On-to function
• Many one function
• In to function
• Bijection
• Composition of functions
• Properties of the composition of functions
• Composition of real function
• Inverse of a function
• Inverse of an element
• Relation between graphs of a function and its inverse
### Exercise 2.1 Page No: 2.31
1. Give an example of a function
(i) Which is one-one but not onto.
(ii) Which is not one-one but onto.
(iii) Which is neither one-one nor onto.
Solution:
(i) Let f: Z → Z given by f(x) = 3x + 2
Let us check one-one condition on f(x) = 3x + 2
Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f (x) = f(y)
⇒ 3x + 2 =3y + 2
⇒ 3x = 3y
⇒ x = y
⇒ f(x) = f(y)
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z(domain).
Let f(x) = y
⇒ 3x + 2 = y
⇒ 3x = y – 2
⇒ x = (y – 2)/3. It may not be in the domain (Z)
Because if we take y = 3,
x = (y – 2)/3 = (3-2)/3 = 1/3 ∉ domain Z.
So, for every element in the co domain there need not be any element in the domain such that f(x) = y.
Thus, f is not onto.
(ii) Example for the function which is not one-one but onto
Let f: Z → N ∪ {0} given by f(x) = |x|
Injectivity:
Let x and y be any two elements in the domain (Z),
Such that f(x) = f(y).
⇒ |x| = |y|
⇒ x = ± y
So, different elements of domain f may give the same image.
So, f is not one-one.
Surjectivity:
Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
⇒ |x| = y
⇒ x = ± y
Which is an element in Z (domain).
So, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.
(iii) Example for the function which is neither one-one nor onto.
Let f: Z → Z given by f(x) = 2x2 + 1
Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
⇒ 2x2+1 = 2y2+1
⇒ 2x2 = 2y2
⇒ x= y2
⇒ x = ± y
So, different elements of domain f may give the same image.
Thus, f is not one-one.
Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f (x) = y
⇒ 2x2+1=y
⇒ 2x2= y − 1
⇒ x2 = (y-1)/2
⇒ x = √ ((y-1)/2) ∉ Z always.
For example, if we take, y = 4,
x = ± √ ((y-1)/2)
= ± √ ((4-1)/2)
= ± √ (3/2) ∉ Z
So, x may not be in Z (domain).
Thus, f is not onto.
2. Which of the following functions from A to B are one-one and onto?
(i) f1 = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}
(ii) f2 = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}
(iii) f3 = {(a, x), (b, x), (c, z), (d, z)}; A = {a, b, c, d,}, B = {x, y, z}.
Solution:
(i) Consider f1 = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}
Injectivity:
f1 (1) = 3
f(2) = 5
f1 (3) = 7
⇒ Every element of A has different images in B.
So, f1 is one-one.
Surjectivity:
Co-domain of f1 = {3, 5, 7}
Range of f1 =set of images = {3, 5, 7}
⇒ Co-domain = range
So, f1 is onto.
(ii) Consider f2 = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}
Injectivity:
f2 (2) = a
f2 (3) = b
f2 (4) = c
⇒ Every element of A has different images in B.
So, f2 is one-one.
Surjectivity:
Co-domain of f2 = {a, b, c}
Range of f2 = set of images = {a, b, c}
⇒ Co-domain = range
So, f2 is onto.
(iii) Consider f3 = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}
Injectivity:
f3 (a) = x
f3 (b) = x
f3 (c) = z
f3 (d) = z
⇒ a and b have the same image x.
Also c and d have the same image z
So, f3 is not one-one.
Surjectivity:
Co-domain of f3 ={x, y, z}
Range of f3 =set of images = {x, z}
So, the co-domain is not same as the range.
So, f3 is not onto.
3. Prove that the function f: N → N, defined by f(x) = x2 + x + 1, is one-one but not onto
Solution:
Given f: N → N, defined by f(x) = x2 + x + 1
Now we have to prove that given function is one-one
Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
⇒ x2 + x + 1 = y2 + y + 1
⇒ (x2 – y2) + (x – y) = 0 `
⇒ (x + y) (x- y ) + (x – y ) = 0
⇒ (x – y) (x + y + 1) = 0
⇒ x – y = 0 [x + y + 1 cannot be zero because x and y are natural numbers
⇒ x = y
So, f is one-one.
Surjectivity:
When x = 1
x2 + x + 1 = 1 + 1 + 1 = 3
⇒ x2 + x +1 ≥ 3, for every x in N.
⇒ f(x) will not assume the values 1 and 2.
So, f is not onto.
4. Let A = {−1, 0, 1} and f = {(x, x2) : x ∈ A}. Show that f : A → A is neither one-one nor onto.
Solution:
Given A = {−1, 0, 1} and f = {(x, x2): x ∈ A}
Also given that, f(x) = x2
Now we have to prove that given function neither one-one or nor onto.
Injectivity:
Let x = 1
Therefore f(1) = 12=1 and
f(-1)=(-1)2=1
⇒ 1 and -1 have the same images.
So, f is not one-one.
Surjectivity:
Co-domain of f = {-1, 0, 1}
f(1) = 12 = 1,
f(-1) = (-1)2 = 1 and
f(0) = 0
⇒ Range of f = {0, 1}
So, both are not same.
Hence, f is not onto
5. Classify the following function as injection, surjection or bijection:
(i) f: N → N given by f(x) = x2
(ii) f: Z → Z given by f(x) = x2
(iii) f: N → N given by f(x) = x3
(iv) f: Z → Z given by f(x) = x3
(v) f: R → R, defined by f(x) = |x|
(vi) f: Z → Z, defined by f(x) = x2 + x
(vii) f: Z → Z, defined by f(x) = x − 5
(viii) f: R → R, defined by f(x) = sin x
(ix) f: R → R, defined by f(x) = x3 + 1
(x) f: R → R, defined by f(x) = x3 − x
(xi) f: R → R, defined by f(x) = sin2x + cos2x
(xii) f: Q − {3} → Q, defined by f (x) = (2x +3)/(x-3)
(xiii) f: Q → Q, defined by f(x) = x3 + 1
(xiv) f: R → R, defined by f(x) = 5x3 + 4
(xv) f: R → R, defined by f(x) = 5x3 + 4
(xvi) f: R → R, defined by f(x) = 1 + x2
(xvii) f: R → R, defined by f(x) = x/(x+ 1)
Solution:
(i) Given f: N → N, given by f(x) = x2
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
f(x) = f(y)
x= y2
x = y (We do not get ± because x and y are in N that is natural numbers)
So, f is an injection.
Surjection condition:
Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).
f(x) = y
x2= y
x = √y, which may not be in N.
For example, if y = 3,
x = √3 is not in N.
So, f is not a surjection.
Also f is not a bijection.
(ii) Given f: Z → Z, given by f(x) = x2
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
x= y2
x = ±y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
x= y
x = ± √y which may not be in Z.
For example, if y = 3,
x = ± √ 3 is not in Z.
So, f is not a surjection.
Also f is not bijection.
(iii) Given f: N → N given by f(x) = x3
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
f(x) = f(y)
x3 = y3
x = y
So, f is an injection
Surjection condition:
Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).
f(x) = y
x3= y
x = ∛y which may not be in N.
For example, if y = 3,
X = ∛3 is not in N.
So, f is not a surjection and f is not a bijection.
(iv) Given f: Z → Z given by f(x) = x3
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y)
f(x) = f(y)
x3 = y3
x = y
So, f is an injection.
Surjection condition:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
x3 = y
x = ∛y which may not be in Z.
For example, if y = 3,
x = ∛3 is not in Z.
So, f is not a surjection and f is not a bijection.
(v) Given f: R → R, defined by f(x) = |x|
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
f(x) = f(y)
|x|=|y|
x = ±y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
|x|=y
x = ± y ∈ Z
So, f is a surjection and f is not a bijection.
(vi) Given f: Z → Z, defined by f(x) = x2 + x
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
x2+ x = y+ y
Here, we cannot say that x = y.
For example, x = 2 and y = – 3
Then,
x+ x = 2+ 2 = 6
y+ y = (−3)– 3 = 6
So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (Z),
such that f(x) = y for some element x in Z (domain).
f(x) = y
x2 + x = y
Here, we cannot say x ∈ Z.
For example, y = – 4.
x2 + x = − 4
x+ x + 4 = 0
x = (-1 ± √-5)/2 = (-1 ± i √5)/2 which is not in Z.
So, f is not a surjection and f is not a bijection.
(vii) Given f: Z → Z, defined by f(x) = x – 5
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
x – 5 = y – 5
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
x – 5 = y
x = y + 5, which is in Z.
So, f is a surjection and f is a bijection
(viii) Given f: R → R, defined by f(x) = sin x
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
Sin x = sin y
Here, x may not be equal to y because sin 0 = sin π.
So, 0 and π have the same image 0.
So, f is not an injection.
Surjection test:
Range of f = [-1, 1]
Co-domain of f = R
Both are not same.
So, f is not a surjection and f is not a bijection.
(ix) Given f: R → R, defined by f(x) = x3 + 1
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
x3+1 = y3+ 1
x3= y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
x3+1=y
x = ∛ (y – 1) ∈ R
So, f is a surjection.
So, f is a bijection.
(x) Given f: R → R, defined by f(x) = x3 − x
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
x– x = y− y
Here, we cannot say x = y.
For example, x = 1 and y = -1
x− x = 1 − 1 = 0
y– y = (−1)3− (−1) – 1 + 1 = 0
So, 1 and -1 have the same image 0.
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
x3 − x = y
By observation we can say that there exist some x in R, such that x– x = y.
So, f is a surjection and f is not a bijection.
(xi) Given f: R → R, defined by f(x) = sin2x + cos2x
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
f(x) = sin2x + cos2
We know that sin2x + cos2x = 1
So, f(x) = 1 for every x in R.
So, for all elements in the domain, the image is 1.
So, f is not an injection.
Surjection condition:
Range of f = {1}
Co-domain of f = R
Both are not same.
So, f is not a surjection and f is not a bijection.
(xii) Given f: Q − {3} → Q, defined by f (x) = (2x +3)/(x-3)
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).
f(x) = f(y)
(2x + 3)/(x – 3) = (2y + 3)/(y – 3)
(2x + 3) (y − 3) = (2y + 3) (x − 3)
2xy − 6x + 3y − 9 = 2xy − 6y + 3x − 9
9x = 9y
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).
f(x) = y
(2x + 3)/(x – 3) = y
2x + 3 = x y − 3y
2x – x y = −3y − 3
x (2−y) = −3 (y + 1)
x = -3(y + 1)/(2 – y) which is not defined at y = 2.
So, f is not a surjection and f is not a bijection.
(xiii) Given f: Q → Q, defined by f(x) = x3 + 1
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Q), such that f(x) = f(y).
f(x) = f(y)
x+ 1 = y+ 1
x3 = y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain).
f(x) = y
x3+ 1 = y
x = ∛(y-1), which may not be in Q.
For example, if y= 8,
x3+ 1 = 8
x3= 7
x = ∛7, which is not in Q.
So, f is not a surjection and f is not a bijection.
(xiv) Given f: R → R, defined by f(x) = 5x3 + 4
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
5x+ 4 = 5y+ 4
5x3= 5y3
x= y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
5x3+ 4 = y
x3 = (y – 4)/5 ∈ R
So, f is a surjection and f is a bijection.
(xv) Given f: R → R, defined by f(x) = 5x3 + 4
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
5x+ 4 = 5y+ 4
5x= 5y3
x= y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
5x+ 4 = y
x3 = (y – 4)/5 ∈ R
So, f is a surjection and f is a bijection.
(xvi) Given f: R → R, defined by f(x) = 1 + x2
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
1 + x= 1 + y2
x= y2
x = ± y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
1 + x= y
x= y − 1
x = ± √-1 = ± i` is not in R.
So, f is not a surjection and f is not a bijection.
(xvii) Given f: R → R, defined by f(x) = x/(x+ 1)
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
x /(x+ 1) = y /(y2 + 1)
x y2+ x = x2y + y
xy− x2y + x − y = 0
−x y (−y + x) + 1 (x − y) = 0
(x − y) (1 – x y) = 0
x = y or x = 1/y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
x /(x2 + 1) = y
y x– x + y = 0
x = (-(-1) ± √ (1-4y2))/(2y) if y ≠ 0
= (1 ± √ (1-4y2))/ (2y), which may not be in R
For example, if y=1, then
(1 ± √ (1-4)) / (2y) = (1 ± i √3)/2, which is not in R
So, f is not surjection and f is not bijection.
6. If f: A → B is an injection, such that range of f = {a}, determine the number of elements in A.
Solution:
Given f: A → B is an injection
And also given that range of f = {a}
So, the number of images of f = 1
Since, f is an injection, there will be exactly one image for each element of f .
So, number of elements in A = 1.
7. Show that the function f: R − {3} → R − {2} given by f(x) = (x-2)/(x-3) is a bijection.
Solution:
Given that f: R − {3} → R − {2} given by f (x) = (x-2)/(x-3)
Now we have to show that the given function is one-one and on-to
Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).
f(x) = f(y)
⇒ (x – 2) /(x – 3) = (y – 2) /(y – 3)
⇒ (x – 2) (y – 3) = (y – 2) (x – 3)
⇒ x y – 3 x – 2 y + 6 = x y – 3y – 2x + 6
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element x in R − {3} (domain).
f(x) = y
⇒ (x – 2) /(x – 3) = y
⇒ x – 2 = x y – 3y
⇒ x y – x = 3y – 2
⇒ x ( y – 1 ) = 3y – 2
⇒ x = (3y – 2)/ (y – 1), which is in R – {3}
So, for every element in the co-domain, there exists some pre-image in the domain.
⇒ f is onto.
Since, f is both one-one and onto, it is a bijection.
8. Let A = [-1, 1]. Then, discuss whether the following function from A to itself is one-one, onto or bijective:
(i) f (x) = x/2
(ii) g (x) = |x|
(iii) h (x) = x2
Solution:
(i) Given f: A → A, given by f (x) = x/2
Now we have to show that the given function is one-one and on-to
Injection test:
Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y)
x/2 = y/2
x = y
So, f is one-one.
Surjection test:
Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain)
f(x) = y
x/2 = y
x = 2y, which may not be in A.
For example, if y = 1, then
x = 2, which is not in A.
So, f is not onto.
So, f is not bijective.
(ii) Given g: A → A, given by g (x) = |x|
Now we have to show that the given function is one-one and on-to
Injection test:
Let x and y be any two elements in the domain (A), such that f(x) = f(y).
g(x) = g(y)
|x| = |y|
x = ± y
So, f is not one-one.
Surjection test:
For y = -1, there is no value of x in A.
So, g is not onto.
So, g is not bijective.
(iii) Given h: A → A, given by h (x) = x2
Now we have to show that the given function is one-one and on-to
Injection test:
Let x and y be any two elements in the domain (A), such that h(x) = h(y).
h(x) = h(y)
x2 = y2
x = ±y
So, f is not one-one.
Surjection test:
For y = – 1, there is no value of x in A.
So, h is not onto.
So, h is not bijective.
9. Are the following set of ordered pair of a function? If so, examine whether the mapping is injective or surjective:
(i) {(x, y): x is a person, y is the mother of x}
(ii) {(a, b): a is a person, b is an ancestor of a}
Solution:
Let f = {(x, y): x is a person, y is the mother of x}
As, for each element x in domain set, there is a unique related element y in co-domain set.
So, f is the function.
Injection test:
As, y can be mother of two or more persons
So, f is not injective.
Surjection test:
For every mother y defined by (x, y), there exists a person x for whom y is mother.
So, f is surjective.
Therefore, f is surjective function.
(ii) Let g = {(a, b): a is a person, b is an ancestor of a}
Since, the ordered map (a, b) does not map ‘a’ – a person to a living person.
So, g is not a function.
10. Let A = {1, 2, 3}. Write all one-one from A to itself.
Solution:
Given A = {1, 2, 3}
Number of elements in A = 3
Number of one-one functions = number of ways of arranging 3 elements = 3! = 6
(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}
11. If f: R → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.
Solution:
Given f: R → R is a function defined by f(x) = 4x3 + 7
Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
⇒ 4x+ 7 = 4y+ 7
⇒ 4x= 4y3
⇒ x= y3
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain)
f(x) = y
⇒ 4x+ 7 = y
⇒ 4x= y − 7
⇒ x3 = (y – 7)/4
⇒ x = ∛(y-7)/4 in R
So, for every element in the co-domain, there exists some pre-image in the domain. f is onto.
Since, f is both one-to-one and onto, it is a bijection.
### Exercise 2.2 Page No: 2.46
1. Find gof and fog when f: R → R and g : R → R is defined by
(i) f(x) = 2x + 3 and g(x) = x2 + 5.
(ii) f(x) = 2x + x2 and g(x) = x3
(iii) f (x) = x2 + 8 and g(x) = 3x3 + 1
(iv) f (x) = x and g(x) = |x|
(v) f(x) = x2 + 2x − 3 and g(x) = 3x − 4
(vi) f(x) = 8x3 and g(x) = x1/3
Solution:
(i) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
Also given that f(x) = 2x + 3 and g(x) = x2 + 5
Now, (gof) (x) = g (f (x))
= g (2x +3)
= (2x + 3)2 + 5
= 4x2+ 9 + 12x +5
=4x2+ 12x + 14
Now, (fog) (x) = f (g (x))
= f (x2 + 5)
= 2 (x2 + 5) +3
= 2 x2+ 10 + 3
= 2x2 + 13
(ii) Given, f: R → R and g: R → R
so, gof: R → R and fog: R → R
f(x) = 2x + x2 and g(x) = x3
(gof) (x)= g (f (x))
= g (2x+x2)
= (2x+x2)3
Now, (fog) (x) = f (g (x))
= f (x3)
= 2 (x3) + (x3)2
= 2x+ x6
(iii) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = x2 + 8 and g(x) = 3x3 + 1
(gof) (x) = g (f (x))
= g (x2 + 8)
= 3 (x2+8)3 + 1
Now, (fog) (x) = f (g (x))
= f (3x3 + 1)
= (3x3+1)2 + 8
= 9x6 + 6x+ 1 + 8
= 9x+ 6x+ 9
(iv) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = x and g(x) = |x|
(gof) (x) = g (f (x))
= g (x)
= |x|
Now (fog) (x) = f (g (x))
= f (|x|)
= |x|
(v) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = x2 + 2x − 3 and g(x) = 3x − 4
(gof) (x) = g (f(x))
= g (x+ 2x − 3)
= 3 (x+ 2x − 3) − 4
= 3x+ 6x − 9 − 4
= 3x+ 6x − 13
Now, (fog) (x) = f (g (x))
= f (3x − 4)
= (3x − 4)+ 2 (3x − 4) −3
= 9x+ 16 − 24x + 6x – 8 − 3
= 9x− 18x + 5
(vi) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = 8x3 and g(x) = x1/3
(gof) (x) = g (f (x))
= g (8x3)
= (8x3)1/3
= [(2x)3]1/3
= 2x
Now, (fog) (x) = f (g (x))
= f (x1/3)
= 8 (x1/3)3
= 8x
2. Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.
Solution:
Given f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}
f : {3, 9, 12} → {1, 3, 4} and g : {1, 3, 4, 5} → {3, 9}
Co-domain of f is a subset of the domain of g.
So, gof exists and gof: {3, 9, 12} → {3, 9}
(gof) (3) = g (f (3)) = g (1) = 3
(gof) (9) = g (f (9)) = g (3) = 3
(gof) (12) = g (f (12)) = g (4) = 9
⇒ gof = {(3, 3), (9, 3), (12, 9)}
Co-domain of g is a subset of the domain of f.
So, fog exists and fog: {1, 3, 4, 5} → {3, 9, 12}
(fog) (1) = f (g (1)) = f (3) = 1
(fog) (3) = f (g (3)) = f (3) = 1
(fog) (4) = f (g (4)) = f (9) = 3
(fog) (5) = f (g (5)) = f (9) = 3
⇒ fog = {(1, 1), (3, 1), (4, 3), (5, 3)}
3. Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.
Solution:
Given f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}
f: {1, 4, 9, 16} → {-1, -2, -3, 4} and g: {-1, -2, -3, 4} → {-2, -4, -6, 8}
Co-domain of f = domain of g
So, gof exists and gof: {1, 4, 9, 16} → {-2, -4, -6, 8}
(gof) (1) = g (f (1)) = g (−1) = −2
(gof) (4) = g (f (4)) = g (−2) = −4
(gof) (9) = g (f (9)) = g (−3) = −6
(gof) (16) = g (f (16)) = g (4) = 8
So, gof = {(1, −2), (4, −4), (9, −6), (16, 8)}
But the co-domain of g is not same as the domain of f.
So, fog does not exist.
4. Let A = {a, b, c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as: f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.
Show that f and g both are bijections and find fog and gof.
Solution:
Given f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.
Also given that A = {a, b, c}, B = {u, v, w}
Now we have to show f and g both are bijective.
Consider f = {(a, v), (b, u), (c, w)} and f: A → B
Injectivity of f: No two elements of A have the same image in B.
So, f is one-one.
Surjectivity of f: Co-domain of f = {u, v, w}
Range of f = {u, v, w}
Both are same.
So, f is onto.
Hence, f is a bijection.
Now consider g = {(u, b), (v, a), (w, c)} and g: B → A
Injectivity of g: No two elements of B have the same image in A.
So, g is one-one.
Surjectivity of g: Co-domain of g = {a, b, c}
Range of g = {a, b, c}
Both are the same.
So, g is onto.
Hence, g is a bijection.
Now we have to find fog,
we know that Co-domain of g is same as the domain of f.
So, fog exists and fog: {u v, w} → {u, v, w}
(fog) (u) = f (g (u)) = f (b) = u
(fog) (v) = f (g (v)) = f (a) = v
(fog) (w) = f (g (w)) = f (c) = w
So, fog = {(u, u), (v, v), (w, w)}
Now we have to find gof,
Co-domain of f is same as the domain of g.
So, fog exists and gof: {a, b, c} → {a, b, c}
(gof) (a) = g (f (a)) = g (v) = a
(gof) (b) = g (f (b)) = g (u) = b
(gof) (c) = g (f (c)) = g (w) = c
So, gof = {(a, a), (b, b), (c, c)}
5. Find fog (2) and gof (1) when f: R → R; f(x) = x2 + 8 and g: R → R; g(x) = 3x3 + 1.
Solution:
Given f: R → R; f(x) = x2 + 8 and g: R → R; g(x) = 3x3 + 1.
Consider (fog) (2) = f (g (2))
= f (3 × 2+ 1)
= f(3 × 8 + 1)
= f (25)
= 252 + 8
= 633
(gof) (1) = g (f (1))
= g (1+ 8)
= g (9)
= 3 × 9+ 1
= 2188
6. Let R+ be the set of all non-negative real numbers. If f: R+ → R+ and g : R+ → R+ are defined as f(x)=x2 and g(x)=+ √x, find fog and gof. Are they equal functions.
Solution:
Given f: R+ → R+ and g: R+ → R+
So, fog: R+ → R+ and gof: R+ → R+
Domains of fog and gof are the same.
Now we have to find fog and gof also we have to check whether they are equal or not,
Consider (fog) (x) = f (g (x))
= f (√x)
= √x2
= x
Now consider (gof) (x) = g (f (x))
= g (x2)
= √x2
= x
So, (fog) (x) = (gof) (x), ∀x ∈ R+
Hence, fog = gof
7. Let f: R → R and g: R → R be defined by f(x) = x2 and g(x) = x + 1. Show that fog ≠ gof.
Solution:
Given f: R → R and g: R → R.
So, the domains of f and g are the same.
Consider (fog) (x) = f (g (x))
= f (x + 1) = (x + 1)2
= x+ 1 + 2x
Again consider (gof) (x) = g (f (x))
= g (x2) = x+ 1
So, fog ≠ gof
### Exercise 2.3 Page No: 2.54
1. Find fog and gof, if
(i) f (x) = ex, g (x) = loge x
(ii) f (x) = x2, g (x) = cos x
(iii) f (x) = |x|, g (x) = sin x
(iv) f (x) = x+1, g(x) = ex
(v) f (x) = sin−1 x, g(x) = x2
(vi) f (x) = x+1, g (x) = sin x
(vii) f(x)= x + 1, g (x) = 2x + 3
(viii) f(x) = c, c ∈ R, g(x) = sin x2
(ix) f(x) = x2 + 2 , g (x) = 1 − 1/ (1-x)
Solution:
(i) Given f (x) = ex, g(x) = loge x
Let f: R → (0, ∞); and g: (0, ∞) → R
Now we have to calculate fog,
Clearly, the range of g is a subset of the domain of f.
fog: ( 0, ∞) → R
(fog) (x) = f (g (x))
= f (loge x)
= loge ex
= x
Now we have to calculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→ R
(gof) (x) = g (f (x))
= g (ex)
= loge ex
= x
(ii) f (x) = x2, g(x) = cos x
f: R→ [0, ∞) ; g: R→[−1, 1]
Now we have to calculate fog,
Clearly, the range of g is not a subset of the domain of f.
⇒ Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}
⇒ Domain (fog) = x: x ∈ R and cos x ∈ R}
⇒ Domain of (fog) = R
(fog): R→ R
(fog) (x) = f (g (x))
= f (cos x)
= cos2 x
Now we have to calculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→R
(gof) (x) = g (f (x))
= g (x2)
= cos x2
(iii) Given f (x) = |x|, g(x) = sin x
f: R → (0, ∞) ; g : R→[−1, 1]
Now we have to calculate fog,
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R→R
(fog) (x) = f (g (x))
= f (sin x)
= |sin x|
Now we have to calculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog : R→ R
(gof) (x) = g (f (x))
= g (|x|)
= sin |x|
(iv) Given f (x) = x + 1, g(x) = ex
f: R→R ; g: R → [ 1, ∞)
Now we have calculate fog:
Clearly, range of g is a subset of domain of f.
⇒ fog: R→R
(fog) (x) = f (g (x))
= f (ex)
= e+ 1
Now we have to compute gof,
Clearly, range of f is a subset of domain of g.
⇒ fog: R→R
(gof) (x) = g (f (x))
= g (x+1)
= ex+1
(v) Given f (x) = sin −1 x, g(x) = x2
f: [−1,1]→ [(-π)/2 ,π/2]; g : R → [0, ∞)
Now we have to compute fog:
Clearly, the range of g is not a subset of the domain of f.
Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}
Domain (fog) = {x: x ∈ R and x2 ∈ [−1, 1]}
Domain (fog) = {x: x ∈ R and x ∈ [−1, 1]}
Domain of (fog) = [−1, 1]
fog: [−1,1] → R
(fog) (x) = f (g (x))
= f (x2)
= sin−1 (x2)
Now we have to compute gof:
Clearly, the range of f is a subset of the domain of g.
fog: [−1, 1] → R
(gof) (x) = g (f (x))
= g (sin−1 x)
= (sin−1 x)2
(vi) Given f(x) = x+1, g(x) = sin x
f: R→R ; g: R→[−1, 1]
Now we have to compute fog
Clearly, the range of g is a subset of the domain of f.
Set of the domain of f.
⇒ fog: R→ R
(fog) (x) = f (g (x))
= f (sin x)
= sin x + 1
Now we have to compute gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R → R
(gof) (x) = g (f (x))
= g (x+1)
= sin (x+1)
(vii) Given f (x) = x+1, g (x) = 2x + 3
f: R→R ; g: R → R
Now we have to compute fog
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R→ R
(fog) (x) = f (g (x))
= f (2x+3)
= 2x + 3 + 1
= 2x + 4
Now we have to compute gof
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R → R
(gof) (x) = g (f (x))
= g (x+1)
= 2 (x + 1) + 3
= 2x + 5
(viii) Given f (x) = c, g (x) = sin x2
f: R → {c} ; g: R→ [ 0, 1 ]
Now we have to compute fog
Clearly, the range of g is a subset of the domain of f.
fog: R→R
(fog) (x) = f (g (x))
= f (sin x2)
= c
Now we have to compute gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→ R
(gof) (x) = g (f (x))
= g (c)
= sin c2
(ix) Given f (x) = x2+ 2 and g (x) = 1 – 1 / (1 – x)
f: R → [ 2, ∞ )
For domain of g: 1− x ≠ 0
⇒ x ≠ 1
⇒ Domain of g = R − {1}
g (x )= 1 – [1/(1 – x)] = (1 – x – 1)/ (1 – x) = -x/(1 – x)
For range of g
y = (- x)/ (1 – x)
⇒ y – x y = − x
⇒ y = x y − x
⇒ y = x (y−1)
⇒ x = y/(y – 1)
Range of g = R − {1}
So, g: R − {1} → R − {1}
Now we have to compute fog
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R − {1} → R
(fog) (x) = f (g (x))
= f (-x/ (1 – x))
= ((-x)/ (1 – x))2 + 2
= (x2 + 2x2 + 2 – 4x) / (1 – x)2
= (3x– 4x + 2)/ (1 – x)2
Now we have to compute gof
Clearly, the range of f is a subset of the domain of g.
⇒ gof: R→R
(gof) (x) = g (f (x))
= g (x2 + 2)
= 1 – 1 / (1 – (x2 + 2))
= – 1/ (1 – (x2 + 2))
= (x2 + 2)/ (x2 + 1)
2. Let f(x) = x2 + x + 1 and g(x) = sin x. Show that fog ≠ gof.
Solution:
Given f(x) = x2 + x + 1 and g(x) = sin x
Now we have to prove fog ≠ gof
(fog) (x) = f (g (x))
= f (sin x)
= sinx + sin x + 1
And (gof) (x) = g (f (x))
= g (x2+ x + 1)
= sin (x2+ x + 1)
So, fog ≠ gof.
3. If f(x) = |x|, prove that fof = f.
Solution:
Given f(x) = |x|,
Now we have to prove that fof = f.
Consider (fof) (x) = f (f (x))
= f (|x|)
= ||x||
= |x|
= f (x)
So,
(fof) (x) = f (x), ∀x ∈ R
Hence, fof = f
4. If f(x) = 2x + 5 and g(x) = x2 + 1 be two real functions, then describe each of the following functions:
(i) fog
(ii) gof
(iii) fof
(iv) f2
Also, show that fof ≠ f2
Solution:
f(x) and g(x) are polynomials.
⇒ f: R → R and g: R → R.
So, fog: R → R and gof: R → R.
(i) (fog) (x) = f (g (x))
= f (x2 + 1)
= 2 (x+ 1) + 5
=2x2 + 2 + 5
= 2x2 +7
(ii) (gof) (x) = g (f (x))
= g (2x +5)
= (2x + 5)2 + 1
= 4x2 + 20x + 26
(iii) (fof) (x) = f (f (x))
= f (2x +5)
= 2 (2x + 5) + 5
= 4x + 10 + 5
= 4x + 15
(iv) f2 (x) = f (x) x f (x)
= (2x + 5) (2x + 5)
= (2x + 5)2
= 4x2 + 20x +25
Hence, from (iii) and (iv) clearly fof ≠ f2
5. If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
Solution:
Given f(x) = sin x and g(x) = 2x
We know that
f: R→ [−1, 1] and g: R→ R
Clearly, the range of f is a subset of the domain of g.
gof: R→ R
(gof) (x) = g (f (x))
= g (sin x)
= 2 sin x
Clearly, the range of g is a subset of the domain of f.
fog: R → R
So, (fog) (x) = f (g (x))
= f (2x)
= sin (2x)
Clearly, fog ≠ gof
Hence they are not equal functions.
6. Let f, g, h be real functions given by f(x) = sin x, g (x) = 2x and h (x) = cos x. Prove that fog = go (f h).
Solution:
Given that f(x) = sin x, g (x) = 2x and h (x) = cos x
We know that f: R→ [−1, 1] and g: R→ R
Clearly, the range of g is a subset of the domain of f.
fog: R → R
Now, (f h) (x) = f (x) h (x) = (sin x) (cos x) = ½ sin (2x)
Domain of f h is R.
Since range of sin x is [-1, 1], −1 ≤ sin 2x ≤ 1
⇒ -1/2 ≤ sin x/2 ≤ 1/2
Range of f h = [-1/2, 1/2]
So, (f h): R → [(-1)/2, 1/2]
Clearly, range of f h is a subset of g.
⇒ go (f h): R → R
⇒ Domains of fog and go (f h) are the same.
So, (fog) (x) = f (g (x))
= f (2x)
= sin (2x)
And (go (f h)) (x) = g ((f(x). h(x))
= g (sin x cos x)
= 2sin x cos x
= sin (2x)
⇒ (fog) (x) = (go (f h)) (x), ∀x ∈ R
Hence, fog = go (f h)
### Exercise 2.4 Page No: 2.68
1. State with reason whether the following functions have inverse:
(i) f: {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Solution:
(i) Given f: {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have:
f (1) = f (2) = f (3) = f (4) = 10
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
(ii) Given g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
from the question it is clear that g (5) = g (7) = 4
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
(iii) Given h: {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Here, different elements of the domain have different images in the co-domain.
⇒ h is one-one.
Also, each element in the co-domain has a pre-image in the domain.
⇒ h is onto.
⇒ h is a bijection.
Therefore h inverse exists.
⇒ h has an inverse and it is given by
h-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}
2. Find f −1 if it exists: f: A → B, where
(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2
Solution:
(i) Given A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}
Here, different elements of the domain have different images in the co-domain.
Clearly, this is one-one.
Range of f = Range of f = B
so, f is a bijection and,
Thus, f -1 exists.
Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}
(ii) Given A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2
So, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}
Here, different elements of the domain have different images in the co-domain.
Clearly, f is one-one.
But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)
⇒ f is not a bijection.
So, f -1does not exist.
3. Consider f: {1, 2, 3} → {a, b, c} and g: {a, b, c} → {apple, ball, cat} defined as f (1) = a, f (2) = b, f (3) = c, g (a) = apple, g (b) = ball and g (c) = cat. Show that f, g and gof are invertible. Find f−1, g−1 and gof−1and show that (gof)−1 = f −1o g−1
Solution:
Given f = {(1, a), (2, b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.
So, f and g are invertible.
Now,
-1 = {(a ,1) , (b , 2) , (3,c)} and g-1 = {(apple, a), (ball , b), (cat , c)}
So, f-1 o g-1= {apple, 1), (ball, 2), (cat, 3)}……… (1)
f: {1,2,3,} → {a, b, c} and g: {a, b, c} → {apple, ball, cat}
So, gof: {1, 2, 3} → {apple, ball, cat}
⇒ (gof) (1) = g (f (1)) = g (a) = apple
(gof) (2) = g (f (2))
= g (b)
= ball,
And (gof) (3) = g (f (3))
= g (c)
= cat
∴ gof = {(1, apple), (2, ball), (3, cat)}
Clearly, gof is a bijection.
So, gof is invertible.
(gof)-1 = {(apple, 1), (ball, 2), (cat, 3)}……. (2)
Form (1) and (2), we get
(gof)-1 = f-1 o g -1
4. Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f: A → B, g: B → C be defined as f(x) = 2x + 1 and g(x) = x2 − 2. Express (gof)−1 and f−1 og−1 as the sets of ordered pairs and verify that (gof)−1 = f−1 og−1.
Solution:
Given that f (x) = 2x + 1
⇒ f= {(1, 2(1) + 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) + 1)}
= {(1, 3), (2, 5), (3, 7), (4, 9)}
Also given that g(x) = x2−2
⇒ g = {(3, 32−2), (5, 52−2), (7, 72−2), (9, 92−2)}
= {(3, 7), (5, 23), (7, 47), (9, 79)}
Clearly f and g are bijections and, hence, f−1: B→ A and g−1: C→ B exist.
So, f−1= {(3, 1), (5, 2), (7, 3), (9, 4)}
And g−1= {(7, 3), (23, 5), (47, 7), (79, 9)}
Now, (f−1 o g−1): C→ A
f−1 o g−1 = {(7, 1), (23, 2), (47, 3), (79, 4)}……….(1)
Also, f: A→B and g: B → C,
⇒ gof: A → C, (gof) −1 : C→ A
So, f−1 o g−1and (gof)−1 have same domains.
(gof) (x) = g (f (x))
=g (2x + 1)
=(2x +1 )2− 2
⇒ (gof) (x) = 4x+ 4x + 1 − 2
⇒ (gof) (x) = 4x2+ 4x −1
Then, (gof) (1) = g (f (1))
= 4 + 4 − 1
=7,
(gof) (2) = g (f (2))
= 4(2)2 + 4(2) – 1 = 23,
(gof) (3) = g (f (3))
= 4(3)2 + 4(3) – 1 = 47 and
(gof) (4) = g (f (4))
= 4(4)2 + 4(4) − 1 = 79
So, gof = {(1, 7), (2, 23), (3, 47), (4, 79)}
⇒ (gof)– 1 = {(7, 1), (23, 2), (47, 3), (79, 4)}…… (2)
From (1) and (2), we get:
(gof)−1 = f−1 o g−1
5. Show that the function f: Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1
Solution:
Given function f: Q → Q, defined by f(x) = 3x + 5
Now we have to show that the given function is invertible.
Injection of f:
Let x and y be two elements of the domain (Q),
Such that f(x) = f(y)
⇒ 3x + 5 = 3y + 5
⇒ 3x = 3y
⇒ x = y
so, f is one-one.
Surjection of f:
Let y be in the co-domain (Q),
Such that f(x) = y
⇒ 3x +5 = y
⇒ 3x = y – 5
⇒ x = (y -5)/3 belongs to Q domain
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have to find f-1:
Let f-1(x) = y…… (1)
⇒ x = f(y)
⇒ x = 3y + 5
⇒ x −5 = 3y
⇒ y = (x – 5)/3
Now substituting this value in (1) we get
So, f-1(x) = (x – 5)/3
6. Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution:
Given f: R → R given by f(x) = 4x + 3
Now we have to show that the given function is invertible.
Consider injection of f:
Let x and y be two elements of domain (R),
Such that f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒ x = y
So, f is one-one.
Now surjection of f:
Let y be in the co-domain (R),
Such that f(x) = y.
⇒ 4x + 3 = y
⇒ 4x = y -3
⇒ x = (y-3)/4 in R (domain)
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have to find f -1
Let f-1(x) = y……. (1)
⇒ x = f (y)
⇒ x = 4y + 3
⇒ x − 3 = 4y
⇒ y = (x -3)/4
Now substituting this value in (1) we get
So, f-1(x) = (x-3)/4
7. Consider f: R → R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1(x) = √ (x-4) where R+ is the set of all non-negative real numbers.
Solution:
Given f: R → R+ → [4, ∞) given by f(x) = x2 + 4.
Now we have to show that f is invertible,
Consider injection of f:
Let x and y be two elements of the domain (Q),
Such that f(x) = f(y)
⇒ x+ 4 = y+ 4
⇒ x= y2
⇒ x = y (as co-domain as R+)
So, f is one-one
Now surjection of f:
Let y be in the co-domain (Q),
Such that f(x) = y
⇒ x2 + 4 = y
⇒ x2 = y – 4
⇒ x = √ (y-4) in R
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have to find f-1:
Let f−1 (x) = y…… (1)
⇒ x = f (y)
⇒ x = y2 + 4
⇒ x − 4 = y2
⇒ y = √ (x-4)
So, f-1(x) = √ (x-4)
Now substituting this value in (1) we get,
So, f-1(x) = √ (x-4)
8. If f(x) = (4x + 3)/ (6x – 4), x ≠ (2/3) show that fof(x) = x, for all x ≠ (2/3). What is the inverse of f?
Solution:
It is given that f(x) = (4x + 3)/ (6x – 4), x ≠ 2/3
Now we have to show fof(x) = x
(fof)(x) = f (f(x))
= f ((4x+ 3)/ (6x – 4))
= (4((4x + 3)/ (6x -4)) + 3)/ (6 ((4x +3)/ (6x – 4)) – 4)
= (16x + 12 + 18x – 12)/ (24x + 18 – 24x + 16)
= (34x)/ (34)
= x
Therefore, fof(x) = x for all x ≠ 2/3
=> fof = 1
Hence, the given function f is invertible and the inverse of f is f itself.
9. Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with
f-1(x) = (√(x +6)-1)/3
Solution:
Given f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x – 5
We have to show that f is invertible.
Injectivity of f:
Let x and y be two elements of domain (R+),
Such that f(x) = f(y)
⇒ 9x+ 6x – 5 = 9y+ 6y − 5
⇒ 9x+ 6x = 9y+ 6y
⇒ x = y (As, x, y ∈ R+)
So, f is one-one.
Surjectivity of f:
Let y is in the co domain (Q)
Such that f(x) = y
⇒ 9x2 + 6x – 5 = y
⇒ 9x2 + 6x = y + 5
⇒ 9x2 + 6x +1 = y + 6 (By adding 1 on both sides)
⇒ (3x + 1)2 = y + 6
⇒ 3x + 1 = √(y + 6)
⇒ 3x = √ (y + 6) – 1
⇒ x = (√ (y + 6)-1)/3 in R+ (domain)
f is onto.
So, f is a bijection and hence, it is invertible.
Now we have to find f-1
Let f−1(x) = y….. (1)
⇒ x = f (y)
⇒ x = 9y+ 6y − 5
⇒ x + 5 = 9y+ 6y
⇒ x + 6 = 9y2+ 6y + 1 (adding 1 on both sides)
⇒ x + 6 = (3y + 1)2
⇒ 3y + 1 = √ (x + 6)
⇒ 3y =√(x +6) -1
⇒ y = (√ (x+6)-1)/3
Now substituting this value in (1) we get,
So, f-1(x) = (√ (x+6)-1)/3
10. If f: R → R be defined by f(x) = x3 −3, then prove that f−1 exists and find a formula for f−1. Hence, find f−1 (24) and f−1 (5).
Solution:
Given f: R → R be defined by f(x) = x3 −3
Now we have to prove that f−1 exists
Injectivity of f:
Let x and y be two elements in domain (R),
Such that, x3 − 3 = y3 − 3
⇒ x3 = y3
⇒ x = y
So, f is one-one.
Surjectivity of f:
Let y be in the co-domain (R)
Such that f(x) = y
⇒ x3 – 3 = y
⇒ x3 = y + 3
⇒ x = ∛(y+3) in R
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding f -1:
Let f-1(x) = y…….. (1)
⇒ x= f(y)
⇒ x = y− 3
⇒ x + 3 = y3
⇒ y = ∛(x + 3) = f-1(x) [from (1)]
So, f-1(x) = ∛(x + 3)
Now, f-1(24) = ∛ (24 + 3)
= ∛27
= ∛33
= 3
And f-1(5) =∛ (5 + 3)
= ∛8
= ∛23
= 2
11. A function f: R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).
Solution:
Given that f: R → R is defined as f(x) = x3 + 4
Injectivity of f:
Let x and y be two elements of domain (R),
Such that f (x) = f (y)
⇒ x3 + 4 = y3 + 4
⇒ x3 = y3
⇒ x = y
So, f is one-one.
Surjectivity of f:
Let y be in the co-domain (R),
Such that f(x) = y.
⇒ x3 + 4 = y
⇒ x3 = y – 4
⇒ x = ∛ (y – 4) in R (domain)
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding f-1:
Let f−1 (x) = y…… (1)
⇒ x = f (y)
⇒ x = y+ 4
⇒ x − 4 = y3
⇒ y =∛ (x-4)
So, f-1(x) =∛ (x-4) [from (1)]
f-1 (3) = ∛(3 – 4)
= ∛-1
= -1
At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio… |
# Class 7 Math NCERT Solutions for Chapter – 11 Perimeter and Area Ex – 11.4
## Perimeter and Area
1. A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.
Solution:-
From the question it is given that,
Length of the garden (L) = 90 m
Breadth of the garden (B) = 75 m
Then,
Area of the garden = length × breadth
= 90 × 75
= 6750 m2
From the figure,
The new length and breadth of the garden when path is included is 100 m and 85 m respectively.
New area of the garden = 100 × 85
= 8500 m2
The area of path = New area of the garden including path – Area of garden
= 8500 – 6750
= 1750 m2
For 1 hectare = 10000 m2
Hence, area of garden in hectare = 6750/10000
= 0.675 hectare
2. A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Solution:-
From the question it is given that,
Length of the park (L) = 125 m
Breadth of the park (B) = 65 m
Then,
Area of the park = length × breadth
= 125 × 65
= 8125 m2
From the figure,
The new length and breadth of the park when path is included is 131 m and 71 m respectively.
New area of the park = 131 × 71
= 9301 m2
The area of path = New area of the park including path – Area of park
= 9301 – 8125
= 1176 m2
3. A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:-
From the question it is given that,
Length of the cardboard (L) = 8 cm
Breadth of the cardboard (B) = 5 cm
Then,
Area of the cardboard = length × breadth
= 8 × 5
= 40 cm2
From the figure,
The new length and breadth of the cardboard when margin is not included is 5 cm and 2 cm respectively.
New area of the cardboard = 5 × 2
= 10 cm2
The area of margin = Area of the cardboard when margin is including – Area of the
cardboard when margin is not including
= 40 – 10
= 30 cm2
4. A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2.
Solution:-
(i)
From the question it is given that,
Length of the room (L) = 5.5 m
Breadth of the room (B) = 4 m
Then,
Area of the room = length × breadth
= 5.5 × 4
= 22 m2
From the figure,
The new length and breadth of the room when verandah is included is 10 m and 8.5 m respectively.
New area of the room when verandah is included = 10 × 8.5
= 85 m2
The area of verandah = Area of the room when verandah is included – Area of the room
= 85 – 22
= 63 m2
(ii)
Given, the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2
Then the cost of cementing the 63 m2 area of floor of the verandah = 200 × 63
= ₹ 12600
5. A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
(i) the area of the path
(ii) the cost of planting grass in the remaining portion of the garden at the rate of ₹ 40 per m2.
Solution:-
(i)
From the question it is given that,
Side of square garden (s) = 30 m
Then,
Area of the square garden = S2
= 302
= 30 × 30
= 900 m2
From the figure,
The new side of the square garden when path is not included is 28 m.
New area of the room when verandah is included = 282
= 28 × 28
= 784 m2
The area of path = Area of the square garden when path is included – Area of the square
Garden when path is not included
= 900 – 784
= 116 m2
(ii)
Given, the cost of planting the grass in the remaining portion of the garden at the rate of
= ₹ 40 per m2
Then the cost of planting the grass in 784 m2 area of the garden = 784 × 40
= ₹ 31360
6. Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Solution:-
From the question it is given that,
Length of the park (L) = 700 m
Breadth of the park (B) = 300 m
Then,
Area of the park = length × breadth
= 700 × 300
= 210000 m2
Let us assume that ABCD is the one cross road and EFGH is another cross road in the park.
The length of ABCD cross road = 700 m
The length of EFGH cross road = 300 m
Both cross road have the same width = 10 m
Then,
= 700 × 10
= 7000 m2
= 300 × 10
= 3000 m2
Area of the IJKL at center = length × breadth
= 10 × 10
= 100 m2
Area of the roads = Area of ABCD + Area of EFGH – Area of IJKL
= 7000 + 3000 – 100
= 10000 – 100
= 9900 m2
We know that, for 1 hectare = 10000 m2
Hence, area of roads in hectare = 9900/10000
= 0.99 hectare
Finally, Area of the park excluding roads = Area of park – Area of the roads
= 210000 – 9900
= 200100 m2
= 200100/10000
= 20.01 hectare
7. Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
(i) the area covered by the roads.
(ii) the cost of constructing the roads at the rate of ₹ 110 per m2.
Solution:-
(i)
From the question it is given that,
Length of the field (L) = 90 m
Breadth of the field (B) = 60 m
Then,
Area of the field = length × breadth
= 90 × 60
= 5400 m2
Let us assume that ABCD is the one cross road and EFGH is another cross road in the park.
The length of ABCD cross road = 90 m
The length of EFGH cross road = 60 m
Both cross road have the same width = 3 m
Then,
= 90 × 3
= 270 m2
= 60 × 3
= 180 m2
Area of the IJKL at center = length × breadth
= 3 × 3
= 9 m2
Area of the roads = Area of ABCD + Area of EFGH – Area of IJKL
= 270 + 180 – 9
= 450 – 9
= 441 m2
(ii)
Given, the cost of constructing the roads at the rate of ₹ 110 per m2.
Then the cost of constructing the 441 m2 roads = 441 × 110
= ₹ 48510
8. Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π = 3.14)
Solution:-
From the question it is given that,
Radius of a circular pipe = 4 cm
Side of a square = 4 cm
Then,
Perimeter of the circular pipe = 2πr
= 2 × 3.14 × 4
= 25.12 cm
Perimeter of the square = 4 × side of the square
= 4 × 4
= 16 cm
So, the length of cord left with Pragya = Perimeter of circular pipe – Perimeter of square
= 25.12 – 16
= 9.12 cm
Yes, 9.12 cm cord is left.
9. The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land (ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed
(iv) the circumference of the flower bed.
Solution:-
(i)
From the figure,
Length of rectangular lawn = 10 m
Breadth of rectangular lawn = 5 m
Area of the rectangular lawn = Length × Breadth
= 10 × 5
= 50 m2
(ii)
From the figure,
Radius of the flower bed = 2 m
Area of the flower bed = πr2
= 3.14 × 22
= 3.14 × 4
= 12.56 m2
(iii)
The area of the lawn excluding the area of the flower bed = Area of rectangular lawn –
Area of flower bed
= 50 – 12.56
= 37.44 m2
(iv)
The circumference of the flower bed = 2πr
= 2 × 3.14 × 2
= 12.56 m
10. In the following figures, find the area of the shaded portions:
(i)
Solution:-
To find the area of EFDC, first we have to find the area of ΔAEF, ΔEBC and rectangle ABCD
Area of ΔAEF = ½ × Base × Height
= ½ × 6 × 10
= 1 × 3 × 10
= 30 cm2
Area of ΔEBC = ½ × Base × Height
= ½ × 8 × 10
= 1 × 4 × 10
= 40 cm2
Area of rectangle ABCD = length × breadth
= 18 × 10
= 180 cm2
Then,
Area of EFDC = ABCD area – (ΔAEF + ΔEBC)
= 180 – (30 + 40)
= 180 – 70
= 110 cm2
(ii)
Solution:-
To find the area of ΔQTU, first we have to find the area of ΔSTU, ΔTPQ, ΔQRU and square PQRS
Area of ΔSTU = ½ × Base × Height
= ½ × 10 × 10
= 1 × 5 × 10
= 50 cm2
Area of ΔTPQ = ½ × Base × Height
= ½ × 10 × 20
= 1 × 5 × 20
= 100 cm2
Area of ΔQRU = ½ × Base × Height
= ½ × 10 × 20
= 1 × 5 × 20
= 100 cm2
Area of square PQRS = Side2
= 20 × 20
= 400 cm2
Then,
Area of ΔQTU = PQRS area – (ΔSTU + ΔTPQ + ΔQRU)
= 400 – (50 + 100 + 100)
= 400 – 250
= 150 cm2
11. Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC
Solution:-
From the it is given that,
AC = 22 cm, BM = 3 cm DN = 3 cm and BM ⊥ AC, DN ⊥ AC
To find the area of quadrilateral ABCD, first we have to find the area of ΔABC, and ΔADC Area of ΔABC = ½ × Base × Height
= ½ × 22 × 3
= 1 × 11 × 3
= 33 cm2
Area of ΔADC = ½ × Base × Height
= ½ × 22 × 3
= 1 × 11 × 3
= 33 cm2
Then,
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC
= 33 + 33
= 66 cm2
#### Class 7 Math NCERT Solutions for Chapter – 12 Algebraic Expressions Ex – 12.1
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Phasellus cursus rutrum est nec suscipit. Ut et ultrices nisi. Vivamus id nisl ligula. Nulla sed iaculis ipsum.
Company Name |
# Additional Math – Trigonometry – Proving Identities
Sunday Additional Math Group Tuition at Woodlands – prove identities trigonometry
$latex \displaystyle \text{LHS}$
$latex \displaystyle {{\sec }^{4}}x-{{\tan }^{4}}x$
Factorizing using the algebraic special products a2-b2=(a+b)(a-b).
$latex \displaystyle {{\left( {{{{\sec }}^{2}}x} \right)}^{2}}-{{\left( {{{{\tan }}^{2}}x} \right)}^{2}}$
$latex \displaystyle \left( {{{{\sec }}^{2}}x+{{{\tan }}^{2}}x} \right)\left( {{{{\sec }}^{2}}x-{{{\tan }}^{2}}x} \right)$
Use the trigonometry formula sec2x=1+tan2x.
$latex \displaystyle \left( {1+{{{\tan }}^{2}}x+{{{\tan }}^{2}}x} \right)\left( {1+{{{\tan }}^{2}}x-{{{\tan }}^{2}}x} \right)$
$latex \displaystyle \left( {1+2{{{\tan }}^{2}}x} \right)\left( 1 \right)$
Change the expression by making cos2x the common denominat.
$latex \displaystyle \frac{{{{{\cos }}^{2}}x}}{{{{{\cos }}^{2}}x}}+2\frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}$
or
$latex \displaystyle \frac{{{{{\cos }}^{2}}x+2{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}$
$latex \displaystyle \frac{{{{{\cos }}^{2}}x+{{{\sin }}^{2}}x+{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}$
Simplify the expression.
$latex \displaystyle \frac{{1+{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}$
$latex \displaystyle \text{LHS=RHS}$
Always start with the side that looks more complicated. For this case, it is the left hand side. Place the N/O Level formula sheet beside you, look at the trigonometry formula while solving the trigonometric identities.
Trigonometric Identities in one of the most challenging sub-topics in Additional Math. But regular practice will give you confidence and help you solve quickly and correctly.
Additional Math (amath) trigonometry identities. Students from Woodlands, Choa Chu Kang, Yew Tee, Sembawang and Yishun.
## Share:
### Math – Statistics – 1st Quartile, Median and 3rd Quartile Ungroup Data
The above two diagrams show you how to find the 1st quartile, median (2nd Quartile) and the 3rd quartile ungroup
### Additional Math – Differentiation – Quotient Rule (Challenging)
Differentiate $latex \displaystyle\ y=\frac{{{{x}^{2}}\sqrt{{x+1}}}}{{x-1}}$ with respect to x. Simplify the Numerator (otherwise you need to use both quotient rule for
### Additional Math – Binomial theorem – Using Normal Expansion vs Binomial Theorem
The above video shows two method of expanding the expression; using the algebraic expansion (rainbow method) versus the binomial theorem. |
Math Is Easy - Use Algebra To Solve Word Problem 5!If 1 is subtracted from seven times a certain number,the result is the same as if 31 is added to three times the number. Find thenumber. (Note 1)If we let n represent the number to be found, the sentenceIf 1 is subtracted from seven times a certain number, the result is the sameas if 31 is added to three times the number. can be translated into thefollowing algebraic equation: 7n1 = 3n + 31Solve this equation7n1 = 3n +31 + 1= + 1 Add 1to both sides7n + 0 = 3n +327n =3n + 32 7n = 3n + 32- 3n =- 3n Subtract 3n from each side 4n= 0n + 32 4n= 32 4n/4 = 32/4 Divide each side by 4 n = 8 Now we substitute the solutionfor n into the original equation to see if it works.7n 1 =3n + 317(8)1 = 3(8) + 3156 1 = 24+ 3155 = 55We must also see if the solution works in theoriginal problem statement. If 1 is subtracted from seven times a certain number,the result is the same as if 31 is added to three times the number.If 1 is subtracted from seven times a certain number (7X 8), the result is the same as if 31 is added to three times the number (3 X8).If 1 is subtracted from seven times a certain number(56), the result is the same as if 31 is added to three times the number(24).If 1 is subtracted from 56, the result is the same as if31 is added to 24.561 = 31 + 2455 = 55 We have just solved a first degree equation.(Note1)The source for the problem statement is:Intermediate AlgebraFor College Students by Jerome E. Kaufman
CloseWindow |
Class 8 Maths Chapter 2 Exercise 2.1 Pdf Notes NCERT Solutions
Class 8 Maths Chapter 2 Linear Equations In One Variable exercise 2.1 pdf notes:-
Exercise 2.2 Class 8 maths Chapter 2 Pdf Notes:-
Ncert Solution for Class 8 Maths Chapter 2 Linear Equations In One Variable Exercise 2.1 Tips:
Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x ā 3, 3x + y, 2xy + 5, xyz + x + y + z, x^2+1, y + y^2
You would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2xy + 5
has two variables. We however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear. This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
We assume that the two sides of the equation are balanced.
We perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution. |
Categories
# Ceil Matlab
In this article, we will learn how the Ceil function works in MATLAB. Ceil function finds its utility in Mathematical problems involving decimal values. Ceil function helps in rounding off these decimal values to the nearest integer. As we will move forward in the article, we will learn how this is achieved for different types of scenarios.
## Ceil Function
The ceil function or ceiling function (also commonly called ‘least integer function) for any real number gives the smallest integer which is greater than the number itself. Now the question is there can be the infinite number of integers that are greater than the given real number. So, how does Ceil function work?
As we can interpret from the definition above, the Ceil function will choose the ‘SMALLEST’ of all those integers. Which clearly will be the immediate next integer. The output of ceil function can be the same as the input if the real number is actually an integer and not a decimal value.
## Example
Real number = 2.134: If this number is provided as input to the Ceil function, the output will be ‘3’, which is the immediate integer greater than 2.134
## What is the Syntax of Ceil Function in MATLAB?
Let us now understand the Syntax of Ceil function in MATLAB
• A = ceil(X): This syntax is used when we have a number as input
• A = ceil (time): This syntax is used when we have TIME as our input
ceil(X) function in MATLAB also helps us in rounding off the complex numbers. For complex value X, the real and imaginary parts are independently rounded off.
## Examples to Implement Ceil Matlab
Let us now understand the use of above-mentioned functions clearly with the help of a few examples:
## Example 1:
To use the ceil function, we will first create an array of numbers:
## Syntax
``X = [1.23 2.34 5.43 6]``
Now, this is a simple example without any other negative value or complex number.
This array of numbers is then passed as an argument to our function ‘ceil(X). This is how our input and output will look like in MATLAB console:
## Code:
``````X = [1.23 2.34 5.43 6];
Y = ceil(X)``````
## Output:
As we can see in our output, all the values are rounded off to their immediate next integer. Also notice that the value ‘6’ stays unchanged, as it is already an integer.
## Example 2:
Now let us take an example with negative values as well, and see how our function works
## Syntax
``X = [-0.56 2.78 -4.56 -3.23]``
This array of both positive and negative numbers is then passed as an argument to our function ‘ceil(X). This is how our input and output will look like in MATLAB console:
## Code:
``````X = [-0.56 2.78 -4.56 -3.23];
Y = ceil(X)``````
## Output:
Note: In the above output how negative values are rounded off. ‘-4.56’ is rounded off to ‘-4’ and not ‘–5’, because ‘-4’ is the integer greater than ‘-4.56’ and falls immediately after it.
## Example 3:
In the next example, we will understand how are ceil function works for Complex values.
## Syntax
``X = [2.9 -6.27.1; 593.1+6.2i]``
In this problem, we have a 2*3 matrix with a complex value ‘3.1 + 6.2i’.
This array of both simple and complex numbers is then passed as an argument to our function ‘ceil(X). This is how our input and output will look like in MATLAB console:
## Code;
``X = [2.9 -6.2 7.1; 5 9 3.1+6.2i] Y = ceil(X)``
## Output:
As we can see in the above output, for the complex number, the real and the imaginary parts are rounded off independently.
## Example 4:
In the next example, we will understand how are ceil function works for TIME input. For this purpose, the syntax used is A = ceil (time). We will use MATLAB command to create an array of ‘time’.
## Code:
``````time = hours(5) + minutes(9:13) + seconds(1.43);
time.Format = ‘hh:mm:ss.SS’``````
The above format is used to create a TIME array in MATLAB.Note: That ‘minutes(9:13)’ will create 5 elements with minutes in the range 9 to 13, both values included.
## Code:
``out = ceil(time)``
## Output:
As we can clearly notice in our output, the time is rounded off to the immediate integer which is greater than the ‘seconds’ value. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.