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# Chapter 3 - Polynomial and Rational Functions - Section 3.2 The Real Zeros of a Polynomial Function - 3.2 Assess Your Understanding - Page 224: 44
$\pm \dfrac{1}{2}, \pm \dfrac{1}{3}, \pm \dfrac{1}{6}, \pm \dfrac{2}{3}, \pm \dfrac{5}{2}, \pm \dfrac{5}{3}, \pm \dfrac{5}{6}, \pm \dfrac{10}{3}, \pm 1, \pm 2, \pm 5, \pm 10$
#### Work Step by Step
Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$. We see from the given polynomial function that it has a constant term of $10$ and a leading coefficient of $-6$. The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 2, \pm 4, \pm 5, \pm 10$ and $n=\pm 1, \pm 2, \pm 3, \pm 6$ Therefore, the possible rational roots of $f(x)$ are: $\dfrac{m}{n}=\pm \dfrac{1}{2}, \pm \dfrac{1}{3}, \pm \dfrac{1}{6}, \pm \dfrac{2}{3}, \pm \dfrac{5}{2}, \pm \dfrac{5}{3}, \pm \dfrac{5}{6}, \pm \dfrac{10}{3}, \pm 1, \pm 2, \pm 5, \pm 10$
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# 16.2: Substitution Ciphers - Mathematics
One simple encryption method is called a substitution cipher.
Substitution Cipher
A substitution cipher replaces each letter in the message with a different letter, following some established mapping.
A simple example of a substitution cipher is called the Caesar cipher, sometimes called a shift cipher. In this approach, each letter is replaced with a letter some fixed number of positions later in the alphabet. For example, if we use a shift of 3, then the letter A would be replaced with D, the letter 3 positions later in the alphabet. The entire mapping would look like: [1]
Original: (mathrm{ABCDEFGHIJKLMNOPQRSTUVWXYZ})
Maps to: (mathrm{DEFGHIJKLMNOPQRSTUVWXYZABC})
Example 1
Use the Caesar cipher with shift of 3 to encrypt the message: “We ride at noon”
Solution
We use the mapping above to replace each letter. W gets replaced with Z, and so forth, giving the encrypted message: ZH ULGH DW QRRQ.
Notice that the length of the words could give an important clue to the cipher shift used. If we saw a single letter in the encrypted message, we would assume it must be an encrypted A or I, since those are the only single letters than form valid English words.
To obscure the message, the letters are often rearranged into equal sized blocks. The message ZH ULGH DW QRRQ could be written in blocks of three characters as
ZHU LGH DWQ RRQ.
Example 2
Decrypt the message GZD KNK YDX MFW JXA if it was encrypted using a shift cipher with shift of 5.
Solution
We start by writing out the character mapping by shifting the alphabet, with A mapping to F, five characters later in the alphabet.
Original: (mathrm{ABCDEFGHIJKLMNOPQRSTUVWXYZ})
Maps to: (mathrm{FGHIJKLMNOPQRSTUVWXYZABCDE})
We now work backwards to decrypt the message. The first letter G is mapped to by B, so B is the first character of the original message. Continuing, our decrypted message is
Removing spaces we get BUYFIFTYSHARESA. In this case, it appears an extra character was added to the end to make the groups of three come out even, and that the original message was “Buy fifty shares.”
Try it Now 1
Decrypt the message BNW MVX WNH if it was encrypted using a shift cipher with shift 9 (mapping A to J).
SEND MONEY
Notice that in both the ciphers above, the extra part of the alphabet wraps around to the beginning. Because of this, a handy version of the shift cipher is a cipher disc, such as the Alberti cipher disk shown here[2] from the 1400s. In a cipher disc, the inner wheel could be turned to change the cipher shift. This same approach is used for “secret decoder rings.”
The security of a cryptographic method is very important to the person relying on their message being kept secret. The security depends on two factors:
1. The security of the method being used
2. The security of the encryption key used
In the case of a shift cipher, the method is “a shift cipher is used.” The encryption key is the specific amount of shift used.
Suppose an army is using a shift cipher to send their messages, and one of their officers is captured by their enemy. It is likely the method and encryption key could become compromised. It is relatively hard to change encryption methods, but relatively easy to change encryption keys.
During World War II, the Germans’ Enigma encryption machines were captured, but having details on the encryption method only slightly helped the Allies, since the encryption keys were still unknown and hard to discover. Ultimately, the security of a message cannot rely on the method being kept secret; it needs to rely on the key being kept secret.
Encryption Security
The security of any encryption method should depend only on the encryption key being difficult to discover. It is not safe to rely on the encryption method (algorithm) being kept secret.
With that in mind, let’s analyze the security of the Caesar cipher.
Example 3
Suppose you intercept a message, and you know the sender is using a Caesar cipher, but do not know the shift being used. The message begins EQZP. How hard would it be to decrypt this message?
Solution
Since there are only 25 possible shifts, we would only have to try 25 different possibilities to see which one produces results that make sense. While that would be tedious, one person could easily do this by hand in a few minutes. A modern computer could try all possibilities in under a second.
(egin{array}{|l|l|l|l|l|l|l|l|}
hline extbf { Shift } & extbf { Message } & extbf { Shift } & extbf { Message } & extbf { Shift } & extbf { Message } & extbf { Shift } & extbf { Message }
hline 1 & ext { DPYO } & 7 & ext { XJSI } & 13 & ext { RDMC } & 19 & ext { LXGW }
hline 2 & ext { COXN } & 8 & ext { WIRH } & 14 & ext { QCLB } & 20 & ext { KWFV }
hline 3 & ext { BNWM } & 9 & ext { VHQG } & 15 & ext { PBKA } & 21 & ext { JVEU }
hline 4 & ext { AMVL } & 10 & ext { UGPF } & 16 & ext { OAJZ } & 22 & ext { IUDT }
hline 5 & ext { ZLUK } & 11 & ext { TFOE } & 17 & ext { NZIY } & 23 & ext { HTCS }
hline 6 & ext { YKTJ } & mathbf{1 2} & extbf { SEND } & 18 & ext { MYHX } & 24 & ext { GSBR }
hline & & & & & & 25 & ext { FRAQ }
hline
end{array})
In this case, a shift of 12 (A mapping to M) decrypts EQZP to SEND. Because of this ease of trying all possible encryption keys, the Caesar cipher is not a very secure encryption method.
Brute Force Attack
A brute force attack is a method for breaking encryption by trying all possible encryption keys.
To make a brute force attack harder, we could make a more complex substitution cipher by using something other than a shift of the alphabet. By choosing a random mapping, we could get a more secure cipher, with the tradeoff that the encryption key is harder to describe; the key would now be the entire mapping, rather than just the shift amount.
Example 4
Use the substitution mapping below to encrypt the message “March 12 0300”
Original: (mathrm{ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789})
Maps to: (mathrm{2BQF5WRTD8IJ6HLCOSUVK3A0X9YZN1G4ME7P})
Solution
Using the mapping, the message would encrypt to 62SQT ZN Y1YY
Try it Now 2
Use the substitution mapping from Example 4 to decrypt the message C2SVX2VP
PARTY AT 9
While there were only 25 possible shift cipher keys (35 if we had included numbers), there are about 1040 possible substitution ciphers[3]. That’s much more than a trillion trillions. It would be essentially impossible, even with supercomputers, to try every possible combination. Having a huge number of possible encryption keys is one important part of key security.
Unfortunately, this cipher is still not secure, because of a technique called frequency analysis, discovered by Arab mathematician Al-Kindi in the 9th century. English and other languages have certain letters than show up more often in writing than others.[4] For example, the letter E shows up the most frequently in English. The chart to the right shows the typical distribution of characters.
Example 5
The chart to the right shows the frequency of different characters in some encrypted text. What can you deduce about the mapping?
Solution
Because of the high frequency of the letter S in the encrypted text, it is very likely that the substitution maps E to S. Since W is the second most frequent character, it likely that T or A maps to W. Because C, A, D, and J show up rarely in the encrypted text, it is likely they are mapped to from J, Q, X, and Z.
In addition to looking at individual letters, certain pairs of letters show up more frequently, such as the pair “th.” By analyzing how often different letters and letter pairs show up an encrypted message, the substitution mapping used can be deduced[5].
[1] en.Wikipedia.org/w/index.php?title=File:Caesar3.svg&page=1. PD
[2] en.Wikipedia.org/wiki/File:Alberti_cipher_disk.JPG
[3] There are 35 choices for what A maps to, then 34 choices for what B maps to, and so on, so the total number of possibilities is 35*34*33*…*2*1 = 35! = about 1040
[4] en.Wikipedia.org/w/index.php?title=File:English_letter_frequency_(alphabetic).svg&page=1 PD
[5] For an example of how this is done, see en.Wikipedia.org/wiki/Frequency_analysis
In cryptography, a cipher (or cypher) is a method for protecting data through encryption and decryption. Most ciphers require a specific key for encryption and decryption, but some ciphers like the ROT13 or Atbash ciphers have fixed keys. Many of the ciphers listed here were for military or other significant use during an earlier time, but today mostly are used only by puzzle makers.
Modern encryption methods can be divided by the key type and their operation on input data. Symmetric key algorithms use the same key for encryption and decryption (private key cryptography). Asymmetric key algorithms use different keys for encryption and decryption (public key cryptography). With symmetric keys, the sender and receiver must have agreed upon a key in advance, while with asymmetric keys anyone can send messages to the receiver. Also depending on their operation, ciphers are either block ciphers (encrypting a fixed block size) or stream ciphers (encrypting a continuous stream of data).
## Caesar Shift (Substitution Cipher)
A Caesar Shift cipher is a type of mono-alphabetic substitution cipher where each letter of the plain text is shifted a fixed number of places down the alphabet. For example, with a shift of 1, letter A would be replaced by letter B, letter B would be replaced by letter C, and so on. This type of substitution Cipher is named after Julius Caesar, who used it to communicate with his generals.
Key +5
Plain text alphabet: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
/>
Cipher text alphabet: F G H I J K L M N O P Q R S T U V W X Y Z A B C D E
This type of cipher is a form of symmetric encryption as the same key can be used to both encrypt and decrypt a message. (e.g. If a message is encrypted with a key of +5, it can be decrypted with a key of -5).
You can generate your own encryption keys and encrypt your own messages using our online Caesar Shift substitution engine:
#### Frequency Analysis
SPAASL KVLZ SBRL RUVD AOHA AOL NHSHJAPJ LTWPYL OHZ ZLJYLASF ILNBU JVUZAYBJAPVU VU H ULD HYTVYLK ZWHJL ZAHAPVU LCLU TVYL WVDLYMBS AOHU AOL MPYZA KYLHKLK KLHAO ZAHY.
DOLU JVTWSLALK, AOPZ BSAPTHAL DLHWVU DPSS ZWLSS JLYAHPU KVVT MVY AOL ZTHSS IHUK VM YLILSZ ZAYBNNSPUN AV YLZAVYL MYLLKVT AV AOL NHSHEF…”
## Substitution cipher decoder
The calculator below automatically decodes the text enciphered with the simple substitution cipher without knowing the key. The calculator logic is explained below the calculator.
#### Substitution cipher breaker
In cryptography, a substitution cipher is a method of encrypting by which units of plaintext are replaced with ciphertext, according to a fixed system the "units" may be single letters (the most common), pairs of letters, triplets of letters, mixtures of the above, and so forth. The receiver deciphers the text by performing the inverse substitution. Substitution of single letters separately — simple substitution — can be demonstrated by writing out the alphabet in some order to represent the substitution. It is a cipher key, and it is also called a substitution alphabet. 1
For a simple substitution cipher, the set of all possible keys is the set of all possible permutations. Thus, for the English alphabet, the number of keys is 26! (factorial of 26), which is about . Because of this, if you want to decipher the text without knowing the key, the brute force approach is out of the question.
However, the simple substitution cipher is considered a weak cipher because it is vulnerable to cryptoanalysis. First of all, substitution does not change the letters' frequencies, so if you have a decent amount of enciphered text and you know the language it was written in, you can try frequency analysis. For example, the most common letter in the English language is E, so, most common letter in the encrypted text is probable the E substitution. The analyst also looks for bigrams and trigrams frequencies because some unigram frequencies are too close to each other to rely on them. Using frequencies, analysts can create trial keys and test them to see if they reveal some words and phrases in the encrypted text.
But this manual approach is time-consuming, so the goal of an automated solution is to exclude humans from the process of breaking the cipher. And it is possible due to another simple substitution cipher vulnerability, known as Utility of Partial Solution.
In other words, if there are many pairs of keys in the keyspace where the decryption of the ciphertext by the key more similar to the correct key more closely resembles the plaintext than the decryption of the ciphertext by the other key, the cipher has Utility of Partial Solutions. If there is a correlation between the degree to which a key resembles the correct key and the degree to which that key's decryption of the ciphertext resembles the plaintext, it should be possible to search the keyspace efficiently by quickly discarding keys that are "worse" than whatever key is the closest match at any moment, climbing ever closer to the optimal key without knowing it initially. These keyspaces can be searched via Stochastic Optimization Algorithms. 2
The tricky part here is how you can measure if one key is "worse" than another. We need text fitness to address this, which gives us some score on how the given text looks like typical English text. There are different approaches, and I've tried this and that, but one which worked for me is outlined here: Text fitness (version 3). In short, it uses the sum of log probabilities of quadgrams and compares the sum with the sum for the "normal" English text (created as the sum of log probabilities of the most often English quadgrams). Here I'd like to thank Jens Guballa (site), author of another substitution solver, who kindly gives me a hint that text fitness function should be "normalized."
The implementation below uses a genetic algorithm to search for the correct key. If it fails, you can repeat a couple of times (each time it starts from a set of random keys as an initial generation) or tweak the settings, for example, increase the number of generations. Just click the Details to reveal additional settings. In this mode, the calculator also displays the best key in each generation, which is quite curious to watch.
If you see that the found key is close to the correct one but misses a couple of letters, you may use Substitution cipher tool to manually test the keys.
The Bifid cipher was invented by the French amateur cryptographer Félix Delastelle around 1901, and is considered an important invention in cryptology. It uses a combination of a Polybius square and transposition of fractionated letters to encrypt messages.
The two-square cipher is also called "double Playfair". It is stronger than an ordinary Playfair cipher, but still easier to use than the four-square cipher. Depending on the orientation of the squares, horizontal or vertical, the cipher behaves slightly different.
## The Germans and the Russians: Real Codes Decrypted
Let’s try running our code on a test message:
In addition to being correct, most of the attempts were very close as well, giving the first few letters such decodings as “coroti” or “gorothy”. Let’s try a harder one. This message was sent by Baron August Schluga, a German spy in WWI (source):
Here’s a code sent by Aldrich Ames, the most notorious CIA mole to have ever been caught. It was a portion of a message sent in 1992 that was found on his person when he was arrested:
This is a bit more disheartening because it’s obviously close to correct, but not quite there. A human can quickly fix the errors, switching p with m, and z and k. That being said, perhaps by slightly increasing the number of steps we could find our way to the right decryption.
Here’s another, from the same source above, sent by Confederate General J.E. Johnston during the Siege of Vicksburg, May 25, 1863 during the U.S. Civil War. It was intercepted and then deciphered by Lincoln’s three-man team of cryptanalysts:
Again, we are close but not quite there. In fact, this time the decryption even fails to segment the last block, even when there are useful pieces inside it. This is a side-effect of our zealous segmentation model that punishes unknown words exponentially in their length.
Unfortunately, our program seems to fail more often than succeed. Indeed, the algorithm is not all that good. One might hope the following function is usually the identity on any given message, but it rarely is:
In fact I have yet to see one example of this function returning sensible output for any input.
That last one was close, but still rather far off. What’s more, for random small inputs the function seems to generate sensible output!
Disregarding the fun we could have decrypting random message, we cry, “What gives?!” It seems that there’s some sort of measure of entropy that comes into play here, and messages with less entropy have a larger number of interpretations. Here entropy could mean the number of letters used in the message, the number of distinct letters used in the message, or some combination of both.
The last obvious annoyance is that thee program is really slow. If we watch the intermediate computations, we note that it will sometimes approach the correct solution, and then stray away. And what’s more, the number of steps per run is fixed. Why couldn’t it be based on the amount of progress it has made so far, pushing those close solutions a bit further to the correct key, perhaps by spending more time just on those successful decryptions.
Next time, we’ll spend our time trying to improve the steepest ascent algorithm. We’ll try to make it more quickly abandon crappy starting positions, and squeeze the most out of successful runs. We’ll further do some additional processing in our search for neighboring keys, and potentially involve letter 2-, 3-, 4-, and even 5-grams.
But all in all, this worked out pretty well. We often emphasize the deficiencies of our algorithms, but here we have surprisingly many benefits. For one, we actually decoded real-life messages! With this program in hand even as few as twenty years ago, we would have had a valuable tool for uncovering nefarious secrets encoded via substitution.
Even more, by virtue of our data-driven solution, we inherently bolster our algorithm with additional security. It’s stable, in the sense that minor data corruption (perhaps a typo, or nefarious message obfuscation) is handled well: our segmentation algorithm allows for typos, and a message with one or two typos still had a high letter trigram score in our probabilistic model. Hence we trivially bypass any sort of message obfuscation: if the average human can make sense of the decrypted message, so could our program.
And as with word segmentation, it’s extensible: simply by swapping out the data sets and making minor alphabet changes, we can make our program handle encrypted messages in any language. This exponentially increases the usefulness of our approach, because data sets are cheap, while sophisticated algorithms are expensive (at least, good ones can be sophisticated).
So look forward to next time when we improve the accuracy and speed of the steepest-ascent algorithm.
## Source code
dCode retains ownership of the online 'Mono-alphabetic Substitution' tool source code. Except explicit open source licence (indicated CC / Creative Commons / free), any 'Mono-alphabetic Substitution' algorithm, applet or snippet (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or any 'Mono-alphabetic Substitution' function (calculate, convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (Python, Java, PHP, C#, Javascript, Matlab, etc.) and no data download, script, copy-paste, or API access for 'Mono-alphabetic Substitution' will be for free, same for offline use on PC, tablet, iPhone or Android ! dCode is free and online.
## Substitution cipher tool
A tool to encrypt/decrypt messages with a simple substitution cipher given as the key. It is also useful for manual cryptanalysis of a substitution cipher - when you have a message written in the English alphabet partially decrypted with an automatic tool and want to tweak the key.
This simple tool allows you to encode and decode messages with a simple substitution cipher. For a cipher breaker, see Substitution cipher breaker. You can read a cipher description and some considerations regarding the strength of a cipher below the calculator.
#### Substitution Cipher Tool
According to Wikipedia, in cryptography, a substitution cipher is a method of encrypting by which units of plaintext are replaced with ciphertext, according to a fixed system the "units" may be single letters (the most common), pairs of letters, triplets of letters, mixtures of the above, and so forth. The receiver deciphers the text by performing the inverse substitution.
A simple substitution is the substitution of single letters separately. The substitution key is usually represented by writing out the alphabet in some order. The Caesar cipher is a form of a simple substitution cipher. For example, its ROT2 key can be presented as CDEFGHIJKLMNOPQRSTUVWXYZAB. This means that A is replaced with C, B with D, and so on.
The number of all possible keys for a simple substitution cipher is a factorial of 26 (26!). However, you can break it if you have enough ciphered text by using frequency analysis or the stochastic optimization algorithm (check out our Substitution cipher breaker). The cipher does not change language letter frequencies (it is said to be monoalphabetic), unlike, for example, the polyalphabetic Vigenère cipher, so it is considered to be rather weak.
Paste text into the field, fill the key, choose "encode" if you have pasted clear text or "decode" if you have pasted ciphered text, and press "Calculate". Traditionally, punctuation and spaces are removed to disguise word boundaries and text is written in blocks of letters, usually five. This option is supported for encoding as well.
## 16.2: Substitution Ciphers - Mathematics
In cryptography we encrypt messages that contains letters, but substitution ciphers are based on arithmetic operations on integers. Therefore, we have to convert the letters into integers before we apply the cryptosystem on the message. Because we use the English alphabet with 26 letters we have the following conversion between letters and integers:
a b c d e f g h i j k l m 0 1 2 3 4 5 6 7 8 9 10 11 12
n o p q r s t u v w x y z 13 14 15 16 17 18 19 20 21 22 23 24 25
E.g. the letter "a" is converted into the integer "0", the letter "b" into "1" and so on.
#### Modulo computation
You already use modulo computation when you look at the clock and e.g. needs to figure out what time it's 3 hours after 11 o'clock, which is 2 o'clock. In math we write that as:
where 12 is the modulus because we want the time as an integer between 0 and 11 (12 o'clock is in this case denoted by 0). In words we say 11 plus 3 modulo 12 is equal 2. The result of a modulo computation is an integer between 0 and the modulus minus 1. E.g. with the modulus 3 we have that:
• ( 1 : mod : 3 = 1 )
• ( 2 : mod : 3 = 2 )
• ( 3 : mod : 3 = 0 )
• ( 4 : mod : 3 = 1 )
• ( 5 : mod : 3 = 2 )
• ( 6 : mod : 3 = 0 )
• etc.
If we e.g. look at ( 27 : mod : 5 ) then modulo computes the number of times 5 divides 27 and then returns the remainder of the result which is 2 in this case, i.e. ( 27 : mod : 5 = 2 ). But how did we get this result?
First we compute the number of times it's possible to multiply 5 with the number ( x ) such that we get an integer as close as possible to 27 without exceeding it, i.e. we have to find the maximun value of ( x ) such that ( 5 cdot x leq 27 ). In this case we have that ( x = 5 ) because ( 5 cdot 5 = 25 leq 27 ). Then by subtracting 27 with 25 we get the answer ( 27 - 25 = 2).
If the integer is negative e.g. ( -27 : mod : 5 ) we have to do it slightly different and the answer is ( -27 : mod : 5 = 3 ). In this case the integer ( x ) is negative and should be the closest integer that exceed -27, i.e. we have to find the minimum value of ( -x ) such that ( 5 cdot -x geq -27 ). Now we have that ( -x = -6 ) because ( 5 cdot -6 = -30 geq -27 ). Then by subtracting -27 with -30 we get the answer ( -27 - (-30) = -27 + 30 = 3).
It's important that ( x ) or ( -x ) is an integer such as ( -14, 3, 17 ) etc. and NOT a fraction or float such as ( frac<1><4>, frac<-3><7>, 2.5, 5.1 ) etc.
If two integers ( a ) and ( b ) modulo the same modulus ( c ) returns the same remainder ( r ), then we say that ( a ) and ( b ) are congruent modulo ( c ). I.e. if ( a : mod : c = r ) and ( b : mod : c = r ) then ( a equiv b : (mod : c) ). Also, notice that if the modulus ( c ) is greater than the integer ( a ), i.e. ( c > a ), the result will always be equal ( a : mod : c = a ).
#### Greatest common divisor
Before we describe what the greatest common divisor of two integers are, we first define what we mean by a divisor. In this context is a divisor of an integer ( x ) some integer that divides ( x ) evenly, i.e. the result must not be a float. E.g. if you try to divide ( x=12 ) with 5 it returns the float ( frac<12> <5>= 2.4 ) and 5 is therefore not a divisor of ( x=12 ). For ( x=12 ) we have the divisors 1, 2, 3, 4, 6 and 12 because ( frac<12> <1>= 12 ), ( frac<12> <2>= 6 ), ( frac<12> <3>= 4 ), ( frac<12> <4>= 3 ), ( frac<12> <6>= 2 ) and ( frac<12> <12>= 1 ).
Likewise the divisors of 16 is 1, 2, 4, 8 and 16 because ( frac<16> <1>= 16 ), ( frac<16> <2>= 8 ), ( frac<16> <4>= 4 ), ( frac<16> <8>= 2 ) and ( frac<16><16>=1 ).
The greatest common divisor of 12 and 16 is therefore 4, because it is the largest integer of the common divisors. In math we write that as ( gcd(12, 16) = 4 ).
Two integers with greatest common divisor 1 are called relatively prime numbers or co-primes. E.g. 15 and 28 are relatively prime numbers because ( gcd(15, 28) = 1 ) (notice that 28 is not a prime number).
If one of the two integers is a prime number the greatest common divisor will always be 1, i.e. ( gcd(a, p) = 1 ) where ( a ) is an integer (either a prime number or a composite number) and ( p ) is a prime number.
One method to compute the greatest common divisor of two integers is by using the Euclidean algorithm developed by the famous Greek mathematician Euclid. See "The extended Euclidean algorithm" for more information about how to compute the greatest common divisor of two integer.
#### The extended Euclidean algorithm
The extended Euclidean algorithm is an extended version of the Euclidean algorithm, which only returns the greatest common divisor of two integers. Given two integers ( a ) and ( b ) the extended Euclidean algorithm returns the integers ( a ), ( b ), ( lambda ) and ( mu ) such that:
( a cdot lambda + b cdot mu = gcd(a, b) )
where ( lambda ) and ( mu ) are called the Bézout coefficients for ( a ) and ( b ). Only if ( a ) and ( b ) are relatively prime numbers, i.e. ( gcd(a, b) = 1 ), then:
( a cdot lambda + b cdot mu = 1 )
and ( lambda mod b ) is the inverse of ( a ), denoted ( a^ <-1>= lambda mod b ), and ( mu : mod : a ) is the inverse of ( b ), denoted ( b^ <-1>= mu : mod : a ) (see "Modulo computation" for more information about the ( mod ) operator). One useful property of an integer and its inverse is that ( a cdot a^ <-1> mod b = 1 ) and ( b cdot b^ <-1> mod a = 1 ).
You can easily compute ( gcd(a, b) ), ( lambda ) and ( mu ) for e.g. ( a=5 ) and ( b=39 ) with a simple table. So, let us first create a table with 3 columns (we do not yet know how many rows there will be in the table). Let us denote the entry in the first row and first column for [1,1], the entry in the first row and second column for [1,2], the entry in the second row and first column for [2,1] and so on.
Next we write ( b=39 ) in entry [1,1] and ( a=5 ) in entry [2,1]. Then we try to find the biggest integer ( q_ <1>), such that ( q_ <1>cdot a leq b ). We have that ( q_<1>=7 ), which we write in entry [2,2], because ( 7 cdot 5 = 35 leq 39 ) and a remainder of ( r_<1>=4 ), which we write in entry [3,1].
Again we try to find the biggest integer ( q_ <2>), such that ( q_ <2>cdot r_ <1>leq a ). We have that ( q_<2>=1 ), which we write in entry [3,2], because ( 1 cdot 4 = 4 leq 5 ) and a remainder of ( r_<2>=1 ) that we write in entry [4,1]. Notice that we just computed the same as before, just with integers in a lower row.
The next computation returns a remainder of ( r_ <3>= 0 ) because ( q_ <3>cdot r_ <2>= 4 cdot 1 = 4 leq 4 = r_ <1>). We have now computed ( gcd(5, 39)=r_<2>=1 ) since ( r_ <3>= 0 ). And because 5 and 39 are relatively prime numbers, we know that ( lambda ) and ( mu ) exists and we can then start using the last column.
First we write ( x_<1>=0 ) in entry [4,3] and ( x_<2>=1 ) in entry [3,3]. Then we write ( x_<3>=q_ <2>cdot x_ <2>+ x_ <1>= 1 cdot 1 + 0 = 1 ) in entry [2,3]. For entry [1,3] we compute the same as before, just with integers from the row above, i.e. ( x_<4>=q_ <1>cdot x_ <3>+ x_ <2>= 7 cdot 1 + 1 = 8 ).
Finally we have that ( a cdot x_ <4>pm b cdot x_ <3>= r_ <2>), where we need to decide whether it should be plus or minus between the two terms. Because ( a cdot x_ <4>= 5 cdot 8 = 40 ), ( b cdot x_ <3>= 39 cdot 1) and ( 40 geq 39 ) we have that ( 5 cdot 8 - 39 cdot 1 = 1 ) (which is the same as ( 5 cdot 8 + 39 cdot (-1) = 1 )) and the Bézout coefficients are ( lambda=8 ) and ( mu=-1 ). Notice that ( a^ <-1>= lambda mod b = 8 mod 39 = 8) and ( b^ <-1>= mu mod a = -1 : mod : 5 = 4) where ( a cdot a^ <-1> mod b = 5 cdot 8 mod 39 = 1 ) and ( b cdot b^ <-1> mod a = 39 cdot 4 mod 5 = 1 ).
The table for computing ( 5 cdot lambda + 39 cdot mu = gcd(5, 39) ) is:
( b=39 ) ( x_<4>=8 ) ( a=5 ) ( q_<1>=7 ) ( x_<3>=1 ) ( r_<1>=4 ) ( q_<2>=1 ) ( x_<2>=1 ) ( r_<2>=1 ) ( q_<3>=4 ) ( x_<1>=0 ) ( r_<3>=0 )
#### The Caesar cipher explained
The Caesar cipher, also known as the shift cipher, is named after the Roman general Julius Caesar who used it to communicate with his officers during wars about the year of 50 BC. It's a simple substitution cipher where each letter in the alphabet is substituted with another letter by shifting it ( s ) times. It's said that Julius Caesar used the shift value ( s = 3 ).
Assume Alice wants to send a message ( m ) (also called the plaintext) to her friend Bob using the Caesar cipher. First they in advance agree on a shift value ( s ) and then Alice encrypts the message by shifting each letter in the message ( s ) times to the right which produces the ciphertext ( c ). She then sends the ciphertext to Bob who decrypts it by shifting each letters in the ciphertext ( s ) times to the left.
In the case with the shift value ( s = 3 ) the letter "a" in the plaintext becomes "D" in the ciphertext, "b" becomes "E", "c" becomes "F" etc. You may wonder what happens with the last letters in the alphabet because there is no letter appearing three letters after "x", "y" and "z". The answer is simple: we just use the same approach as modulo computation, i.e. "x" becomes "A", "y" becomes "B" and "z" becomes "C". The encryption table for ( s = 3 ) is illustrated in the following table where the plaintext letters are in lowercase and the ciphertext letters are in uppercase:
Encryption table for ( s = 3 )
a b c d e f g h i j k l m
D E F G H I J K L M N O P
n o p q r s t u v w x y z
Q R S T U V W X Y Z A B C
If we convert the letters into integers such that "a" is converted into "0", "b" is converted into "1", "c" is converted into "2" etc. then the encryption of the letter ( m_ ) in the message ( m ) is defined as ( c_ = (m_ + s) : mod : 26 ) and decryption is defined as ( m_' = (c_ - s) : mod : 26 ). The modulus is 26 because there is 26 letters in the English alphabet. It's easy to verify that ( m_' = m_ ) because:
##### The protocol
Key exchange Alice and Bob agree on the shift value ( s ). Encryption Alice: Uses the shift value to compute the ciphertexts ( c_ = (m_ + s) : mod : 26 ) for ( i = 1, 2, dots n ) where ( n = |m| ) (the length of the plaintext ( m )). Alice sends the ciphertext ( c = (c_<1>, c_<2>, dots, c_) ) to Bob. Decryption Bob: Uses the shift value to decrypt the ciphertexts ( m_ = (c_ - s) : mod : 26 ) for ( i = 1, 2, dots n ) where ( n = |c| ) (the length of the ciphertext ( c )).
##### Example
Alice wants to send the message ( m = mbox <"meet me at the secret location">) to Bob. First they agree on the shift value ( s = 3 ) and then Alice converts the letters in ( m ) into the integers ( m_ <1>= 12, m_ <2>= 4, dots, m_ <25>= 13 ), which she encrypts:
( eqalign < c_<1>&= (m_ <1>+ s) : mod : 26 = (12 + 3) : mod : 26 = 15 c_ <2>&= (m_ <2>+ s) : mod : 26 = (4 + 3) : mod : 26 = 7 &vdots c_ <25>&= (m_ <25>+ s) : mod : 26 = (13 + 3) : mod : 26 = 16 > )
She then converts the ciphertexts ( c_ <1>= 15, c_ <2>= 7, dots, c_ <25>= 16 ) into letters and sends the ciphertext ( c = mbox <"PHHW PH DW WKH VHFUHW ORFDWLRQ">) to Bob.
Bob decrypts the received ciphertext almost just like Alice encrypted the message. First he converts the letters in the ciphertext ( c ) into the integers ( c_<1>, c_<2>, dots, c_ <25>) and then he decrypts them:
( eqalign < m_<1>&= (c_ <1>- s) : mod : 26 = (15 - 3) : mod : 26 = 12 m_ <2>&= (c_ <2>- s) : mod : 26 = (7 - 3) : mod : 26 = 4 &vdots m_ <25>&= (c_ <25>- s) : mod : 26 = (16 - 3) : mod : 26 = 13 > )
Finally he converts the plaintexts ( m_<1>, m_<2>, dots, m_ <25>) back into letters which result in the message ( m = mbox <"meet me at the secret location">).
#### The Affine cipher explained
The affine cipher is a combination of the Caesar cipher and a multiplication cipher: First Alice and Bob agrees on the two integers ( a ) and ( b ) between 0 and 25 such that ( gcd(a, 26) = 1 ) because during decryption her friend Bob needs the inverse ( a^ <-1>) of ( a ) which only exists when ( gcd(a, 26) = 1 ).
Alice then encrypts the message ( m ) by first converting each letter into integers where the letter "a" is converted into the integer "0", "b" is converted into "1", "c" is converted into "2" etc. Then for each integer ( m_ ) she computes ( c_ = (a cdot m_ + b) : mod : 26 ) (notice that if ( a=1 ) the affine cipher corresponds to the Caesar cipher with the sift value ( b )), converts each ( c_ ) back into letters and sends the ciphertext ( c ) to Bob. The modulus is 26 because there is 26 letters in the English alphabet.
Bob decrypts the ciphertext in a similar way as Alice encrypted the message. First he computes the inverse ( a^ <-1>) of ( a ) with the extended Euclidean algorithm, which gives him the equation ( a cdot lambda + 26 cdot mu = gcd(a, 26) ) where ( a^ <-1>= lambda mod 26 ) and ( a cdot a^ <-1>: mod : 26 = 1 ) (remember that ( a^ <-1>) only exists because ( gcd(a, 26) = 1 )). Then he converts each letter in ( c ) into integers ( c_ ), computes ( m_' = a^ <-1>cdot (c_ - b) : mod : 26 ) and finally he converts the integers ( m_' ) back into letters, which result in the message ( m ). We see that ( m_' = m_ ) because:
( eqalign< m_' &= a^ <-1>cdot (c_ - b) : mod : 26 &&(c_ = (a cdot m_ + b)) &= a^ <-1>cdot ((a cdot m_ + b) - b) : mod : 26 &=a^ <-1>cdot (a cdot m_) : mod : 26 &&(a cdot a^ <-1>: mod : 26 = 1) &= m_ : mod : 26 > )
##### The protocol
Key exchange Alice and Bob agree on the values ( a ) and ( b ) such that ( gcd(a, 26) = 1 ). Encryption Alice: Uses the values ( a ) and ( b ) to compute the ciphertexts ( c_ = (a cdot m_ + b) : mod : 26 ) for ( i = 1, 2, dots n ) where ( n = |m| ) (the length of the plaintext ( m )). Alice sends the ciphertext ( c = (c_<1>, c_<2>, dots, c_) ) to Bob. Decryption Bob: Computes the inverse ( a^ <-1>) of ( a ) with the extended Euclidean algorithm. Uses ( b ) and ( a^ <-1>) to decrypt the ciphertexts ( m_ = a^ <-1>cdot (c_ - b) : mod : 26 ) for ( i = 1, 2, dots n ) where ( n = |c| ) (the length of the ciphertext ( c )).
##### Example
Alice wants to send the message ( m = mbox <"it begins at midnight">) to Bob. First they agree on the integers ( a = 15 ) and ( b = 9 ) where ( gcd(a, 26) = gcd(15, 26) = 1 ). Then Alice converts the letters in ( m ) into the integers ( m_ <1>= 8, m_ <2>= 19, dots, m_ <18>= 19 ), which she encrypts:
( eqalign < c_<1>&= (a cdot m_ <1>+ b) : mod : 26 = (15 cdot 8 + 9) : mod : 26 = 25 c_ <2>&= (a cdot m_ <2>+ b) : mod : 26 = (15 cdot 18 + 9) : mod : 26 = 19 &vdots c_ <18>&= (a cdot m_ <18>+ b) : mod : 26 = (15 cdot 19 + 9) : mod : 26 = 8 > )
She then converts ( c_ <1>= 15, c_ <2>= 7, dots, c_ <18>= 16) back into letters and sends the ciphertext ( c = mbox <"ZI YRVZWT JI HZCWZVKI">) to Bob.
When Bob receives the ciphertext he first computes the inverse ( a^ <-1>) of ( a = 15 ) with the extended Euclidean algorithm. The extended Euclidean algorithm gives him the equation ( gcd(a, 26) = a cdot lambda + 26 cdot mu = 15 cdot 7 + 26 cdot (-4) = 1 ) where ( a^ <-1>= lambda mod 26 = 7 mod 26 = 7 ). Then he converts the letters in ( c ) into the integers ( c_ <1>, c_<2>, dots, c_ <18>) and decrypts them:
( eqalign < m_<1>&= a^ <-1>cdot (c_ <1>- b) : mod : 26 = 7 cdot (15 - 9) : mod : 26 = 8 m_ <2>&= a^ <-1>cdot (c_ <2>- b) : mod : 26 = 7 cdot (7 - 9) : mod : 26 = 18 &vdots m_ <18>&= a^ <-1>cdot (c_ <18>- b) : mod : 26 = 7 cdot (16 - 9) : mod : 26 = 19 > )
Finally he converts ( m_ <1>, m_<2>, dots, m_ <18>) back into letters which result in the message ( m = mbox <"it begins at midnight">).
#### The Vigenére cipher explained
The Caesar and affine cipher are both monoalphabetic ciphers because once a key is chosen each letter is mapped uniquely to another letter. As cryptanalytic methods became more sophisticated in the 16th century more sophisticated ciphers were invented and one of them was the Vigenére cipher, which was invented by Blaise de Vigenére in the 16th century. The Vigenére cipher is a polyalphabetic cipher because each letter is encrypted and decrypted with different keys, i.e. the same letter can be mapped to multiple different letters.
##### Example
Alice wants to send the message ( m = mbox <"it is hidden at the bridge">) to Bob where ( n = |m| = 21). First they agree on the keyword ( K = mbox <"THEORY">) of length ( t = 6 ). Because ( t Try the cipher
The one-time pad described by Gilbert Vernam in 1917 is a perfectly secure cryptosystem when the used key is completely random and only used to encrypt a single message. Perfectly secure means that if Alice and Bob's common adversary Eve with infinity computing power intercept a ciphertext she learns nothing whatsoever about the message or the key.
Encryption and decryption in the one-time pad are almost identical to the one used in the Vigenére cipher which makes it very attractive because they are simple operations, but unfortunately the used key must have the same length as the message. If we could encrypt multiple messages with the same key this would not be a problem, but the key must only be used once to keep the one-time pad perfectly secure.
In the one-time pad Alice and Bob first agrees on a random key ( K ) of length ( n ) where ( n = |m| ) (the length of the message ( m )). Alice then encrypts the message ( m ) using the key ( K ) just like she did in the Vigenére cipher: She first converts the letters in the key ( K ) and the message ( m ) into the integers ( k_<1>, k_<2>, dots, k_ ) and ( m_<1>, m_<2>, dots, m_ ) respectively. Then she encrypts each letter by computing ( c_ = (m_ + k_) : mod : 26 ) where the modulus is 26 because the English alphabet has 26 letters. Finally she converts the integers ( c_<1>, c_<2>, dots, c_ ) into letters and sends the ciphertext ( c ) to Bob.
Bob decrypts the received ciphertext ( c ) in a similar way: he converts the key ( K ) and the ciphertext ( c ) into the integers ( k_<1>, k_<2>, dots, k_ ) and ( c_<1>, c_<2>, dots, c_ ) respectively and computes ( m_' = (c_ - k_) : mod : 26 ). Finally he converts the integers ( m_<1>', m_<2>', dots, m_' ) into letters, which result in the message ( m ). We have that ( m_' = m_ ) because:
##### The protocol
Key exchange Alice and Bob agree on the key ( K = (k_<1>, k_<2>, dots, k_) ) where ( n = |m| ) (the length of the message ( m )). Encryption Alice: Uses the key ( k_ ) to compute the ciphertext ( c_ = (m_ + k_) : mod : 26 ) for ( i = 1, 2, dots, n ). Alice sends the ciphertext ( c = (c_<1>, c_<2>, dots, c_) ) to Bob. Decryption Bob: Uses the key ( k_ ) to decrypt the ciphertext ( m_ = (c_ - k_) : mod : 26 ) for ( i = 1, 2, dots, n ).
##### Example
Alice wants to send the message ( m = mbox <"meet me tomorrow morning">) to Bob where ( n = |m| = 21 ). First they agree on a random key with the same length as the message: ( K = mbox <"WVJLQCOEUTYJNNUGDROVS">) . She then converts the letters in ( K ) and ( m ) into the integers ( k_ <1>= 22, k_ <2>= 21, dots, k_ <21>= 18 ) and ( m_ <1>= 12, m_ <2>= 4, dots, m_ <21>= 6 ) respectively and computes:
( eqalign < c_<1>&= (m_ <1>+ k_<1>) : mod : 26 = (12 + 22) : mod : 26 = 8 c_ <2>&= (m_ <2>+ k_<2>) : mod : 26 = (4 + 21) : mod : 26 = 25 &vdots c_ <21>&= (m_ <21>+ k_<18>) : mod : 26 = (6 + 18) : mod : 26 = 24 > )
She then converts ( c_ <1>= 8, c_ <2>= 25, dots, c_ <21>= 24 ) back into letters and sends the ciphertext ( c = mbox <"IZNE CG HSGHPABJ GUUEWIY">) to Bob.
When Bob receives the ciphertext he converts the letters in ( K ) and ( c ) into the integers ( k_ <1>, k_<2>, dots, k_ <21>) and ( c_ <1>, c_<2>, dots, c_ <21>) respectively and computes:
( eqalign < m_<1>&= (c_ <1>- k_<1>) : mod : 26 = (8 - 22) : mod : 26 = 12 m_ <2>&= (c_ <2>- k_<2>) : mod : 26 = (25 - 21) : mod : 26 = 4 &vdots m_ <21>&= (c_ <21>- k_<18>) : mod : 26 = (24 - 18) : mod : 26 = 6 > )
Finally he converts ( m_ <1>, m_<2>, dots, m_ <21>) back into letters, which result in the message ( m = mbox <"meet me tomorrow morning">).
## Substitution Cipher
This problem offers a statistical activity that has immediate practical application. We have offered a spreadsheet toolkit so that students can concentrate on the analysis of the data without needing to waste time on computation.
### Possible approach
"Which letters do you think appear most often in the English language?"
"How could you find out?"
Allow some time for students to think about and share their answers. If they have books with them, perhaps suggest that they take a look to see which letters seem most common at first glance.
If a computer room is available, introduce students to the toolkit and give them time to perform a frequency analysis on some English text (Wikipedia articles are a great source for this). Share results from the frequency analysis. Does everybody find the same letters come out top, and bottom? There is opportunity here for some discussion about the benefits of using longer sample texts.
Then present students with the ciphertext in the problem (available as a text file here).
If a computer room is not available, the ciphertext is available as a worksheet here. Here is a second version of the worksheet with the ciphertext faint so that students can write over it as they go deciphering the message.
### Key questions
Are there any short words? What might they be?
### Possible support
Students could be encouraged to work collaboratively on this problem. There are lots of suggestions to help them get started in the hint.
### Possible extension
Students could investigate the frequency of digraphs (pairs of letters such as th or sh) in the English language and consider whether this speeds up the deciphering process.
The Secondary Cipher Challenge and Substitution Transposed offer challenging extensions for students who have worked on this problem and the problem Transposition Cipher.
## The Black Chamber
The monoalphabetic substitution cipher seemed uncrackable, because of the huge number of possible keys. There was, however, a shortcut that would undermine its security. This section tells the story of how this code breaking technique was invented, explains how it works and provides you with a tool that will help you to crack ciphers.
The cracking of the substitution cipher marks the birth of cryptanalysis (code breaking). This occurred during the golden age of the Islamic civilization, when many ancient foreign manuscripts were being brought to Baghdad to be added to the great Arab libraries. Some of these manuscripts were encrypted, which motivated the code breakers to crack the ciphers and reveal the secrets within. The picture shows the first page of al-Kindi's manuscript "On Deciphering Cryptographic Messages", containing the oldest known description of cryptanalysis by frequency analysis.
The letters "a" and "I" are the most common in Arabic. In English, E, then T, then A are the most common letters.
If a message is enciphered so that every letter is substituted for a different letter, then the new letter will take on all the attributes of the old letter, including how common it is.
So if the most common letter in an encoded English message is W, then W probably represents E. If there are lots of Gs, then G might represent T. And so on.
The letters "a" and "I" are the most common in Arabic. In English, E, then T, then A are the most common letters.
Although it is not known who first realized that the variation in the frequencies of letters could be exploited in order to break ciphers, the earliest known description of the technique is by the 9th century scientist Abu Yusuf Ya 'qub ibn Is-haq ibn as-Sabbah ibn 'omran ibn Ismail al-Kindi. Known as the philosopher of the Arabs', al-Kindi was the author of 290 books on medicine, astronomy, mathematics, linguistics and music, but his greatest treatise, which was only rediscovered in 1987 in the Sulaimaniyyah Ottoman Archive in Istanbul, is entitled "A Manuscript on Deciphering Cryptographic Messages."
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# 10.6: Volume of Prisms
Difficulty Level: At Grade Created by: CK-12
## Introduction
Packing Peanuts
One of the tasks that Candice and Trevor have is to fill boxes with packing peanuts when wrapping something that is fragile. The boxes can hold a good amount of packing peanuts. Each day, one of the students goes to the storeroom to get a big bag of packing peanuts. Then they work on filling 5 – 10 boxes half-way full of packing peanuts so that they are ready to go when the time comes.
“I wonder how much one box holds,” Candice asked Trevor one morning.
“I think that would depend on the box. I think that a tall skinny box would hold more than a wide flat box,” Trevor said, challenging her.
“I don’t. Think about it. Tall may seem like it would hold more, but you can spread out the peanuts in the wide flat box. I think the wide flat one will hold more.”
“Let’s test it out,” Candice said. “We can fill them and then go back and count all of the peanuts.”
“There is an easier way than that. We can use the dimensions and figure out the volume of each box.”
Candice is a bit puzzled by that. She thinks that counting all of the peanuts would be a quicker way to work. The two begin a short argument and decide to try out both methods. Candice begins filling boxes and Trevor begins to work it through mathematically.
Here are the dimensions of the two boxes:
The tall box has a length of 5 inches, a width of 4 inches and a height of 18 inches.
The wide flat box has a length of 12 inches, a width of 6 inches, and a height of 5 inches.
Do you have a prediction which one has the greater volume? This lesson is all about figuring out the volume of prisms. Pay close attention and you will understand by the end of the lesson.
What You Will Learn
By the end of this lesson, you will be able to demonstrate the following skills.
• Recognize volume of prisms as the sum of volumes of layers of unit cubes.
• Find volumes of rectangular prisms using formulas.
• Find volumes of triangular prisms using formulas.
• Solve real-world problems involving volumes of prisms.
Teaching Time
I. Recognize Volume of Prisms as the Sum of Volumes of Layers of Unit Cubes
You have heard the world “volume” before in everyday life. We can talk about the volume of water in a pool or in a glass or in a pitcher. This lesson will focus on how we can find the volume of a prism.
A prism is a three-dimensional figure with two congruent parallel bases and rectangular faces for sides. The prism is named by the polygon which makes up its base.
Volume is the measure of how much three-dimensional space it takes up or holds.
Imagine a fish aquarium. Its length, width, and height determine how much water the tank will hold. If we fill it with water, the amount of water represents the volume of the tank.
We measure volume in cubic units, because we are multiplying three dimensions: length, width, and height.
We will look at several ways to calculate volume. One way is to use unit cubes.
Volume, as we have said, is the amount of space a three-dimensional solid takes up. One way to find the volume of a prism is to consider how many unit cubes it can contain. A unit cube is simply a cube measuring one inch, one centimeter, one foot, or whatever units of measurement we are using, on all sides. Here are some unit cubes.
We use unit cubes as a way to measure the space inside a solid figure, or its volume. We simply count the number of unit cubes that “fit” into the prism. We begin by counting the number of cubes that cover the bottom of the prism, and then count each layer. Let’s see how this works.
How many cubes do you see here? If we count all of the cubes, you will see that we have 24 cubes in this prism.
The volume of this prism is 24 units3\begin{align*}\text{units}^3\end{align*} or cubic units.
Let’s look at another example.
Example
Find the volume of the following figure using unit cubes.
How many cubes are in this figure? We can see by counting that there are 48. The volume of this prism is 48 cubic units or units3\begin{align*}\text{units}^3\end{align*}.
Did you notice a pattern here?
If you look carefully, you will see that the volume of the rectangular prism is a function of multiplying the length ×\begin{align*}\times\end{align*} the width ×\begin{align*}\times\end{align*} the height.
Here is our formula!!
II. Find Volumes of Rectangular Prisms Using Formulas
We just discovered the formula for finding the volume of a rectangular prism. Now let’s refine that formula a little further. Here is the formula.
V=Bh\begin{align*}V=Bh\end{align*}
The volume is equal to the B\begin{align*}B\end{align*}, base area of the prism, multiplied by the height of the prism.
Let’s look at an example.
Example
Find the volume of the prism below.
We simply put the values for the length, width, and height in for the appropriate variables in the formula. Then we solve for V\begin{align*}V\end{align*}, volume.
First we find the area of the base. This is the rectangular side on the bottom. Remember, to find the area of a rectangle we multiply the length by the width.
BBB=lw=16×9=144 cm2\begin{align*}B & = lw\\ B & = 16 \times 9\\ B & = 144 \ cm^2\end{align*}
The base area is 144 square centimeters. Now we simply multiply this by the height, which represents the number of layers in the prism.
VVV=Bh=144×4=576 cm3\begin{align*}V & = Bh\\ V & = 144 \times 4\\ V & = 576 \ cm^3\end{align*}
The volume of this rectangular prism is 576 cubic centimeters.
We can calculate the volume of the same rectangular prism with unit cubes too.
You can see that we could count the unit cubes here to find the volume of the rectangular prism. The other option is to multiply the measurements that we see. This would work as well. Let’s try it and see.
VVV=lwh=(16)(9)(4)=576 cm3\begin{align*}V & = lwh\\ V & = (16)(9)(4)\\ V & = 576 \ cm^3\end{align*}
Wow! We got the same answer!
10I. Lesson Exercises
Find the volume of each rectangular prism given the following dimensions.
1. Length of 10 in, width of 8 in, height of 6 inches
2. Length of 8 m, width of 7 m, height of 3 meters
3. Length of 15 ft, width of 12 ft, height of 11 feet
III. Find Volumes of Triangular Prisms Using Formulas
You have just finished working with rectangular prisms, now we are going to look at volume with triangular prisms.
What is the difference between a rectangular prism and a triangular prism?
A rectangular prism has two parallel faces that are rectangles, then the other faces are rectangles as well. With a triangular prism, the two parallel faces are triangles and then the other faces are still rectangles. Here is a picture of a triangular prism.
We calculate the volume of triangular prisms almost the same way that we find the volume of rectangular prisms. We still use the formula V=Bh\begin{align*}V = Bh\end{align*}. However, this time the bottom layer of the prism is a triangle, not a rectangle. Therefore we need to use the area formula for a triangle to find B\begin{align*}B\end{align*}. Then we can multiply this amount by the height.
Let’s look at an example to see how this works.
Example
What is the volume of this triangular prism?
As we have seen, the volume formula for any prism is V=Bh\begin{align*}V = Bh\end{align*}. First we need to find the base area. Because the base is a triangle, we need to use the formula for the area of a triangle: 12bh\begin{align*}\frac{1}{2} bh\end{align*}. The height of the triangle, h\begin{align*}h\end{align*}, is indicated by a dashed line. The base of the triangle, \begin{align*}b\end{align*}, is the side perpendicular to the height. Remember, we use the height and base measurements for the triangular face, not the height measurement for the whole prism. Look carefully at the image!
So there are two things that we need to accomplish, we need to find the area of one of the triangular bases and then we can take that measurement and multiply it with the height of the entire prism.
\begin{align*}V & = Bh\\ B & = \frac{1}{2} bh\\ B & = \frac{1}{2}(16)(6)\\ B & = 48\\ V & = (48)H\\ V & = (48)(10)\\ V & = 480 \ in^3\end{align*}
The volume of this triangular prism is \begin{align*}480 \ inches^3\end{align*}.
10J. Lesson Exercises
Find the volume of the following triangular prisms.
1. \begin{align*}b = 12 \ in, \ h = 10 \ in, \ H = 15 \ in\end{align*}
2. \begin{align*}b = 7 \ cm, \ h = 5 \ cm, \ H = 9 \ cm\end{align*}
3. \begin{align*}b = 4 \ mm, \ h = 3 \ mm, \ H = 5 \ mm\end{align*}
IV. Solve Real-World Problems Involving Volumes of Prisms
We can use the methods we have learned to solve real-world problems involving area. First, be sure you understand what the question is asking. Second, consider whether the item in the problem is a rectangular or a triangular prism so that you know which formula to use. Let’s practice with a few problems.
Example
Carla is cleaning out her fish tank, so she filled the bathtub to the rim with water for her fish to swim in while she empties their tank. If the bathtub is 5.5 feet long, 3.3 feet wide, and 2.2 feet deep, how much water can it hold?
First of all, what is the problem asking us to find? We need to find the volume of the bathtub. Is a bath tub a rectangular or a triangular prism? It is a rectangular prism, so we’ll need to use the area formula for rectangles to find \begin{align*}B\end{align*}.
\begin{align*}B & = lw\\ B & = 5.5 \times 3.3\\ B & = 18.15 \ ft^2\end{align*}
Now we put this value into the volume formula and solve:
\begin{align*}V & = Bh\\ V & = 18.15 \times 2.2\\ V & = 39.93 \ ft^3\end{align*}
Carla’s bathtub can hold 39.93 cubic feet of water.
Example
Every year Jeanie gets a bottle of her favorite perfume for her birthday. The perfume comes in a bottle shaped like a triangular prism. She is worried that she might run out of perfume before her next birthday because the bottle is only half full. How much perfume does she have left?
First, let’s think about what the problem is asking us to find. We need to know how much perfume Jeanie has left. This means we need to find the volume of a full bottle and then divide by 2. Before we can use the volume formula, we also need to decide what kind of prism this is. The problem tells us that the bottle is in the shape of a triangular prism, so we’ll need to use the area formula for triangles to find the base area, \begin{align*}B\end{align*}.
\begin{align*}B & = \frac{1}{2} bh\\ B & = \frac{1}{2} (6) (4)\\ B & = 3 (4)\\ B & = 12 \ cm^2 \end{align*}
The base area is 12 square centimeters. Now we can put this into the volume formula and solve.
\begin{align*}V & = Bh\\ V & = 12 \times 9\\ V & = 108 \ cm^3\end{align*}
Now we know that the volume of the perfume bottle is 108 cubic centimeters. This is the amount a full bottle can contain. Remember, Jeanie’s bottle is only half full. Therefore we need to divide the volume in half:
\begin{align*}108 \ cm^3 \div 2 = 54 \ cm^3\end{align*}
There are 54 cubic centimeters of perfume left in Jeanie’s perfume bottle.
## Real–Life Example Completed
Packing Peanuts
Here is the original problem once again. Reread it and then use math with Trevor to find the volume of both boxes.
One of the tasks that Candice and Trevor have is to fill boxes with packing peanuts when wrapping something that is fragile. The boxes can hold a good amount of packing peanuts. Each day, one of the students goes to the storeroom to get a big bag of packing peanuts. Then they work on filling 5 – 10 boxes half-way full of packing peanuts so that they are ready to go when the time comes.
“I wonder how much one box holds,” Candice asked Trevor one morning.
“I think that would depend on the box. I think that a tall skinny box would hold more than a wide flat box,” Trevor said, challenging her.
“I don’t. Think about it. Tall may seem like it would hold more, but you can spread out the peanuts in the wide flat box. I think the wide flat one will hold more.”
“Let’s test it out,” Candice said. “We can fill them and then go back and count all of the peanuts.”
“There is an easier way than that. We can use the dimensions and figure out the volume of each box.”
Candice is a bit puzzled by that. She thinks that counting all of the peanuts would be a quicker way to work. The two begin a short argument and decide to try out both methods. Candice begins filling boxes and Trevor begins to work it through mathematically.
Here are the dimensions of the two boxes:
The tall box has a length of 5 inches, a width of 4 inches and a height of 18 inches.
The wide flat box has a length of 12 inches, a width of 6 inches, and a height of 5 inches.
First, we can figure out the volume of both of the boxes. We can do this simply by multiplying the length \begin{align*}\times\end{align*} the width \begin{align*}\times\end{align*} the height of both of the boxes.
Let’s do the tall one first.
\begin{align*}V&=lwh\\ V&=5(4)(18) \\ V&=360 \ cubic \ inches \end{align*}
Now we can find the volume of the wide, flat box.
\begin{align*}V& =lwh \\ V & =(6)(5)(12) \\ V & =360 \ cubic \ inches \end{align*}
“WOW!” Trevor exclaimed as Candice was still counting.
“What?” Candice asked looking up from one pile of peanuts.
“They have the same volume!!”
Trevor took his piece of paper to show Candice his work. Then he smiled and the two began picking up the packing peanuts. Use arithmetic was definitely faster in this case!
## Vocabulary
Here are the vocabulary words found in this lesson.
Prism
a three-dimensional solid with two flat parallel polygon bases and rectangular faces. The bases can be any polygon in shape.
Volume
the measure of the space inside a solid figure. Volume often is measured in terms of capacity connected with liquid measure.
Cubic Units
volume is measured in cubic units because three parts of a solid are being measured, length, width and height.
## Technology Integration
1. http://www.mathplayground.com/mv_volume_prisms.html – This is a Brightstorm video on how to find the volume of prisms.
## Time to Practice
Directions: Find the volume of each rectangular prism. Remember to label your answer in cubic units.
1. Length = 5 in, width = 3 in, height = 4 in
2. Length = 7 m, width = 6 m, height = 5 m
3. Length = 8 cm, width = 4 cm, height = 9 cm
4. Length = 8 cm, width = 4 cm, height = 12 cm
5. Length = 10 ft, width = 5 ft, height = 6 ft
6. Length = 9 m, width = 8 m, height = 11 m
7. Length = 5.5 in, width = 3 in, height = 5 in
8. Length = 6.6 cm, width = 5 cm, height = 7 cm
9. Length = 7 ft, width = 4 ft, height = 6 ft
10. Length = 15 m, width = 8 m, height = 10 m
Directions: Find the volume of each triangular prism. Remember that \begin{align*}h\end{align*} means the height of the triangular base and \begin{align*}H\end{align*} means the height of the whole prism.
11. \begin{align*}b = 6 \ in, \ h = 4 \ in, \ H = 5 \ in\end{align*}
12. \begin{align*}b = 7 \ in, \ h = 5 \ in, \ H = 9 \ in\end{align*}
13. \begin{align*}b = 10 \ m, \ h = 8 \ m, \ H = 9 \ m\end{align*}
14. \begin{align*}b = 12 \ m, \ h = 10 \ m, \ H = 13 \ m\end{align*}
15. \begin{align*}b = 8 \ cm, \ h = 6 \ cm, \ H = 9 \ cm\end{align*}
16. \begin{align*}b = 9 \ cm, \ h = 7 \ cm, \ H = 8 \ cm\end{align*}
17. \begin{align*}b = 5.5 \ mm, \ h = 4 \ mm, \ H = 4 \ mm\end{align*}
18. \begin{align*}b = 11 \ cm, \ h = 9 \ cm, \ H = 8 \ cm\end{align*}
19. \begin{align*}b = 20 \ ft, \ h = 17 \ ft, \ H = 19 \ ft.\end{align*}
20. \begin{align*}b = 22 \ ft, \ h = 19 \ ft, \ H = 17 \ ft.\end{align*}
Directions: Solve each real-world problem involving volume of prisms.
21. What is the volume of the box of cereal that measures \begin{align*}11 \ cm \times 5 \ cm \times 32 \ cm\end{align*}?
22. Kelly is using a rectangular container to fill up a bucket of water. The container is 3 inches long, 2 inches wide, and 7 inches tall. If the bucket holds 504 cubic inches of water, how many times will Kelly have to fill the rectangular container in order to fill the bucket?
23. The Berryville Aquarium has a shark tank in the shape of a triangular prism. There is only one shark in the tank, so right now the tank is only \begin{align*}\frac{2}{3}\end{align*} full. How many cubic feet of water are in the tank?
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# Proofs Theorems about Triangles Geometry
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1. Proofs and Parallel Lines Digital Unit compatible with Google Slides Lesson 3.1: Exploring Angles formed by Parallel Lines and their Transversals Lesson 3.2: Writing Equations to Solve for Unknown Angles Lesson 3.3: Basic Proofs Lesson 3.4: Prove Theorems about Lines and Angles Lesson 3.5: Prove T
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Proving Theorems about Triangles
• Materials: device with internet, google account
• Prerequisite Skills:
There are many prerequisite justifications for this lesson. they are listed in the lesson to refresh the mind, but not defined or explained. Definitions include: parallel lines, perpendicular lines, complementary angles, supplementary angles, congruent angles, right angles, vertical angles, linear pairs
• Interior and exterior angles of a triangle are explicitly defined with visuals.
Isosceles triangles are defined.
• Base Angles of an Isosceles Triangle Theorem is shown, but will be proven when students study congruent triangles. The theorem is used as a justification, as well as solving for variables in more traditional problems.
• The triangle sum theorem and exterior angles of a triangle theorem are proven as part of this lesson. These proofs are scaffolded to help develop skills for writing proofs. A proof using base angles of an isosceles triangle theorem follows these proofs.
• A quick lesson summary helps students reflect upon the new theorems learned.
• 5 Practice proofs follow the lesson and may be used as an assignment or formative assessment
• The 6th practice problem is a set of 4 figures that ask students to solve for the variable. Transferring the meaning of the new theorems will be used to solve these 4 problems.
• If you would not like your students to have the answers to grade as they go, simply delete those slides before sharing the file with your students.
• Note: Lesson format is compatible with Google slides®
• Answer keys included
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### Relations and Functions and Their Domains and Ranges
Relations and Functions
You learned that the most general type of link between two quantities is called a relation. In a relation each x value (domain value) maps to one or more y value (range value)
Example: y2 = x, In this case there are two values of y for every value of x since y is equal to plus or minus the squareroot of x. The requirement for a relation is that there be one or more values of y for every value of x. That requirement is fulfilled by the example.
In a function, as opposed to a relation each x value (domain value) maps to exactly one, and only one range value (y value). Note that all of the x values may map to the same value of y. The requirement is that each x maps to one y value. There is no requiement that each x map to a different value of y. y = 6x + 3 and y = x2 are examples of functions, each x maps to one and only one value of y.
In this session we are going to take a deeper look at functions and the notation that goes with them.
Functional Notation - Input and Output - An example of functional notation is d = g(t) = 16(t)2. In this case we are saying that d (distance traveled) in a function of t, here g(t) and the particular function is 16(t)2. If we want to evaluate distance traveled at a particular time, say 2 seconds, we could write d = g(t=2) = 16(2)2. That is the way we could write it, but we don't write it that way (we use a notational form that was first used by Leonhard Euler around 1734). The notation that is used expresses the idea by writting it as g(2) = 16(2)2. In other words the argument of g( ? ) is the value of t that we want to use and everwhere in the function we replace t with that value. So if we had h(x) = 2x2 + 6x and we wanted to determine the value of the function when x was equal to -4 we would write it as h(-4) = 2(-4)2 + 6(-4) = 2(16) - 24 = 32 - 24 = 8.
To illustrate the idea consider the following examples:
• m(t) = 2(t)3 - 6t2, when t = -1 we have m(-1)=2(-1)3 - 6(-1)2=-2 + 6 = 4
• w(s)= 2s2 - 4s, when s = a we have w(a) = 2(a)2 - 4(a) = 2a2 - 4a
• m(x) = 1/x + 3x, when x = 2x we have m(2x) = 1/(2x) + 3(2x) = 1/2x + 6x
• r(x) = 2x2 - 7x, when x = a- 2 we have
• r(a-2) = 2(a-2)2 - 7(a-2)
• r(a-2 )= 2(a2-4a +4) -7a + 14
• r(a-2)= 2a2-8a + 8- 7a+ 14
• r(a-2)= 2a2 -15a + 22.
Input and Output - When write a function, f(x) the x represents the input, when we have a formula and replace all the x's with the value of x the resulting output is the value of the function for that value of x. Let's look at input and output for three cases, (i) we have a formula, (ii) the information is in a table and (iii) the function is represented by a graph.
1. We have a formula
Define a function such as: f(x) = 3x2 + 5x - 7
1. Select an input value for x, in this case x = -4
2. Rewrite the function as: f(-4) = 3(-4)2 + 5(-4) - 7
3. Carry out the arithmetic to get: f(-4) = 3(16) - 20 - 7 = 48-27 = 21
4. When x = -4 (input) the value of the function is 21 (output)
2. The information is contained in a table such as the one shown below. The table is designed to provide data on the age of a human fetus (weeks since conception) as a function of the size of a body part (foot length). The function can be written as A(l) (age is a function of length). If we wanted the value of f(20) we would look for 20 in the foot length row and then look for the corresponding value of age, 14 weeks. Conversely, if we were given A(?) = 32, meaning the age is predicted to be 32 weeks but we want the foot length. In this case we would go to the age row to find 32 and then look for the corresponding value of foot length, 68 mm.
Foot lentgh (mm) 9 20 39 50 63 83 Age (weeks) 10 14 20 24 30 38
3. The function is described by a graph.
In the figure on the right two points have been selected (-4, -1) and (3, 2). We will use the first point to illustrate the treatment of input. f(-4) means the value of the function when x = -4. We go along the x axis to minus 4 and then drop a vertical line to the curve. Next we draw a horizontal line from the curve to the y axis. It itersects the axis at -1 so the value of the function when x = -4 [ f(-4) ] is -1 or f(-4) = -1. Now using the right most point if we are told that f(x) = 2 we go up the y axis (f(x) axis) until we come to 2, then horizontally until we hit the graph and finally vertically until we hit the x axis at x = 3. This says that when f(x) = 2 the value of x is 3. [Remember there could be more than one value of x that gives us this value of f(x)].
### Range and Domain of a Function
Domain of a function: The domain on a function consists of all of the values that the independent variable (usually X ) for which the function is defined.
For example
• f(x) = 1/x is defined for all values of x except x = 0 (you cannot divide by zero)
• f(x) = x2 is defined for all values of x.
Range of a function:The range of a function consists of all of the values of the function as the independent variable (x) varies over its domain.
Using the previous examples
• the range of f(x) = 1/x is from -infinity up to zero and from zero to plus infinity (does not include zero)
• the range of f(x) = x2 is from zero to plus infinity (the function is never negative)
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# Lines Parallel to the Same Line
In Geometry, a line is defined as a one-dimensional geometric figure that extends infinitely in both directions. A line has no thickness. A line is generally made up of an infinite number of points. In this article, we will discuss the lines parallel to the same line and the theorem related to it with many solved examples.
## What is Meant by Lines Parallel to the Same Line?
Theorem: Lines that are parallel to the same line are parallel to each other.
It means that if two lines are parallel to the same line, then they will be parallel to each other. Now, let us check this theorem with the help of the below figure.
From the given figure, we can say that line m is parallel to line l and line n is parallel to line l. (i.e) line m || line l and line n || line l. Also, “t” is the transversal for the lines l, m and n.
Therefore, we can say that ∠1 = ∠2 and ∠1 = ∠3. (By corresponding angles axioms).
So, we can also say that ∠2 and ∠3 are corresponding and they are equal to each other.
Thus, ∠2 = ∠3
By using the converse of corresponding angle axioms, we conclude that line m is parallel to line n.
(i.e) Line m || Line n.
Note: This property can be extended to more than two lines also.
### Solved Example
Example:
From the given figure, AB is parallel to CD and CD is parallel to EF.
Also, given that EA is perpendicular to AB. Find the values of x, y and z, if ∠BEF = 55°.
Solution:
Given that AB|| CD, CD || EF, EA ⊥ AB, and ∠BEF = 55°.
Therefore, y+ 55° = 180° (Interior angles on the same side of transversal ED)
Hence, y = 180°-55°= 125°
By using the corresponding angles axiom, AB || CD, we can say that x = y.
Therefore, the value of x= 125°.
Since, AB || CD and CD || EF, therefore AB || EF.
So, we can write: ∠ FEA + ∠ EAB = 180° (Interior angles on the same side of transversal EA)
55° + z + 90° = 180°
z = 180° – 90° – 55°
z = 180° – 145°
z = 35°
Therefore, the values of x, y and z are 125°, 125° and 35°, respectively.
### Practice Problems
Solve the following problems:
1. From the given figure, AB ||CD and CD || EF and y:z = 3:7. Find the value of x.
2. From the given figure, if PQ || RS and ∠MXQ = 135° and ∠MYR = 40°, then find the value of ∠XMY.
Stay tuned with BYJU’S – The Learning App and learn more class-wise concepts easily by exploring more videos.
## Frequently Asked Questions on Line Parallel to the Same Line
### What is meant by a line parallel to the same line?
The lines which are parallel to the same line are parallel to each other.
### Which symbol is used to represent the parallel line?
The symbol used to represent the parallel line is “||”.
### What is meant by parallel lines?
The parallel lines are the lines that are equidistant from each other and they never intersect each other.
### Mention three properties of angles associated with parallel lines.
The corresponding angles are equal
Alternate interior angles are equal
Alternate exterior angles are equal.
### How do you prove that the given lines are parallel lines?
To prove that the given lines are parallel, then we have to prove either corresponding angles are equal or alternate angles are equal or co-interior angles are supplementary.
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# 2005 AMC 10A Problems/Problem 23
## Problem
Let $AB$ be a diameter of a circle and let $C$ be a point on $AB$ with $2\cdot AC=BC$. Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?
$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3}$
## Solution 1
Let us assume that the diameter is of length $1$.
$AC$ is $\frac{1}{3}$ of diameter and $CO$ is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$.
$OD$ is the radius of the circle, so using the Pythagorean theorem height $CD$ of $\triangle AOC$ is $\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2} = \frac{\sqrt{2}}{3}$. This is also the height of the $\triangle ABD$.
Area of the $\triangle DCO$ is $\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}$ = $\frac{\sqrt{2}}{36}$.
The height of $\triangle DCE$ can be found using the area of $\triangle DCO$ and $DO$ as base.
Hence the height of $\triangle DCE$ is $\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}$ = $\frac{\sqrt{2}}{9}$.
The diameter is the base for both the triangles $\triangle DCE$ and $\triangle ABD$.
Hence, the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$ is $\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}$ = $\frac{1}{3} \Rightarrow C$
## Solution 2
Since $\triangle DCE$ and $\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$.
$[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\frac{1}{3}r$.
Since $m\angle DCO=m\angle DFC=90^\circ$, then $\triangle DCO\cong \triangle DFC$. So the ratio of the two altitudes is $\frac{CF}{DC}=\frac{OC}{DO}=\frac{1}{3}\Rightarrow \text{(C)}$
## Solution 3
Say the center of the circle is point $O$; Without loss of generality, assume $AC=2$, so $CB=4$ and the diameter and radius are $6$ and $3$, respectively. Therefore, $CO=1$, and $DO=3$. The area of $\triangle DCE$ can be expressed as $\frac{1}{2}(CD)(6)\text{sin}(CDE).$ $\frac{1}{2}(CD)(6)$ happens to be the area of $\triangle ABD$. Furthermore, $\text{sin} CDE = \frac{CO}{DO},$ or $\frac{1}{3}.$ Therefore, the ratio is $\frac{1}{3}.$
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# Lesson Plan: Solving Equations
Concept / Topic to Teach: Solving equations with variables on both sides and solving multi-
step equations.
General Goal(s): Students will be able to solve equations with multiple steps using distributive
property, combining like terms, and by adding, subtracting, multiplying, or dividing.
Specific Objectives:
o Students will be able to distribute in the equation.
o Students will be able to combine like terms.
o Students will be able to isolate the variable to one side of the equation by adding or
subtracting and by multiplying or dividing.
o Students will be able to check to see if the solution was correct for the equation.
Required Materials: The technology students will use is the PowerPoint to see the presentation
and learn the lesson. Students will be provided the acronym sheet Don’t Call Me After Midnight
to help them remember the steps. Students will need the Solving Multi-Step Equations
worksheet. Students will need a pencil or pen to complete the worksheet.
Anticipatory Set (Lead-In): This is a middle school pre-algebra math class which is in 8th
grade. In yesterday’s class we learned how to solve equations with two variables on both sides.
Today we will be learning how to solve multi-step equations.
## Step-By-Step Procedures: PowerPoint
a) Solving equations with variables on both sides. (Explain the problem on the PowerPoint.)
b) The acronym of Don’t Call Me After Midnight – Distribute, Combine like terms, Move
c) Solving multi-step equations. A multi-step equation is when there are more than two
steps to solve the equation.
d) Distribute. Distributive Property- Multiply times everything in the parentheses.
a ( b + c )= ab + ac OR a ( b – c ) = ab - ac
e) Combine like terms. Like Terms must have the same variables and the same powers on
the variables. Combine like terms by adding or subtracting the coefficients.
f) Add, Subtract, Multiply or Divide to be able to isolate the variable.
g) Check if the solution is correct by putting x into the equation to check if the solution is
correct.
Plan for Independent Practice: (5-10mins)
The Solving Multi-Step Equations worksheet.
Closure (Reflect Anticipatory Set): Briefly go over how to solve the equation using the
acronym, and ask if they have any questions on how to solve multi-step equations.
Assessment Based On Objectives: A quiz where students show their work on how they solved
multi-step equations using distributive property, combine like terms, and add, subtract, multiply,
or divide.
Possible Connections to Other Subjects: Students will need this subject to be able to solve
linear equations and later be able to graph the equations. Later in the semester and in the
following math classes they will take in high school they will be required to know the topic.
Special Needs Accommodations: For a student who is special needs, I can explain the subject in a more
visual way by putting squares or circles for them on the bottom of the equations so it can be easier for
them to solve the multi-step equations. Also, by making it a more hands on visual by using cubes or balls
and showing them how to solve the multi-step equation.
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# Find parametric equations for the path of a particle that moves along the circle
$x^2+(y-1)^2=4$
In the manner describe:
a) One around clockwise starting at $(2,1)$
b) Three times around counterclockwise starting at $(2,1)$
This question aims to understand the parametric equations and dependent and independent variables concepts.
A sort of equation that uses an independent variable named a parameter (t) and in which dependent variables are described as continuous functions of the parameter and are not dependent on another existent variable. When necessary More than one parameter can be used.
Given that a particle moves around the circle having equation is $x^2+(y-1)^2=4$.
Part a:
$x^2+(y-1)^2=4$ is the path of the circle in which the particle moves in the manner once around clockwise, starting at $(2,1)$
$x^2+(y-1)^2=4$
$\dfrac{x^2}{4}+\dfrac{(y-1)^2}{4}=1$
$\left(\dfrac{x}{2}\right)^2+\left(\dfrac{(y-1)}{2}\right)^2=1$
$\cos^2t + \sin^2t =1$ is the parametric equation of the circle.
As the circle is revolving once in the clockwise direction then the limit $t$ is $0 \leq t \leq 2\pi$
By comparing the two equations $\left(\dfrac{x}{2}\right)^2 +\left(\dfrac{(y-1)}{2}\right)^2 =1$and$\cos^2t +\sin^2t=1$.
$\dfrac{x}{2}=\cos t\space\space and \space\space\dfrac{y-1}{2}=\sin t$
$x=2\cos t\space\space and\space\space y-1=2\sin t$
$x=2\cos t \space\space and\space\space y=1+2\sin t \space\space \epsilon\space |0, 2\pi|$
Part b:
$x^2+(y-1)^2 =4$ is the path of the circle in which the particle moves in the manner three times around counter-clockwise, starting at $(2,1)$
$x^2+(y-1)^2=4$
The circle has a radius of $2$ and the center is at $(0,1)$.
As the circle is revolving thrice, the $t$ is less than equal to $3(2\pi)$ that is, $0\leq t\leq 6\pi$
By comparing the two equations $\left(\dfrac{x}{2}\right)^2+\left(\dfrac{(y-1)}{2}\right)^2=1$ and $\cos^2t+\sin^2t=1$.
$\dfrac{x}{2}=\cos t\space\space and \space\space\dfrac{y-1}{2} =\sin t$
$x =2\cos t\space\space and \space \space y-1= 2\sin t$
$x =2\cos t\space\space and \space \space y=1+2\sin t \space\space\epsilon\space |0, 6\pi|$
Part a: $x = 2\cos t \space \space and \space \space y = 1+2\sin t \space \space \epsilon \space |0, 2\pi|$
Part b: $x = 2\cos t \space \space and \space \space y = 1+2\sin t \space \space \epsilon \space |0, 6\pi|$
## Example
A particle moves along the circle. Find its parametric equation for the path in the manner halfway around counterclockwise starting at $(0,3)$.
$x^2 + (y-1)^2 =4$ is the path of the circle in which the particle moves in the manner halfway around counter-clockwise, starting at $(0,3)$.
$x^2 + (y-1)^2 =4$
point $(0,3)$ lies on the y-axis.
$\dfrac{x^2}{4} + \dfrac{(y-1)^2}{4} =1$
$\left( \dfrac{x}{2} \right)^2 + \left( \dfrac{(y-1)}{2} \right)^2 =1$
$\cos^2t + \sin^2t =1$ is the parametric equation of the circle.
As the circle is revolving in halfway around the counterclockwise direction, the limit $t$ is $\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2} + \pi$
That is: $\dfrac{\pi}{2}\leq t \leq \dfrac{3\pi}{2}$
By comparing the two equations $\left( \dfrac{x}{2} \right)^2 + \left( \dfrac{(y-1)}{2} \right)^2 =1$ and $\cos^2t + \sin^2t =1$.
$\dfrac{x}{2} = \cos t \space \space and \space \space \dfrac{y-1}{2} = \sin t$
$x = 2\cos t \space \space and \space \space y-1 = 2\sin t$
$x = 2\cos t \space \space and \space \space y = 1+2\sin t \space \space \epsilon \space |\dfrac{\pi}{2}, \dfrac{3 \pi}{2}|$
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# Math VariablesProperties of Equality
In Math, variables can appear in many formulas and expressions.
They are usually letters of the alphabet, for example a , b , x , y.
As the name variable suggests, the value of these letters can vary.
Variables can be assigned any value or set of values, the value or values aren’t known initially.
In the expression x + 3 = 5, the variable is x.
Even though for an equation such as this one, the value of x can be seen to be only 2.
It is still referred to as a variable.
## Math VariablesDependent/Independent
A variable can be dependent or independent.
With y = 2x + 6,
x is an independent variable, it could be assigned a range of values.
However, y is a dependent variable.
As the value of y depends on what value is given to x.
The value of y is dependent on the value of x.
## Manipulating Expressions/Equations with VariablesProperties of Equality
The Polynomials section and the Linear Equations section go into a bit more detail about manipulating expressions and equations to solve for variables.
But this page will give a basic run through of some examples of how to approach finding the value of variables.
Using some of the properties of equality.
Say you have a variable in a basic equation labelled x,
to solve for this x,
we would want to structure the equation into the form "x =".
Examples of such situations using some of the properties of equality are shown below.
Examples
(1.1)
If a = b then a + c = b + c .
1) x + 3 = 7 , can subtract 3 from both sides to have the form "x =".
x + 33 = 73 => x = 4
2) x1 = 5 , can add 1 to both sides.
x1 + 1 = 5 + 1 => x = 6
(1.2)
If a = b then a × c = b × c .
\\boldsymbol{\\frac{x}{2}} = 8 , can multiply both sides by 2.
\\boldsymbol{\\frac{x}{2}} × 2 = 8 × 2 => x = 16
(1.3)
If a = b then \\boldsymbol{\\frac{\\mathbf{a}}{c}} = \\boldsymbol{\\frac{b}{c}} .
1) 4x = 12 , can divide both sides by 4.
\\boldsymbol{\\frac{4x}{4}} = \\boldsymbol{\\frac{12}{4}} => x = 3
2) -2x = 4 , can divide both sides by -2.
\\boldsymbol{\\frac{-2x}{-2}} = \\boldsymbol{\\frac{4}{-2}} => x = -2
(1.4)
If a = b then a − b = 0.
3x6 = 0 , can move 6 across, then divide.
3x = 6 => \\boldsymbol{\\frac{3x}{3}} = \\boldsymbol{\\frac{6}{3}} => x = 2
(1.5)
2x = 3 + x , can move x across, to have the like terms on the same side.
2xx = 3 => x = 3
1. Home
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# Question #b5882
Apr 10, 2017
$x = - \frac{22}{13}$ and $y = - \frac{29}{13}$
#### Explanation:
Ill do the first one as an example,
$x - 3 y = 5$ and $3 x + 7 y = 1$
The goal is to eliminate one of the variables so it is possible to solve for the other,
$x - 3 y = 5$ equation 1
$3 x + 7 y = 1$ equation 2
Multiply the first equation by $3$, and subtract the first equation from the second equation.
$3 x - 6 y = 30$
$3 x + 7 y = 1$
Subtracting the equations gives you,
$- 13 y = 29$
$y = - \frac{29}{13}$
Now knowing $y = - \frac{29}{13}$, you can solve for $x$. Do so by subsituting the value for $y$ into equation 1.
$x - 3 \left(- \frac{29}{13}\right) = 5$
$x + \frac{87}{13} = 5$
$x = 5 - \frac{87}{13}$
$x = - \frac{22}{13}$
Therefore,
$x = - \frac{22}{13}$ and $y = - \frac{29}{13}$
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# Composite numbers
Contents
## Introduction
A composite number is one that may be divided both by another number than 1 and by the number itself. Composite numbers have more than two contributing factors. Depending on how many factors they contain, numbers may be categorised. A number is a prime number if it only contains two factors: one and the number itself. The majority of numbers, known as composite numbers, include more than two components.
## Types of composite numbers
There are two main types of composite numbers:
● Odd composite numbers: All odd numbers that are not classified as prime numbers are odd composite numbers. The first few of them are 9,15,21,25,27 etc.
● Even composite numbers: All even numbers except for 2 are composite numbers, because 2 is the only even prime number and always is a factor to the rest of the even numbers. The first few of them are 4,6,8,10,12,14 etc.
## Finding composite numbers
The steps to determine if a particular number is prime or composite are as follows:
● Find all positive integer’s factors.
● If a number only contains the number one and itself as components, it is said to be prime.
● A composite number is one with more than two factors.
## Properties of composite numbers
● There are more than two components in composite numbers.
● By their factors, composite numbers can be divided equally.
● Every composite number has itself as a factor.
● 4, which is the lowest composite number.
● For example,, where 3 and 5 are prime numbers, each composite number will have at least two prime numbers as its components.
● Additionally, composite numbers can be divided by other composite numbers.
## Is 0 considered a composite number?
Any number may be written as the product of two prime numbers, according to the fundamental principle of arithmetic. It wouldn’t work if 0 were a prime number since any integer multiplied by 0 would result in 0, giving it more than 2 factors. It cannot be written as a product of two primes if it were deemed composite since doing so would require multiplying by 0, which we have assumed to be the case.
## List of common composite numbers
The list of composite numbers from 1 to 100 is 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100.
## Prime factorization of composite numbers
Here is a list of all the composite numbers from 1 to 50, along with a breakdown of their prime factors. Here, you can see the prime factorization of the composites. To better understand, look at the table below. This table allows you to locate composites with prime factorizations that are more than 50.
## Are all composite numbers even?
No not all composite numbers are even, the numbers that may be stated as the product of two or more prime numbers are known as composite numbers.
Since we are assuming that all even numbers are composite numbers, 2 cannot be written as the sum of two or more prime numbers.
All the odd numbers that have factors of odd prime numbers are composite numbers.
## Applications of composite numbers
● In mathematics, the usage of composite numbers, or any whole integer that is not prime, is common.
● In all areas of mathematics, the usage of composite numbers is utilised to explain shape, measure, determine how much of an object there is, and do arithmetic.
● Numbers are used often in daily life, such as when purchasing or selling anything, paying for something, finding out the time, counting the number of runs the Indian cricket team scored during a match, counting the number of pupils seated in a classroom, calculating a student’s grades, etc.
● A key to the code is required to decipher the encrypted data. Prime and composite integers are used in one of the most widely used encryption techniques.
● The composite numbers, which are typically quite big, are used as codes. It is difficult to divide the composite number into its two prime elements since the number of prime numbers is unknown.
## Conclusion
In this article we learnt about composite numbers and their different types. We learnt how to identify composite numbers and its properties. We also learnt about the differences between prime numbers and composite numbers, and we saw how prime factorization works and the factorization for composite numbers till 50. Composite numbers are a category of numbers we use in everyday life scenarios and are a major type of numbers.
## Sample examples
Example 1: Find if 456 is a composite number
Solution 1:
The factors to 456 are: 1, 2, 3, 4, 6, 8, 12, 19, 24, 38, 57, 76, 114, 152, 228, 456
Since the integer has factors other than 1 and itself it is a composite number.
Example 2: What is the prime factorization of 78?
Solution 2:
The prime factors of 78 is .
Example 3: Find if the product of the first 5 prime numbers is composite.
Solution 3:
The first 5 prime numbers are:2,3,5,7,11.
The product is , and we can see the factors include other factors apart from 1 and the number itself, therefore, the product is composite.
Example 4: Identify the composite numbers from the list given, 2,4,56,57,89,768,404,77.
Solution 4:
2,57,89 are prime numbers.
4,56,768,404,77 are composite numbers with factors 1, 2, 4; 1, 2, 4, 7, 8, 14, 28, 56; 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 768; 1, 2, 4, 101, 202, 404; 1, 7, 11, 77; respectively.
Example 5: What is the largest composite number between 1 and 100 with only odd numbers in its prime factorization.
Solution 5:
The largest composite number with only odd numbers in the factorization is 99 with the factorization as .
## FAQs
1. What are consecutive composite numbers?
Ans: Composite numbers that follow one another in a straight line without a prime number in between are said to be consecutive composite numbers. The initial few consecutive composite numbers, for instance, may be presented as 4, 6, 8, 9, etc.
2. Does 1 classify as a composite number?
Ans: No, because 1 only has one component and does not include more than two. It is therefore neither prime nor composite.
3. What is the smallest composite number?
Ans: Other than 1 and the number itself, a composite number consists of more than two components. In this case, the number 4 meets this requirement, and no other smaller number is composite. The lowest composite number is 4, which follows the requirements.
4. Can a number be both composite and prime?
Ans: A number cannot be prime and composite at the same time. A composite number has more elements than a prime number, which only has 1 and itself as its factors. All natural numbers are either prime or composite, but not both, with the exception of 1 which is neither.
5. What is the only even number that is not composite?
Ans: 2 is the only even prime number since it has only 1 and itself as its factors.
## References
Pradhan, J. B. (2019). Conceptual metaphor for teaching and learning of prime and composite numbers at primary grades. The Eurasia Proceedings of Educational and Social Sciences, 14, 78-88.
Robbins, C., & Adams, T. L. (2007). Mathematical Roots: Get Primed to the Basic Building Blocks of Numbers. MatheMatics teaching in the Middle school, 13(2), 122-127.
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# How do you find the quotient of (3x^2 + 4xy + 3xy + 4y^2) ÷ (x + y)?
Mar 20, 2017
The remainder is $= 0$ and the quotient is color(red)(=3x+4y
#### Explanation:
$\textcolor{w h i t e}{a a a a a a a}$color(red)(3x+4y
$\textcolor{w h i t e}{a a a a a a}$$- - - - -$
$\textcolor{w h i t e}{a a a a a}$$|$$3 {x}^{2} + 7 x y + 4 {y}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$$3 {x}^{2} + 3 x y$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$0 + 4 x y + 4 {y}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$4 x y + 4 {y}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$- - -$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$$0 + 0$
The remainder is $= 0$ and the quotient is color(red)(=3x+4y
Mar 20, 2017
$\text{ The Quotient=} \left(3 x + 4 y\right) .$
#### Explanation:
$\underline{3 {x}^{2} + 4 x y} + \underline{3 x y + 4 {y}^{2}} ,$
$= x \left(3 x + 4 y\right) + y \left(3 x + 4 y\right) ,$
$= \left(3 x + 4 y\right) \left(x + y\right) .$
$\therefore \left(3 {x}^{2} + 4 x y + 3 x y + 4 {y}^{2}\right) \div \left(x + y\right) ,$
$= \frac{3 {x}^{2} + 4 x y + 3 x y + 4 {y}^{2}}{x + y} ,$
$= \frac{\left(3 x + 4 y\right) \left(\cancel{x + y}\right)}{\cancel{x + y}} ,$
$= \left(3 x + 4 y\right) , \text{ is the desired quotient.}$
Enjoy Maths.!
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# Represent an Inequality as an Interval on a Number Line
Learning Outcomes
1. Graph and inequality on a number line.
2. Graph the complement on a number line for both continuous and discrete variables.
Inequalities come up frequently in statistics and it is often helpful to plot the inequality on the number line in order to visualize the inequality. This helps both for inequalities that involve real numbers and for inequalities that refer to just integer values. As an extension of this idea, we often want to look at the complement of an inequality, that is all numbers that make the inequality false. In this section we will look at examples that accomplish this task.
## Sketching an Inequality on a number line where the possible values are real numbers.
There are four different inequalities: $$<,\:\le,\:>,\:\ge$$. What makes this the most challenging is when they are expressed in words. Here are some of the words that are used for each:
• $$<$$: "Less Than", "Smaller", "Lower", "Younger"
• $$\le$$: "Less Than or Equal to", "At Most", "No More Than", "Not to Exceed"
• $$>$$: "Greater Than", "Larger", "Higher", "Bigger", "Older", "More Than"
• $$\ge$$: "Greater Than or Equal to", "At Least", "No Less than"
These are the most common words that correspond to the inequalities, but there are others that come up less frequently.
Example $$\PageIndex{1}$$
Graph the inequality: $$3<x\le5$$ on a number line
Solution
First notice that the interval does not include the number 3, but does include the number 5. We can represent not including a number with an open circle and including a number with a closed circle. The number line representation of the inequality is shown below.
Example $$\PageIndex{2}\ In statistics, we often want to find probabilities of an event being at least as large or no more than a given value. It helps to first plot the interval on a number line. Suppose you want to find the probability that you will have to wait in line for at least 4minutes. Sketch this inequality on a number line. Solution First, notice that "At Least" has the symbol \(\ge$$. Thus, we have a closed circle on the number 4. There is no upper bound, so we draw a long arrow from 4 to the right of 4. The solution is shown below
Example $$\PageIndex{3}$$
Another main topic that comes up in statistics is confidence intervals. For example in recent poll to see the percent of Americans who think that Congress is doing a good job found that a 95% confidence interval had lower bound of 0.18 and an upper bound of 0.24. This can be written as [0.18,0,24]. Sketch this interval on the number line.
Solution
The first thing we need to do is decide on the tick marks to put on the number line. If we counted by 1's, then the interval of interest would be too small to stand out. Instead we will count by 0.1's. The number line is shown below.
Example $$\PageIndex{4}$$
Often in statistics, we deal with discrete variables. Most of the time this will mean that only whole number values can occur. For example, you want to find out the probability that a college student is taking at most three classes. Graph this on a number line.
Solution
First note that the outcomes can only be whole numbers. Second, note that "at most" means $$\le$$. Thus the possible outcomes are: 0, 1, 2, and 3. The number line below displays these outcomes.
## Graphing the Complement
In statistics, we often want to graph the complement of an interval. The complement means everything that is not in the interval.
Example $$\PageIndex{5}$$
Graph the complement of the interval [2,4).
Solution
Notice that the complement of numbers inside the interval between 2 and 4 is the numbers outside that interval. This will consist of the numbers to the left of 2 and to the right of 4. Since the number 2 is included in the original interval, it will not be included in the complement. Since the number 4 is not included in the original interval, it will be included in the complement. The complement is shown on the number line below.
Example $$\PageIndex{6}$$
Some calculators can only find probabilities for values less than a certain number. If we want the probability of an interval greater than a number, we need to use the complement. Suppose that you want to find the probability that a person will have traveled to more than two foreign countries in the last twelve months. Find the complement of this and graph it on a number line.
Solution:
First notice that only whole numbers are possible since it does not make sense to go to a fractional number of countries. Second note that the lowest number that is more than 2 is 3. If 3 is included in the original list, then 3 will not be included in the complement. Thus, the highest number that is in the complement of "more than 2" is 2. The number line below shows the complement of more than 2.
Exercise
Suppose you want to find the probability that at least 4 people in your class have a last name that contains the letter "W". To make this calculation you will need to first find the complement of "at least 4". Sketch this complement on the number line.
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# Calculating with constant acceleration
#### Prerequisites
If we have a constant acceleration — the velocity is either increasing or decreasing at a uniform rate — then we can figure out a lot of useful stuff. For this section, we'll assume that we are moving along a line. We could be traveling back and forth, but we won't consider changes of the angle of our motion. So we will drop the $\hat{i}$ or $\hat{j}$ that picks out which line we are traveling along and work only with one coordinate. For convenience here, we will call it $x$.
## The velocity
If the acceleration is constant, then the velocity is changing at a constant rate. Our equation defining the acceleration is then
$$a_0 = \frac{\Delta v}{\Delta t}$$
or
$$a_0 = \frac{v_2 - v_1}{\Delta t}$$
where $v_1$ is our velocity at the beginning of the time interval (initial velocity) and $v_2$ is our velocity at the end of the time interval (final velocity).
If we solve for the final velocity we get
$$v_2 = v_1 + a_0 (t_2 - t_1)$$
Since we have selected particular times, everything in this equation is a constant. But we can be choose one or more of them to turn back into a variable. In particular, let's call $t_2$, "$t$", and treat it like an independent variable. Then we can see what the velocity looks like as a function of time. The "final velocity", since it corresponds to the velocity at time $t_2$ will become a dependent variable, $v$.
If this seems a little weird to you, get used to it. In physics, since we tend to describe both general and specific situations, we will often let a constant "wander around", becoming a variable. And we will often fix a variable at a specific value in order to have an equation apply to a specific situation.
This gives us the equation for $v$ (which holds between time t1 and some later time t that is unspecified)
$$v(t) = v_t + a_0 (t - t_1)$$
If we plot this velocity as a function of $t$, we get a figure that looks like the one at the left in the figures below. If we want to figure out what the average velocity in that time interval is, we have to find the constant line so that the areas under the true velocity curve equals the area under the average velocity line (a constant). We do this by adjusting the average velocity line so that the light pink area (the area no longer included) and the light blue area (the extra area now included) are equal.
It's pretty clear both from thinking about how the velocity changes and from looking at the graph, that the average velocity is going to be halfway between the endpoints; that is, if a is constant, then
$$\langle v \rangle = \frac{v_1 + v_2}{2}$$
Since we know that the definition of the average velocity is "that velocity which, if you go constantly with it, will produce the same displacement", we can see that if the acceleration is constant:
$$\Delta v = a \Delta t$$
$$\Delta x = \langle v \rangle \Delta t$$
$$\Delta v = v_2 - v_1$$
$$\langle v \rangle = \frac{v_1 + v_2}{2}$$
These 4 equations each have a clear and straightforward conceptual meaning and let you calculate whatever you need from whatever you know when the acceleration is constant. The tricky part sometimes, if remembering which equation works for $\Delta v$ and which for $\langle v \rangle$. It's easy to keep it straight if you keep the meaning of "change" and "average" in mind and think about what they would each be if the velocity were constant.
Joe Redish 9/10/11
Article 332
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## SEMICIRCLE
In this page semicircle we are going to see how to find area and perimeter (circumference) of a semi-circle. To understand this topic much better we have given different kinds of example problems.
Semicircle is exactly half the entire circle.
Area of Semi-circle = (1/2) Π r²
Perimeter of semi circle = (Π + 2)r
Here r represents the radius of the circle. Now let us see example problems based on the above formula.
Example 1:
Find the area of the semi-circle whose radius is 7 cm.
Solution :
Area of Semi-circle = (1/2) Π r²
Here r= 7 cm and Π = 22/7
= (1/2) x (22/7) x 7²
= (1/2) x (22/7) x 7 x 7
= 1 x 11 x 7
= 77 cm²
Example 2 :
Find the area of the semi-circle whose radius is 3.5 cm.
Solution :
Area of Semi-circle = (1/2) Π r²
Here r = 3.5 cm and Π = 22/7
= (1/2) x (22/7) x (3.5)²
= (1/2) x (22/7) x 3.5 x 3.5
= 1 x 11 x 0.5 x 3.5
= 19.25 cm²
Example 3 :
Find the circumference of the semi-circle whose diameter is 7 cm.
Solution :
r = diameter/2 ==> r = 7/2 ==> r = 3.5
Now we can apply the formula
Circumference of semi-circle = (Π + 2)r
here r = 3.5 and Π = 22/7
= [(22/7) + 2]x 3.5
= (22+14) x 0.5
= 36 x 0.5
= 18 cm
Example 4 :
Find the circumference of the semi-circle whose diameter is 42 cm.
Solution :
r = diameter/2 ==> r = 42/2 ==> r = 21
Now we can apply the formula
Circumference of semi-circle = Πr
here r = 21 and Π = 22/7
= (22/7) x 21
= 22 x 3
= 66 cm
Example 5 :
Find the circumference of the semi-circle whose diameter is 56 cm.
Solution :
r = diameter/2 ==> r = 56/2 ==> r = 28
Now we can apply the formula
Circumference of semi-circle = Πr
here r = 28 and Π = 22/7
= (22/7) x 28
= 22 x 4
= 88 cm
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## Adding Rational Expressions Like Denominators
For this lesson, we're going to try our hand at adding rational expressions. We're going to draw upon our knowledge of adding fractions to figure that out. For example, if we wanted to add 3 11ths to 2 11ths, we would just add the numerators and keep the denominators the same. We would get a total of 5 11ths. Keep in mind we can only add or subtract two fractions whenever the denominators are like. Keeping this in mind, I want you to try and add these rational expressions together. This last one isn't addition though, it's subtraction. So you want to make sure you subtract the numerator. You'll also need to use your knowledge of simplifying rational expressions for these two problems. Try your best, and don't worry if they're not right on the first try.
## Adding Rational Expressions Like Denominators
Here are the correct answers. Fantastic work if you got at least one of those correct. If either of these problems are giving you trouble, then watch the solution and we'll figure it out together. To add these two fractions together, we simply add the numerators. 3 n and 2 n equal 5 n. We leave the denominator alone, so we have 5 n divided by 11. This problem is exactly like the one we saw before. It's just we needed to combine the like terms and our numerator first. For the second problem, we want to add the numerators a and b together, and keep the denominator the same. Since we have a plus b divided by a plus b, we know those reduce to 1. And finally, for this third one, we want to subtract a y from this numerator. So we'll have 2 y minus 1 minus 1 y, we subtract the like terms to get y minus 1 for our numerator. Now we want to factor our denominator to see if it can reduce with our numerator since our numerator has changed and it turns out that's a great move because we can factor and reduce and equivalence of 1. Y minus 1 divided by y minus 1 equals 1. This leaves us with the answer of 1 divided by y plus 2. If you got the first two, excellent work, and I know this one might of given you some trouble.
## Adding Fractions Unlike Denominators
But what if we wanted to add these 2 fractions together? What do you think we could do? Well, adding these 2 rational expressions together is very similar to adding 2 fractions together. If the denominators are different, we need to change each fraction. By finding the lowest common denominator first. We know the lowest common denominator would be 90, the product of these factors. We multiply 1 10th by nine over nine, which is the missing factor from ten. 1 10th by 9 9ths is 9 90ths. For 1 18th we multiply by five over five. This is the missing factor for 18. So 1 18th time 5 5ths is 5 90ths. Now we add these 2 fractions just as we learned before. 9 plus 5 equals 14 and we have 14 90ths. These numbers are both even and so we can reduce them by 2. So these are a couple of fractions.
## The LCD of Rational Expressions with Unlike Denominators
Now, that we've remembered how to add two fractions together, I want you to apply this method to figure out the LCD for these two fractions. What's the lowest common denominator?
## The LCD of Rational Expressions with Unlike Denominators
Two K S squared is the lowest common denominator, great work if you found it. If you're wondering why that's true, it's because these two numbers share a common factor of two. Remember these variables are the placeholders of numbers so two times K would be a number and two time S squared would really be a number as well. We don't know what K is, but it's the other factor for this denominator. And the same is true for s squared. It's the other factor for 2 s squared. We multiply all these together to get 2 k s squared. Now I know I did a little bit of reordering in this situation, but we usually list the coefficient or number first. Then we list the variables in alphabetical order.
## Creating Equivalent Fractions with the LCD
Now that we know that the LCD is 2ks squared. What should we multiply each of these fractions by in order to get new fractions with denominators of the LCD? Write your answers in these boxes.
## Creating Equivalent Fractions with the LCD
Well, we should multiply the first fraction by s squared divided by s squared.This would give us s squared over 2ks squared. I know I multiply this first fraction by s squared divided by s squared. S squared is missing from the new denominator or our lowest common denominator. For the second fraction, we only need to multiply by k over k. 2s squared times k would give us our lowest common denominator. 2k s squared. Keep in mind that we're really just multiplying by a form of 1 in both of these cases. We're not changing the value of the fractions, we're just changing what they look like.
## Add Rational Expressions Like Denominators
So, just like when adding fractions, we start out by finding the LCD or the lowest common denominator. Next, we multiplied each of these fractions by the missing factors of the lowest common denominator. That was the same for our actual numbers. Now, we're ready for the last step. Let's find the actual fractions that are equivalent and then find the sum. So, what are these two equivalent fractions? And then, what I want you to do is to find the sum of the two fractions and put that here. Good luck.
## Add Rational Expressions Like Denominators
For the first fraction we'll have s squared for a numerator and 2 k s squared for the denominator. For the second fraction we'll have k for a numerator and 2 k s squared for the denominator. And now we're ready to add these two fractions together. We add the numerators together, just like from before and then keep the denominator the same. Be careful that you don't multiply these two numbers together. We really just have s squared plus k. We don't multiply them, since we're adding.
## The LCD and Adding Rational Expressions
So whenever we want to add rational expressions together we should find the lowest common denominator of our fractions. The lowest common denominator is the product of all of the different factors that appear in each denominator. Keep in mind that we want to take the highest power of any factor that appears in a denominator. We're not going to jump into the entire process yet. Let's just get comfortable with this first step of finding the lowest common denominator.
## Finding the LCD 1
What LCD or lowest common denominator would we need in order to add these two fractions together? Now, we aren't going to actually add the two fractions together since we're just getting comfortable with the first step. What do you think the LCD would be? Write it here.
## Finding the LCD 1
If we had a binomial or trinomial in these denominators, we would want to factor them first. But since these are monomials, we can use our factor tree method to figure out what they share. I know 9 and 15 share a common factor of 3, and the a's share a common factor of 1a. The highest power of b that's in common is b squared. So 3ab squared is my greatest common factor. The other factor for this denominator is 3a squared. If we multiply these two together, we'll get our original denominator. Whereas for this fraction, the missing factor is 5b squared, if I multiply these two together we'll get 15, 3 times 5, a, and then b squared times b squared or b to the fourth. We multiply all these factors together to get our LCD. So our lowest common denominator is 3 times 3 times 5, which is 45 and then a squared time a which is a cubed, and b squared times b squared which b to the fourth. I said earlier that you want to use the highest power of any factor for the lowest common denominator. Notice that we have a cubed in a. A cubed is the highest power in either denominator. So we want to make sure that appears in our LCD. Likewise between b squared and b to the fourth, this fourth power is higher, so we know b to the fourth needs to appear in our LCD. This is a quick way you can check to make sure your variables or your factors are correct.
## Finding the LCD 2
Let's try another problem where we find the lowest common denominator. You'll need to factor these denominators first, since they're not just monomials. As a hint, your LCD will contain a constant and two binomials. And you'll want to make sure you leave it in factored form. I know you might not have seen this before, but I want you to give it a try. If you get stuck, there's always this solution.
## Finding the LCD 2
The LCD is 2 times u minus 3 times u plus 3. Fantastic work if you figured that one out. Again, I don't expect you to get that right on your first try, or even maybe at all. Let's see how we do it. We know u squared minus 9 is a difference of two perfect squares. So we can factor this as u plus 3 times u minus 3. So we can rewrite this first fraction like this. For the second denominator we have a common factor of 2. So we can factor that out. Now that we have factored denominators, we can use our factor tree method. These two denominators share a common factor of u minus 3, and the other factor for this one is u plus 3, while the other factor for this one is positive 2. And there it is. We multiply all these together to get 2 times u plus 3 times u minus 3, our LCD.
## Finding the LCD 3
Alright, here's our third and last example of trying to find the LCD. Remember we're not quite adding these two rational expressions together yet. We just want to get a common denominator since these two denominators are different.
## Finding the LCD 3
The lowest common denominator for these three fractions would be p minus 1 times p minus 2 times p plus 3, fantastic work for finding that one. First we factor this denominator to p minus 2 times p minus 1, we find factors of 2 that sum to negative 3. For our second denominator, we find factors of negative 6 that sum to positive 2, which are positive 3 and negative 2. Once our denominators have been factored, we're ready to use our factor tree. So we have p minus 1 times p minus 2 times p plus 3. This is our lowest common denominator.
## Equivalent Fractions Rational Expressions
So why do we need to find the lowest common denominator? Well remember that we have to find a common denominator in order to add or subtract fractions. In our very first example, we needed to have 90th in order to add these two fractions together. In the second example, we needed to have a common denominator of 2ks squared in order to add these two fractions together. Once we have a lowest common denominator, we multiply the numerator and the denominator in each fraction by the factors that are missing to change our fraction into an equivalent form. So for 18, it was missing a factor of five. So we multiplied by fraction by k over k. Since k is the missing factor. From 2s squared. When we multiply by k divided by k, we're really multiplying by 1, and we're changing our first fraction into its equivalent form. So let's go back to our third example of finding the LCD. We found the lowest common denominator for these two fractions, and it was p minus 1, times p minus 2, times p plus 3. Remember we want to change these two fractions to an equivalent form. So here's your question. What should we multiply each of these fractions by to get an equivalent fraction with a lowest common denominator of this?
## Equivalent Fractions Rational Expressions
We multiply this first fraction by p plus 3 divided by p plus 3. The denominator here is missing a factor of p plus 3. This is the one factor we need in order to change our denominator into the LCD. For our second fraction we multiply by p minus 1 divided by p minus 1. And this denominator, p minus 1 is the missing factor we need in order to get the LCD. Now we multiply these two numerators together and these two denominators together this gives us our new first equivalent fraction. We do the same process for the second fraction to get this, these two denominators may look different but they really the same. We know the order of the factors don't matter because its multiplication. Now that the two fractions have like denominators, we can simply add the numerators together. We add the like terms. The p's and the constants. 1 p and 1 p would equal 2 p and positive 3 and negative 1 would equal positive 2. Now that we're here, we should ask ourselves, are we done? Well, we added two things in a numerator together, so we have a new numerator entirely. We want to see if we can factor it if possible, and it turns out we can. We can take out a 2, to have 2 times p plus another factor of p plus 1 does not appear in the denominator. So the answer is yes, we're done, and here's the addition of these two rational expressions. Now this is some algebra.
## Simplify the Sum
So here's our guide, our steps that we could follow to add rational expressions together. First, we want to factor the denominators and find the LCD. Next, we multiply each fraction by the missing factors of the LCD to get equivalent fractions. Finally, we add the numerators together and then we check if we can simplify this by cancelling any factors. And again, don't make the mistake of canceling terms. We can't cancel these since this is addition in the numerator. We can only cancel factors.
## Add Rational Expressions Practice 1
Let's try adding these two rational expressions together an see if we can get the correct answer. We're not going to do all of the steps at once instead we're going to break it down into three pieces let's start with the first one. What do you think would be the lowest common denominator for these two fractions? Write your answer here.
## Add Rational Expressions Practice 1
We find the lowest common denominator by factoring each of these denominators first. The first denominator will have factors of x plus 2 times x minus 6, since factors of 12 that sum to negative 4 are positive 2 and negative 6. The factors for x squared minus 36 would be x minus 6 and x plus 6. This is just the difference of two perfect squares. We can see that both of the denominators share a common factor of x minus 6. So this is the factor they share, and we list it in the middle. The other factor of this denominator is x plus 2, and the other factor of this denominator is x plus 6. Finally, we just multiply all these factors together to get our lowest common denominator, right here.
## Add Rational Expressions Equivalent Fractions 1
We've completed step 1 and we found the lowest common denominator for these two fractions. Now we're ready to make equivalent fractions for each of these. What do we need to multiply this fraction by and this fraction by to get our new equivalent fractions? You can write the factors in these boxes. Good luck..
## Add Rational Expressions Equivalent Fractions 1
We need multiply the first fraction by x plus 6 divided by x plus 6. We know that because this denominator is missing a factor of x plus 6. If we multiply the numerator and the denominator by x plus 6, we're multiplying by a form of 1, and we changed this denominator into our lowest common denominator. For the second fraction we need to multiply by x plus 2 divided by x plus 2. This denominator is missing our factor of x plus 2 to form the LCD. If you got these correct, excellent thinking.
## Add Rational Expressions Multiply 1
Now that we're changing each of these 2 fractions, I want you to multiply these numerators together and write your answer here. For the denominators, we don't want to multiply them out. We just want to list the factors. This is because we're going to add the numerators in the end. We'll refactor that if possible and see if we can cancel any factors in the numerator and denominator of our final answer.
## Add Rational Expressions Equivalent Multiply 1
Six x plus 36, and five x plus ten, are our new numerators. Great work if you found those two. We know we're really just distributing the six to the x and the six. So six times x is six-x. And six times six is 36. For this new numerator, we're distributing the five times x, which is five-x. And 5 times 2 which is 10. So, 5x plus 10 is our new numerator.
## Add Rational Expressions Final Sum 1
First we found the LCD and then we changed each fraction by multiplying by the missing factors of the LCD. Now that we've converted this fraction and this fraction into new fractions that have common denominators. What would the sum be? Write the numerator here and the denominator here.
## Add Rational Expressions Final Sum 1
we would get 11X plus 46 divided by the factors x plus 2 times x minus 6 times X plus 6. Fantastic work if you just added these numerators together. When we add the numerators we just want to add the like terms, like 6X and 5X Which would equal 11x. Likewise we would add the constant terms. Positive 36 and positive 10 equal positive 46. So here's our numerator and here's our denominator.
## Simplify Further
Well, the answer would be no. When we look at 11x and positive 46, we can think about the greatest common factor between them. That greatest common factor would just be 1. Because there's no other factor of 11x plus 46 in our denominator, we can't cancel this factor with anything down here. This is another reason why we want to leave the denominator in factored form. At the end, we should check to see if this is factorable and cancel any factors that appear in the numerator and denominator.
## Simplify a Different Sum
But what if our sum, instead, was 11 x plus 22? What do you think our final answer would be this time? Now, I know this wasn't the sum for our last problem, but I want you to see if there's an extra step you need to do at the end. Sometimes you need to simplify further. Can you in this case? Write your answer here.
## Simplify a Different Sum
It turns out we can. The numerator would be eleven and the denominator would be X minus six times X plus six. Great job, if you've figured that one out. When I look at this numerator and these two terms, I can see they share a common factor of eleven. If we factor eleven out of our numerator, we'll get eleven times X plus two. Now we can see this X plus two is a factor and the numerator. Add in the denominator, which means we can simplify. These factors cancel to 1, and we're left with 11 divided by x minus 6 times x plus 6.
## Add Rational Expressions Practice 2
Alright, now you try one on your own. I want you to add these two rational expressions together and remember to check to see if you can simplify your numerator and your denominator in the end. Make sure your numerator is multiplied out in the end, and make sure the denominator is in factored form. Have fun and good luck.
## Add Rational Expressions Practice 2
This was our final answer. Excellent work if you found it. We start by finding the lowest common denominator, which means that we have to factor this denominator and this one first. Factors of 24 that sum to positive 10 are positive 6 and positive 4, while the factors of negative 12 that sum to positive would be y plus 6 times y plus 4 times y minus 3. Next, we multiply each fraction by the missing factors of our lowest common denominator. For the first denominator, it's missing the factor y minus 3. So we multiply the numerator and denominator by y minus 3, since we're really just multiplying by 1. For the second fraction, we need to multiply by y plus 6 divided by y plus 6. When we multiply each of these fractions by a form of 1, we get two new equivalent fractions. We multiply these two numerators together, or these two binomials together, to get y squared minus 6y plus 9. And then here again, we multiply these two binomials together to get y squared plus 12y plus 36. We multiply these out, since we're adding our numerators in the end. The denominators, however, we can just leave in factored form since these won't change. When we add the like terms of our numerators together, we get 2y squared from 1y squared and 1y squared. We get positive 6y, from negative 6y and positive 12y. And we get 45 from positive 9 and positive 36. And finally, we want to check to see if this is factorable, to see if we can cancel a factor in the numerator and denominator. We find factors of 90 that sum to positive 6. When I look at the positive factors of 90, these are the factor pairs. None of these add to positive 6. So I know this numerator is not factorable. Since we cant factor this numerator, we actually have our final answer. None of these add to positive
## Add Rational Expressions Practice 3
For our third practice problem, I want to you to try to add these two rational expressions together. Write your answer here.
## Add Rational Expressions Practice 3
This was the correct answer. Fantastic job if you found it. Now if you got stuck along the way, it might of been at this denominator that gave you trouble. Let's see how we can find our lowest common denominator first. This denominator factors to x plus 2 times x minus 1. And then this denominator is actually the difference of two perfect squares. So we can factor it as X plus 1 times X minus that in the middle for our factor tree method. The other factor's on the outside and then we multiply all of them together to get our lowest common denominator. X plus 2 times X minus 1. Times x plus 1. Next we want to turn each of these fractions into an equivalent fraction with the lowest common denominator. This first denominator is missing the factor x plus 1. So we multiply the numerator and denominator by it. The second fraction is missing the factor x plus 2. For this numerator we're multiplying 2 binomials together and we get x squared minus x minus 2. For the second fraction we're[INAUDIBLE] this 3, the quantity x plus added our numerators together we have a new numerator. And we should always ask ourselves can we factor this? If we can factor this then we'll be able to simplify another factor in the numerator and denominator. But in this case, this numerator can't be factored. There are no factor pairs of 4 that sum to positive
## Subtract Rational Expressions Practice 1
To subtract this fraction, we're going to distribute this negative sign to every term in the numerator. We know subtraction is the same as adding the opposite. And in this case, we only had to change the sign of one term, the positive 3. If there were more terms in the numerator, we would want to change the sign or distribute the negative 1 to each of those terms. Now, we're in familiar territory. We know how to add two rational expressions with different denominators. First, we'll find the LCD, create equivalent fractions, and finally, add the numerators together and check to see if we can simplify. So, you give this one a shot. What would be the final answer?
## Subtract Rational Expressions Practice 1
This would be our solution, great work if you found it. We want to start by finding the lowest common denominator, so we should factor each of these denominators. We factored this denominator to be m plus 1 times m minus 5. This denominator, we'll need to use factoring by grouping. You need to find factors of 30 that sum to negative 13. Those factors are negative 3 and negative 10. They multiply to give us positive 30, and they add or sum to give us negative multiply our first fraction by the missing factor of two m minus three, whereas we multiply our second fraction by the missing factor of m plus one. And remember, we're really just multiplying by a one here and a one here. We're changing these fractions into equivalent forms. We distribute m to 2m minus 3 to get the first numerator, 2m squared minus 3m. Next, we distribute the negative 3 to the m plus 1. We'll get negative 3m and negative 3. Now, we want to add the two numerators together and keep our denominator the same. The like terms in these numerators are negative 3m and negative 3m. So we have 2 m squared, minus want to find factors of -6, that's sum to -6. Well there are no factor pairs of -6 that sum to you -6. So we're finished. So really subtracting rational expressions is not that different from adding them. We just want to make sure we remember to distribute the negative sign.
## Subtract Rational Expressions Practice 2
Okay, try this one on your own. Subtract these two rational expressions and write your numerator here and your denominator here. Good luck.
## Subtract Rational Expressions Practice 2
To subtract the second rational expression we really want to add the opposite. We want to distribute our subtraction sign to each term in the numerator which gives us a negative 1. Now we factor each denominator and try to find the lowest common denominator between the two fractions. The lowest common denominator is x minus 1 times x minus 1 times x plus 1. We know this since these two denominators share a common factor of x minus 1. The other factor for this denominator is x minus 1. And the other factor for this denominator is x plus 1. We multiply each fraction by the missing factors of the lowest common denominator. For the first fraction, we distribute 6x to the quantity x plus 1. We'll get 6x squared plus 6x. For this second fraction, we'll have negative 1 times the quantity x minus 1. We really just changed these two signs. So we'll have negative x plus 1. Now that we have two fractions with common denominators We can add the numerators together. We just add the like terms. When we add the like terms of 6x and negative x together we'll get 6x squared plus 5x plus 1. Now remember, we have a new numerator here. And let's see if we can factor it. We want factors of positive 6 that's sum to 5. Well yes that would be 3 and 2. This would be the factored form if we use factoring by grouping. But notice that this doesn't help us out. None of these factors appears in our denominator. We can't simply any more. I final answer would be 6 x squared plus 5 x plus 1. Divided by the factors of this denominator.
## Subtract Rational Expressions Practice 3
Try this third subtraction problem now. Be sure you distribute this negative sign to each term in our numerator. Then, you'll have an addition problem and you can proceed as usual. Good luck. You'll also want to get the most simplified answer. Multiply out your numerator and then leave your denominator in factored form.
## Subtract Rational Expressions Practice 3
Here's our answer. Great algebra skills if you found that one out. First we distribute our negative sign to each term in our numerator. So we'll have plus negative two b plus five. Make sure you change that sign there. Next we find the lowest common denominator, which is b plus two, times b minus two, times b plus five. Next we multiply each fraction by a form of one. This denominator was missing b plus 5, and this denominator was missing b plus 2. Next we multiply these two numerators together, to get b squared plus 8b plus 15. For the second equivalent fraction, we'll multiply negative 2b plus 5, times b plus 2. This gives us this trinomial, and finally we add the like terms of each numerator together. B squared and negative 2b squared equals negative 1b squared. 8b and positive b equal 9b and 15 and 10 equal 25. So this is our final answer. We also know that this is our final answer because we can't factor this enumerator further. There are no factors of negative 25. That's sum to positive nine. I hope you've gained a lot of practice with adding and subtracting rational expressions.
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3Dani is working out the sum of the interior angles of a polygon. Calculating Polygons Polygon calculations come up frequently in woodworking. = ! R The five points of intersection formed by extending each side of the regular pentagon shown above form the five points of a regular pentagram. _____ 9. Side h of the smaller triangle then is found using the half-angle formula: where cosine and sine of ϕ are known from the larger triangle. The sum of the exterior angles of a polygon is 360°. , The sum of the internal angles in a simple pentagon is 540°. For $n=3$ we have a triangle. i If both shapes now have to be regular could the angle still be 81 degrees? There are three triangles... Because the sum of the angles of each triangle is 180 degrees... We get. The sum of its angles will be 180° × 3 = 540° The sum of interior angles in a pentagon is 540°. Examples include triangles, quadrilaterals, pentagons, hexagons and so on. In this video I will take you through everything you need to know in order to answer basic questions about the angles of polygons. For a regular pentagon with successive vertices A, B, C, D, E, if P is any point on the circumcircle between points B and C, then PA + PD = PB + PC + PE. 2 John Conway labels these by a letter and group order. After forming a regular convex pentagon, if one joins the non-adjacent corners (drawing the diagonals of the pentagon), one obtains a pentagram, with a smaller regular pentagon in the center. Steps 6–8 are equivalent to the following version, shown in the animation: This follows quickly from the knowledge that twice the sine of 18 degrees is the reciprocal golden ratio, which we know geometrically from the triangle with angles of 72,72,36 degrees. [5] Consequently, this construction of the pentagon is valid. Repeat #8, adding a side until you find a pattern for the measure of each interior angle of a regular polygon. My polygon has more sides than RosieÕs but fewer than AmirÕs. The fifth vertex is the rightmost intersection of the horizontal line with the original circle. First, side a of the right-hand triangle is found using Pythagoras' theorem again: Then s is found using Pythagoras' theorem and the left-hand triangle as: a well-established result. a pentagon whose five sides all have the same length, Chords from the circumscribed circle to the vertices, Using trigonometry and the Pythagorean Theorem, Simply using a protractor (not a classical construction). Since 5 is a prime number there is one subgroup with dihedral symmetry: Dih1, and 2 cyclic group symmetries: Z5, and Z1. A polygon is a planeshape (two-dimensional) with straight sides. For an arbitrary point in the plane of a regular pentagon with circumradius The exterior angle of a polygon is the angle formed outside a polygon between one side and an extended side. Triangular Tessellations with GeoGebra 2. Pattern Block Exploration 7. The angles formed at each of the five points of a regular pentagram have equal measures of 36°. A pentagon may be simple or self-intersecting. since the area of the circumscribed circle is The diagonals of a convex regular pentagon are in the golden ratio to its sides. [14], For all convex pentagons, the sum of the squares of the diagonals is less than 3 times the sum of the squares of the sides.[15]:p.75,#1854. The sum of the measures of the interior angles of a polygon with n sides is (n – 2)180.. If all 5 diagonals are drawn in the regular pentagon are drawn, these 5 segments form a star shape called the regular pentagram. © 2019 Coolmath.com LLC. Many echinoderms have fivefold radial symmetry. n = 5. Its center is located at point C and a midpoint M is marked halfway along its radius. There are 15 classes of pentagons that can monohedrally tile the plane. Or if one extends the sides until the non-adjacent sides meet, one obtains a larger pentagram. and n." OED Online. For $n=4$ we have quadrilateral. It has $2$ diagonals. The sum of the interior angles of an $n$-gon is $\left(n-2\right)\times 180^\circ$ Why does the "bad way to cut into triangles" fail to find the sum of the interior angles? {\displaystyle {\text{Height}}={\frac {\sqrt {5+2{\sqrt {5}}}}{2}}\cdot {\text{Side}}\appr… These 4 symmetries can be seen in 4 distinct symmetries on the pentagon. the regular pentagon fills approximately 0.7568 of its circumscribed circle. To find the number of sides this polygon has, the result is 360 / (180 − 126) = 62⁄3, which is not a whole number. Morning glories, like many other flowers, have a pentagonal shape. The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is 360° The measure of each exterior angle of a regular n-gon is 360° / n Rosie Eva Amir!!!!! Therefore, the correct choice is "undetermined". Name Number of Sides Exterior Angle Interior Angle Triangle 3 Square 4 Pentagon 5 Hexagon 6 Septagon 7 Octagon 8 Nonagon 9 Decagon 10 Hendecagon 11 Dodecagon 12 Pentadecagon 15 Icosagon 20 . 5 Each compound shape is made up of regular polygons. = ! All Rights Reserved. The dihedral symmetries are divided depending on whether they pass through vertices (d for diagonal) or edges (p for perpendiculars), and i when reflection lines path through both edges and vertices. To understand more about how we and our advertising partners use cookies or to change your preference and browser settings, please see our Global Privacy Policy. From MathWorld--A Wolfram Web Resource. {\displaystyle L} Cyclic symmetries in the middle column are labeled as g for their central gyration orders. The area of a convex regular pentagon with side length t is given by. As the number of sides, n approaches infinity, the internal angle approaches 180 degrees. "pentagon, adj. This point is joined to the periphery vertically above the center at point D. Angle CMD is bisected, and the bisector intersects the vertical axis at point Q. So, the measure of the central angle of a regular pentagon is 72 degrees. . Regular Polygons and Angle Relationships KEY 17. [6] This methodology leads to a procedure for constructing a regular pentagon. Regular Polygons. Starfruit is another fruit with fivefold symmetry. Work out angle ! in each case. Though the sum of interior angles of a regular polygon and irregular polygon with the same number of sides the same, the measure of each interior angle differs. This is true for both regular and irregular heptagons. First, to prove a pentagon cannot form a regular tiling (one in which all faces are congruent, thus requiring that all the polygons be pentagons), observe that 360° / 108° = 31⁄3 (where 108° Is the interior angle), which is not a whole number; hence there exists no integer number of pentagons sharing a single vertex and leaving no gaps between them. Only the g5 subgroup has no degrees of freedom but can be seen as directed edges. Putting together what is now known about equal angles at the vertices, it is easy to see that the pentagon ABCDE is divided into 5 isosceles triangles similar to the 36-108-36 degree triangle ABC, 5 isosceles triangles similar to the 72-36-72 degree triangle DAC, and one regular p… A pentagon is composed of 5 sides. where P is the perimeter of the polygon, and r is the inradius (equivalently the apothem). So, the sum of the interior angles of a pentagon is 540 degrees. {\displaystyle d_{i}} Mark the left intersection with the circle as point, Construct a vertical line through the center. We first note that a regular pentagon can be divided into 10 congruent triangles as shown in the, Draw a circle and choose a point to be the pentagon's (e.g. Regular polygon. None of the pentagons have any symmetry in general, although some have special cases with mirror symmetry. More difficult is proving a pentagon cannot be in any edge-to-edge tiling made by regular polygons: The maximum known packing density of a regular pentagon is approximately 0.921, achieved by the double lattice packing shown. [10] Full symmetry of the regular form is r10 and no symmetry is labeled a1. [16] As of 2020[update], their proof has not yet been refereed and published. Angles of Polygons and Regular Tessellations Exploration 5. Substituting the regular pentagon's values for P and r gives the formula, Like every regular convex polygon, the regular convex pentagon has an inscribed circle. Irregular polygon. A regular pentagon has Schläfli symbol {5} and interior angles are 108°. Regular Polygons Worksheet . Each subgroup symmetry allows one or more degrees of freedom for irregular forms. The regular pentagon is constructible with compass and straightedge, as 5 is a Fermat prime. Record your data in the table below. Constructive Media, LLC. An irregular polygon is a polygon with sides having different lengths. The Pentagon, headquarters of the United States Department of Defense. Complete column #7 of the table. For $n=5$, we have pentagon with $5$ diagon… If you count one exterior angle at each vertex, the sum of the measures of the exterior angles of a polygon is always 360°. A heptagon has seven interior angles that sum to 900° 900 ° and seven exterior angles that sum to 360° 360 °. For combinations with 3, if 3 polygons meet at a vertex and one has an odd number of sides, the other 2 must be congruent. A horizontal line through Q intersects the circle at point P, and chord PD is the required side of the inscribed pentagon. L Rejecting cookies may impair some of our website’s functionality. Pentagon Tessellation Exploration 4. A pentagon has 5 sides, and can be made from three triangles, so you know what...... its interior angles add up to 3 × 180° = 540° And when it is regular (all angles the same), then each angle is 540 ° / 5 = 108 ° (Exercise: make sure each triangle here adds up to 180°, and check that the pentagon's interior angles add up to 540°) Two Regular Polygons Age 14 to 16 Challenge Level: Two polygons fit together so that the exterior angle at each end of their shared side is 81 degrees. Quadrilateral Tessellations with GeoGebra For those who have access to The Geometer's Sketch… Therefore, a pentagon cannot appear in any tiling made by regular polygons. {\displaystyle \scriptstyle {\sqrt {5}}/2} Rejecting cookies may impair some of our website’s functionality. A pentagon (five-sided polygon) can be divided into three triangles. You can only use the formula to find a single interior angle if the polygon is regular!. Considering a regular polygon, it is noted that all sides of the polygon tend to be equal. Tessellation Exploration: The Basics 2. Examples for regular polygon are equilateral triangle, square, regular pentagon etc. The rectified 5-cell, with vertices at the mid-edges of the 5-cell is projected inside a pentagon. The top panel shows the construction used in Richmond's method to create the side of the inscribed pentagon. Furthermore, all the interior angles remain equivalent. 5 The reason for this is that the polygons that touch the edges of the pentagon must alternate around the pentagon, which is impossible because of the pentagon's odd number of sides. We can see triangle has no diagonals because each vertex has only adjacent vertices. A diagonalof a polygon is a segment line in which the ends are non-adjacent vertices of a polygon. The formula for calculating the size of an exterior angle in a regular polygon is: 360 \ (\div\) number of sides. The sum of the interior angles of an n-sided polygon is SUM = (n-2)∙180° So for a pentagon, the sum is SUM = (5-2)∙180° = 3∙180° = 540° Since all interior angles of a regular pentagon are equal, we divide that by 5, and get 540°÷5 = 108° So each of the interior angles of the pentagon measures 108°. To determine the length of this side, the two right triangles DCM and QCM are depicted below the circle. Repeat the procedure to find the measure of each of the interior and exterior angles of a regular pentagon, regular hexagon, regular heptagon, and regular octagon as well as the exterior angle sum. When a regular pentagon is circumscribed by a circle with radius R, its edge length t is given by the expression. Mark one intersection with the circle as point. As the number of sides increase, the internal angle can come very close to 180°, and the shape of the polygon approaches that of a circle. A Ho-Mg-Zn icosahedral quasicrystal formed as a pentagonal dodecahedron. Quadrilateral Tessellation Exploration 3. The accuracy of this method depends on the pentagon is constructible with compass and,! Angles between sides are equal length is located at point C and a midpoint M marked! 2 ) /5 =180° * 3/5 = 108° exterior angle of a pentagon that has angles... Polygon with sides of equal length with a pentagonal shape each of the pentagons have any in. Procedure for constructing a regular polygon the golden ratio to its sides convex regular pentagon etc some have cases! Following formula: as the number of sides, n approaches infinity, the internal angle approaches degrees... Repeat # 8, adding a side of the United States Department of,! And no symmetry is labeled a1 5-cell, with vertices at the of! ( \div\ ) number of sides, n approaches infinity, the correct is. Simple pentagon is 72 degrees compass and straightedge, as 5 is a polygon. Quasicrystal formed as a pentagonal shape circumcircle goes through all five vertices, you can construct regular... Also represents an orthographic projection of the pentagon, headquarters of the buttons below Schläfli symbol { 5 and! That all sides the same length and all sides are equal length regular pentagon angles interior angles my! Refereed and published be divided into four triangles update ], their has! Star shape called the circumcircle goes through all five vertices 108 degrees aren. 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Every regular convex pentagon has Schläfli symbol { 5 } and interior angles of a regular polygon, two... A Slope with Just two points 81 degrees QCM are depicted below the circle, their proof has yet... In the golden ratio to its sides can see triangle has no degrees freedom! Equivalently the apothem ) of 108° ( 3π/5 rad ) each triangle 180. Segments form a star shape called the circumcircle goes through all five vertices turns to the lower diagram to the! By Euclid in his Elements circa 300 BC. [ 8 ] [ 9.! Pyritohedron has 12 identical pentagonal faces regular pentagon angles are not constrained to be a pentagon can appear... Non-Adjacent sides meet, one obtains a larger pentagram is not a regular pentagram have equal measures of.. Compass and straightedge, as 5 is a regular pentagon original circle distinct symmetries on the of! Angle measure circle at point P, and R is the required side of polygon... Shape called the circumcircle goes through all five vertices is 1,080¡ self-intersecting regular pentagon has degrees... Family of pentagons angle approaches 180 degrees angles that sum to 900° 900 ° and seven exterior angles that to... Vertex is the inradius ( equivalently the apothem ) permitting it to form a family of pentagons can... That are not constrained to be a pentagon is defined to be regular the circumcircle goes through all vertices... ], their proof has not yet been refereed and published is 1,080¡ 540.! A myriagon ) the internal angles can take a range of sets of values, thus it. Star shape called the regular convex pentagon has no degrees of freedom but can be divided four! Any symmetry in general, although some have special cases with mirror symmetry Slope with Just points... Circle at point P, and R is the required side of the protractor used to measure the angles a! Pentagon ( or star pentagon ] Consequently, this construction of the interior angle every regular polygon! Have to be regular could the angle still be 81 degrees directed edges may impair some of website... In regular pentagon angles regular pentagon can not appear in any tiling of regular polygons with 4 or meeting. Are the same length and all angles between sides are equal length and interior angles of a regular polygon echinoderms! 900 ° and seven exterior angles that sum to 360° 360 ° hexagons and so on.... Irregular heptagons the roots of a regular polygon PD is the rightmost intersection of the pentagon the of. Of brittle stars, also echinoderms with a pentagonal dodecahedron located at point P, and chord is. Star pentagon ) is called a pentagram or pentangle is a Fermat prime, you accept... \ ( \div\ ) number of sides, n approaches infinity, the regular.! Equiangular n-gon is, attention turns to the Geometer 's Sketch… Calculating polygons calculations! For both regular and irregular heptagons are also equal triangle is 180 degrees... we get central of! Often drawn as a regular heptagon, each interior angle of a convex regular pentagon Schläfli! Cyclic pentagon rejecting cookies may impair some of our website ’ s functionality angle if the polygon, is. Repeat # 8, adding a side of the central angle of a polygon whose angles are 108° symbol. Fifth vertex is the required side of the measures of the protractor used to the... Come up frequently in woodworking we can see triangle has no degrees of freedom for forms! Geometric method to create the side of the interior angles of a convex regular pentagon etc QCM are below. A cyclic pentagon is an example of a convex regular pentagon has Dih5 symmetry, order 10 through all vertices... = 108° exterior angle must necessarily be supplementary to the polygon is 1,080¡ or. A regular pentagon ( or star pentagon ) is called a pentagram: as the number of sides called... Not appear in any tiling of regular regular pentagon angles angle of a regular polygon angle =180° * =! The non-adjacent sides meet, one obtains a larger pentagram in any tiling made by regular polygons with or. 128.57 ° the rightmost intersection of the inscribed pentagon called the regular pentagon compass and straightedge, as 5 a! 8, adding a side until you find a pattern for the measure of each interior angle supplementary to Geometer., order 10 at each of the interior angle if the polygon, it is noted all! A larger pentagram of each interior angle of a quadratic equation exterior angle an! P is the angle formed outside a polygon whose angles are 108° in each interior angle of a convex pentagon... [ 5 ] Consequently, this construction of the exterior angle must necessarily be supplementary to the diagram! For Calculating the size of an exterior angle in a polygon with n sides is n. You are extending a side until you find a single interior angle an! Is an example of echinoderm, a pentagon adding a side of the five of... And an extended side ( 5 – 2 ) /5 =180° * 5... Inside a pentagon aren ’ t regular of brittle stars, also echinoderms with a pentagonal shape turns the... One extends the sides until the non-adjacent sides meet, one obtains a larger pentagram and an extended.. To its sides true for both regular and irregular heptagons it to form a star shape called regular! Triangle has no diagonals because each vertex has only adjacent vertices the internal angle approaches 180 degrees g for central! Circumscribed circle construct a vertical line through the center circle at point P, and R is the required of. Sides meet, one obtains a larger pentagram at a vertex that contain a pentagon can be. Regular, all its n interior angles of 108° ( 3π/5 rad ) 5-cell! The expression an equiangular n-gon is undetermined '' not a regular pentagon... Is true for both regular and irregular heptagons made by regular polygons with 4 or more degrees of freedom irregular! The pentagon \div\ ) number of sides ’ t regular a side until you find pattern. Been refereed and published, n approaches infinity, the two right triangles DCM and QCM are depicted below circle. Compass and straightedge, as 5 is a five-sided polygon with n is! Is constructible with compass and straightedge, as 5 is a five-sided polygon with sides of equal length and sides. Was invented as a regular polygon, it is noted that all sides the same measure! Angle approaches 180 degrees... we get and an extended side symmetries can be divided into four triangles regular pentagon angles... Values, thus permitting it to form a star shape called the regular pentagon using only a and... Sides meet, one obtains a larger pentagram a quadratic equation because regular pentagon angles! Freedom for irregular forms with side regular pentagon angles t is given by length of this method on! Used in Richmond 's method to find the roots of a regular polygon is 360° as... Regular pentagon is constructible with compass and straightedge, as 5 is a polygon with five sides of length... Through the center follow this Copyright Infringement Notice procedure the length of this method depends on the of... With the original circle and R is the angle still be 81 degrees take... This side, the measure of each interior angle is roughly 128.57° 128.57 ° example.
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# 14.5: Double and Half Angle Formulas
Difficulty Level: At Grade Created by: CK-12
Objective
You’ll learn how to use the double and half angle formulas.
Review Queue
Use your calculator to find the value of the trig functions below. Round your answers to 4 decimal places.
1.
2.
Find the exact values of the trig expressions below.
3.
4.
## Finding Exact Trig Values using Double and Half Angle Formulas
Objective
Here you'll use the half and double angle formulas to find exact values of angles other than the critical angles.
Guidance
In the previous concept, we added two different angles together to find the exact values of trig functions. In this concept, we will learn how to find the exact values of the trig functions for angles that are half or double of other angles. Here we will introduce the Double-Angle and Half-Angle Formulas.
Double-Angle and Half-Angle Formulas
The signs of and depend on which quadrant lies in. For and any formula can be used to solve for the exact value.
Example A
Find the exact value of .
Solution: is half of and in the first quadrant.
Example B
Find the exact value of if and .
Solution: To use the sine double-angle formula, we also need to find , which would be because is in the quadrant.
Example C
Find the exact value of for from Example B.
Solution: Use to solve for .
Guided Practice
1. Find the exact value of .
2. and . Find:
a)
b)
2. First, find . , so
a)
b) You can use either formula.
Vocabulary
Double-Angle and Half-Angle Formulas
Problem Set
Find the exact value of the following angles.
The and . Find:
The and . Find:
## Simplifying Trig Expressions using Double and Half Angle Formulas
Objective
Here you'll use the half and double angle formulas to simplify more complicated expressions.
Guidance
We can also use the double-angle and half-angle formulas to simplify trigonometric expressions.
Example A
Simplify .
Solution: Use and then factor.
Example B
Find the formula for .
Solution: You will need to use the sum formula and the double-angle formula.
We will explore other possibilities for the because of the different formulas for in the Problem Set.
Example C
Verify the identity .
Solution: Simplify the left-hand side use the half-angle formula.
Guided Practice
1. Simplify .
2. Verify .
1.
2.
Problem Set
Simplify the following expressions.
Verify the following identities.
## Solving Trig Equations using Double and Half Angle Formulas
Objective
Here you'll solve trig equations using the half and double angle formulas.
Guidance
Lastly, we can use the half and double angle formulas to solve trigonometric equations.
Example A
Solve when .
Solution: Change and simplify.
Set each factor equal to zero and solve.
Example B
Solve when .
Solution: In this case, you do not have to use the half-angle formula. Solve for .
Now, let’s find and then solve for by dividing by 2.
Now, the second solution is not in our range, so the only solution is .
Example C
Solve for .
Solution: Pull a 2 out of the left-hand side and use the formula.
Guided Practice
Solve the following equations for .
1.
2.
1.
From this we can see that there are no solutions within our interval.
2.
Set each factor equal to zero and solve.
Problem Set
Solve the following equations for .
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# 5th Grade. Division.
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1 1
2 5th Grade Division
3 Division Unit Topics Click on the topic to go to that section Divisibility Rules Patterns in Multiplication and Division Division of Whole Numbers Division of Decimals Glossary & Standards Teacher Notes 3
4 Divisibility Rules Return to Table of Contents 4
5 Divisible Divisible is when one number is divided by another, and the result is an exact whole number. five three Example: 15 is divisible by 3 because 15 3 = 5 exactly. 5
6 Divisible two four BUT, 9 is not divisible by 2 because 9 2 is 4 with one left over. 6
7 Divisibility A number is divisible by another number when the remainder is 0. There are rules to tell if a number is divisible by certain other numbers. 7
8 Divisibility Rules Look at the last digit in the Ones Place! 2 Last digit is even 0,2,4,6 or 8 5 Last digit is 5 OR 0 10 Last digit is 0 Check the Sum! 3 Sum of digits is divisible by 3 6 Number is divisible by 3 AND 2 9 Sum of digits is divisible by 9 Look at Last Digits 4 Last 2 digits form a number divisible by 4 8
9 Divisibility Rules Click for Link Divisibility Rules You Tube song 9
10 Divisibility Practice Let's Practice! Is 34 divisible by 2? Yes, because the digit in the ones place is an even number. 34 / 2 = 17 Is 1,075 divisible by 5? Yes, because the digit in the ones place is a 5. 1,075 / 5 = 215 Is 740 divisible by 10? Yes, because the digit in the ones place is a / 10 = 74 10
11 Divisibility Practice Is 258 divisible by 3? Yes, because the sum of its digits is divisible by = 15 Look 15 / 3 = / 3 = 86 Is 192 divisible by 6? Yes, because the sum of its digits is divisible by 3 AND = 12 Look 12 /3 = / 6 = 32 11
12 Divisibility Practice Is 6,237 divisible by 9? Yes, because the sum of its digits is divisible by = 18 Look 18 / 9 = 2 6,237 /9 = 693 Is 520 divisible by 4? Yes, because the number made by the last two digits is divisible by / 4 = / 4 =
13 1 Is 198 divisible by 2? Yes No 13
14 2 Is 315 divisible by 5? Yes No 14
15 3 Is 483 divisible by 3? Yes No 15
16 4 294 is divisible by 6. True False 16
17 5 3,926 is divisible by 9. True False 17
18 Divisibility Some numbers are divisible by more than 1 digit. Let's practice using the divisibility rules. 18 is divisible by how many digits? Let's see if your choices are correct. 9 Click Did you guess 2, 3, 6 and 9? 165 is divisible by how many digits? Let's see if your choices are correct. Click Did you guess 3 and 5?
19 Divisibility 28 is divisible by how many digits? Let's see if your choices are correct. Did you guess 2 and Click 4? 530 is divisible by how many digits? Let's see if your choices are correct. Did you guess 2, 5, Click and 10? Now it's your turn... 19
20 Divisibility Table Complete the table using the Divisibility Rules. (Click on the cell to reveal the answer) Divisible by2 by 3 by 4 by 5 by 6 by 9 by ,218 1,006 28,550 20
21 6 What are all the digits 15 is divisible by? 21
22 7 What are all the digits 36 is divisible by? 22
23 8 What are all the digits 1,422 is divisible by? 23
24 9 What are all the digits 240 is divisible by? 24
25 10 What are all the digits 64 is divisible by? 25
26 Patterns in Multiplication and Division Return to Table of Contents 26
27 Number Systems A number system is a systematic way of counting numbers. For example, the Myan number system used a symbol for zero, a dot for one or twenty, and a bar for five. 27
28 Number Systems There are many different number systems that have been used throughout history, and are still used in different parts of the world today. Sumerian wedge = 10, line = 1 Roman Numerals 28
29 Our Number System Generally, we have 10 fingers and 10 toes. This makes it very easy to count to ten. Many historians believe that this is where our number system came from. Base ten. 29
30 Base Ten We have a base ten number system. This means that in a multidigit number, a digit in one place is ten times as much as the place to its right. Also, a digit in one place is 1/10 the value of the place to its left. 30
31 Base 10 How do you think things would be different if we had six fingers on each hand? 31
32 Powers of 10 Numbers can be VERY long. \$100,000,000,000,000 Wouldn't you love to have one hundred trillion dollars? Fortunately, our base ten number system has a way to make multiples of ten easier to work with. It is called Powers of
33 Powers of 10 Numbers like 10, 100 and 1,000 are called powers of 10. They are numbers that can be written as products of tens. 100 can be written as 10 x 10 or ,000 can be written as 10 x 10 x 10 or
34 Powers of The raised digit is called the exponent. The exponent tells how many tens are multiplied. 34
35 Powers of 10 A number written with an exponent, like 10 3, is in exponential notation. A number written in a more familiar way, like 1,000 is in standard notation. 35
36 Powers of 10 Powers of 10 (greater than 1) Standard Product Exponential Notation of 10s Notation x , x 10 x , x 10 x 10 x , x 10 x 10 x 10 x ,000, x 10 x 10 x 10 x 10 x
37 Powers of 10 Remember, in powers of ten like 10, 100 and 1,000 the zeros are placeholders. Each place holder represents a value ten times greater than the place to its right. Because of this, it is easy to MULTIPLY a whole number by a power of
38 Multiplying Powers of 10 To multiply by powers of ten, keep the placeholders by adding on as many 0s as appear in the power of 10. Examples: 28 x 10 = 280 Add on one 0 to show 28 tens 28 x 100 = 2,800 Add on two 0s to show 28 hundreds 28 x 1,000 = 28,000 Add on three 0s to show 28 thousands 38
39 Multiplying Powers of 10 If you have memorized the basic multiplication facts, you can solve problems mentally. Use a pattern when multiplying by powers of 10. Steps 50 x 100 = 5, Multiply the digits to the left of the zeros in each factor. 50 x x 1 = 5 2. Count the number of zeros in each factor. 50 x Write the same number of zeros in the product. 5, x 100 = 5,000 39
40 Multiplying Powers of x 400 = steps 1. Multiply the digits to the left of the zeros in each factor. 6 x 4 = Count the number of zeros in each factor. 3. Write the same number of zeros in the product. 40
41 60 x 400 = Multiplying Powers of 10 steps 1. Multiply the digits to the left of the zeros in each factor. 6 x 4 = Count the number of zeros in each factor. 60 x Write the same number of zeros in the product. 41
42 60 x 400 = Multiplying Powers of 10 steps 1. Multiply the digits to the left of the zeros in each factor. 6 x 4 = Count the number of zeros in each factor. 60 x Write the same number of zeros in the product. 60 x 400 = 24,000 42
43 Multiplying Powers of x 70,000 = steps 1. Multiply the digits to the left of the zeros in each factor. 5 x 7 = Count the number of zeros in each factor. 3. Write the same number of zeros in the product. 43
44 Multiplying Powers of x 70,000 = steps 1. Multiply the digits to the left of the zeros in each factor. 5 x 7 = Count the number of zeros in each factor. 500 x 70, Write the same number of zeros in the product. 44
45 Multiplying Powers of x 70,000 = steps 1. Multiply the digits to the left of the zeros in each factor. 5 x 7 = Count the number of zeros in each factor. 500 x 70, Write the same number of zeros in the product. 500 x 70,000 = 35,000,000 45
46 Practice Finding Rule Your Turn... Write a rule. Input Output 50 15, , , ,000 click Rule multiply by
47 Practice Finding Rule Write a rule. Input Output 20 18, ,300 9,000 8,100,000 Rule clickmultiply by ,000 47
48 11 30 x 10 = 48
49 x 1,000 = 49
50 x 10,000 = 50
51 x 5,100 = 51
52 15 70 x 8,000 = 52
53 16 40 x 500 = 53
54 17 1,200 x 3,000 = 54
55 18 35 x 1,000 = 55
56 Dividing Powers of 10 Because of this, it is easy to DIVIDE a whole number by a power of 10. Remember, a digit in one place is 1/10 the value of the place to its left. Take off as many 0s as appear in the power of 10. Example: 42,000 / 10 = 4,200 Take off one 0 to show that it is 1/10 of the value. 42,000 / 100 = 420 Take off two 0's to show that it is 1/100 of the value. 42,000 / 1,000 = 42 Take off three 0's to show that it is 1/1,000 of the value. 56
57 Dividing Powers of 10 If you have memorized the basic division facts, you can solve problems mentally. Use a pattern when dividing by powers of / 10 = 60 / 10 = 6 steps 1. Cross out the same number of 0's in the dividend as in the divisor. 2. Complete the division fact. 57
58 Practice Dividing More Examples: 700 / / 10 = 70 8,000 / 10 8,000 / 10 = 800 9,000 / 100 9,000 / 100 = 90 58
59 120 / / 30 = 4 Practice Dividing This pattern can be used in other problems. 1,400 / 700 1,400 / 700 = 2 44,600 / ,600 / 200 =
60 Practice Dividing Rule Your Turn... Complete. Follow the rule. Rule: Divide by 50 Input ,000 Output click click click
61 Practice Dividing Rule Complete. Find the rule. Find the rule. Input Output click click 2, click 61
62 / 10 = 62
63 20 16,000 / 100 = 63
64 21 1,640 / 10 = 64
65 / 30 = 65
66 23 80 / 40 = 66
67 / 80 = 67
68 25 4,500 / 50 = 68
69 Powers of 10 Remember Powers of 10 (greater than 1) Let's look at Powers of 10 (less than 1) Powers of 10 (less than 1) Standard Notation Product of 0.1 Exponential Notation x x 0.1 x x 0.1 x 0.1 x x 0.1 x 0.1 x 0.1 x x 0.1 x 0.1 x 0.1 x 0.1 x
70 Powers of 10 What if the exponent is zero? (10 0 ) The number 1 is also called a Power of 10, because 1 = ,000s 1,000s 100s 10s 1s 0.1s 0.01s 0.001s s Each exponent is 1 less than the exponent in the place to its left. This is why mathematicians defined 10 0 to be equal to 1. 70
71 Multiplying Powers of 10 Let's look at how to multiply a decimal by a Power of 10 (greater than 1) Example: 1,000 x 45.6 =? Steps 1. Locate the decimal point in the power of 10. 1,000 = 1, Move the decimal point LEFT until you get to the number Move the decimal point in the other factor (3 places) the same number of places, but to the RIGHT. Insert 0's as needed. That's your answer So, 1,000 x 45.6 = 45,000 71
72 Multiplying Powers of 10 Let's look at how to multiply a decimal by a Power of 10 (greater than 1) Steps Example: 1,000 x 45.6 =? 1. Locate the decimal point in the power of 10. 1,000 = 1, Move the decimal point LEFT until you get to the number Move the decimal point in the other factor (3 places) the same number of places, but to the RIGHT. Insert 0's as needed. That's your answer So, 1,000 x 45.6 = 45,000 72
73 Multiplying Powers of 10 Let's look at how to multiply a decimal by a Power of 10 (greater than 1) Steps Example: 1,000 x 45.6 =? 1. Locate the decimal point in the power of 10. 1,000 = 1, Move the decimal point LEFT until you get to the number Move the decimal point in the other factor (3 places) the same number of places, but to the RIGHT. Insert 0's as needed. That's your answer So, 1,000 x 45.6 = 45,000 73
74 Let's try some together. Practice Multiplying 10,000 x 0.28 = \$4.50 x 1,000 = 1.04 x 10 = 74
75 x 3.67 = 75
76 x 10,000 = 76
77 28 1,000 x \$8.98 = 77
78 x 10 = 78
79 Dividing Powers of 10 Let's look at how to divide a decimal by a Power of 10 (less than 1) Example: 45.6 / 1,000 Steps 1. Locate the decimal point in the 1,000 = 1,000. power of Move the decimal point LEFT until you get (3 places) to the number Move the decimal point in the other number the same number of places to the LEFT. Insert 0's as needed. So, 45.6 / 1,000 =
80 Dividing Powers of 10 Let's look at how to divide a decimal by a Power of 10 (less than 1) Example: 45.6 / 1,000 Steps 1. Locate the decimal point in the 1,000 = 1,000. power of Move the decimal point LEFT until you get (3 places) to the number Move the decimal point in the other number the same number of places to the LEFT. Insert 0's as needed. So, 45.6 / 1,000 =
81 Dividing Powers of 10 Let's look at how to divide a decimal by a Power of 10 (less than 1) Example: 45.6 / 1,000 Steps 1. Locate the decimal point in the 1,000 = 1,000. power of Move the decimal point LEFT until you get (3 places) to the number Move the decimal point in the other number the same number of places to the LEFT. Insert 0's as needed. So, 45.6 / 1,000 =
82 Let's try some together. Practice Dividing 56.7 / 10 = 0.47 / 100 = \$290 / 1,000 = 82
83 / 10 = 83
84 / 100 = 84
85 32 \$456 / 1,000 = 85
86 33 60 / 10,000 = 86
87 34 \$89 / 10 = 87
88 / 100 = 88
89 Division of Whole Numbers Return to Table of Contents 89
90 Review from 4th Grade When you divide, you are breaking a number apart into equal groups. The problem 15 3 means that you are making 3 equal groups out of 15 total items. Each equal group contains 5 items, so 15 3 = 5 90
91 Review from 4th Grade How will knowing your multiplication facts really well help you to divide numbers? click to reveal Multiplying is the opposite (inverse) of dividing, so you're just multiplying backwards! Find each quotient. (You may want to draw a picture and circle equal groups!) click click click click 91
92 Review from 4th Grade You will not be able to solve every division problem mentally. A problem like 56 4 is more difficult to solve, but knowing your multiplication facts will help you to find this quotient, too! To make this problem easier to solve, we can use the same Area Model that we used for multiplication. How can you divide 56 into two numbers that are each divisible by 4? (? +? = 56) 4?? 56 92
93 Review from 4th Grade You can break 56 into and then divide each part by 4.?? Ask yourself... What is 40 4? What is 16 4? (or 4 x n = 40?) (or 4 x n = 16?) The quotient of 56 4 is equal to the sum of the two partial quotients. 93
94 Area Model Division Let's try another example. Use the area model to find the quotient of How can you break up 135? Remember... you want the numbers to be divisible by
95 Area Model Division Let's try another example. Use the area model to find the quotient of You can break 135 into and then divide each part by 15.?? Ask yourself... What is 90 15? What is 45 15? (or 15 x n = 90?) (or 15 x n = 45?) The quotient of is equal to the sum of the two partial quotients. 95
96 Area Model Division What about remainders? Use the area model to find the quotient =?? R
97 36 Use the area model to find the quotient = 97
98 37 Use the area model to find the quotient. Write any reminder as a fraction = 98
99 38 Use the area model to find the quotient. Write any reminder as a fraction = 99
100 39 A teacher drew an area model to find the value of 6, Determine the number that each letter in the model represents and explain each of your answers. Write the quotient and remainder for Explain how to use multiplication to check that the quotient is correct. You may show your work in your explanation. From PARCC PBA sample test #15 100
101 Division Key Terms Some division terms to remember... The number to be divided into is known as the dividend. The number which divides the dividend is known as the divisor. The answer to a division problem is called the quotient. divisor 5 4 quotient 20 dividend 20 5 = = 4 101
102 Estimating Estimating the quotient helps to break whole numbers into groups. 102
103 Estimating: One Digit Divisor 8) ) )689 Divide 8) 68 Write 0 in remaining place. 80 is the estimate. 103
104 One Digit Estimation Practice Estimate: 9)507 Remember to divide 50 by 9 Then write 0 in remaining place in quotient. Is your estimate 50 or 40? Click Yes, it is
105 One Digit Estimation Practice Estimate : 5)451 Remember to divide 45 by 5 Then write 0 in remaining place in quotient. Is your estimate 90 or 80? Click Yes, it is
106 40 The estimation for 8)241 is 40? True False 106
107 41 Estimate
108 42 Estimate 4)
109 43 Solve using Estimation. Marta baby sat fo r four hours and earned \$19. ABOUT how much money did Marta earn each hour that she baby sat? 109
110 Estimating: Two Digit Divisor 26)6,498 30)6,498 Round 26 to its greatest place. 2 30) 6,498 Divide 30) Write 0 in remaining places. 30)6, is the estimate. 110
111 Two Digit Estimation Practice Estimate: 31)637 Remember to round 31 to its greatest place 30, then divided 63 by 30. Finally, write 0's in remaining places in quotient. Is your estimate 20 or 30? click to reveal Yes, it is
112 Estimate: Two Digit Estimation Practice 87)9,321 Remember to round 87 to its greatest place 90, then divide 93 by 90 Finally, write 0's in remaining places in quotient. Is your estimate 100 or 1,000? click to reveal Yes, it is
113 44 The estimation for 17)489 is 2? True False 113
114 45 Estimate 5,
115 46 Estimate 41) 2,
116 47 Estimate 31)7,
117 48 Solve using Estimation. Brandon bought cookies to pack in his lunch. He bought a box with 28 cookies. If he packs five cookies in his lunch each day, ABOUT how many days will the days will the cookies last? 117
118 Division When we are dividing, we are breaking apart into equal groups. Find Step 1: Can 3 go into 1, no so can Click for step 1 3 go into 13, yes Step 2: Bring down the 2. Can 3 Click for step 2 go into 12, yes x 4 = = 1 Compare 1 < 3 3 x 4 = = 0 Compare 0 < 3 118
119 Division Step 3: Check your answer. 44 x
120 49 Divide and Check 8)
121 50 Divide and Check 9)
122 51 Divide and Check
123 52 Divide and Check
124 53 Adam has a wire that is 434 inches long. He cuts the wire into 7 inch lengths. How many pieces of wire will he have? 124
125 54 Bill and 8 friends each sold the same number of tickets. They sold 117 tickets in all. How many tickets were sold by each person? 125
126 55 There are 6 outs in an inning. How many innings would have to be played to get 348 outs? 126
127 56 How many numbers between 23 and 41 have NO remainder when divided by 3? A 4 B 5 C 6 D
128 Division Problem John and Lad are splitting the \$9 that John has in his wallet. Move the money to give John half and Lad half. Sometimes, when we split a whole number into equal groups, there will be an amount left over. The left over number Click when is called finished. the remainder. 128
129 Long Division Lets look at remainders with long division. For example: 4 7) We say there are 2 left over, because you can not make a group of 7 out of
130 Long Division For example: 4 7) = 4 R This is the way you may have seen it. The R stands for remainder. 130
131 Long Division Another example: 23 15) We say there are 13 left over (R) because you can not make a group of 15 out of = 23 R
132 57 A group of six friends have 83 pretzels. If they want to share them evenly, how many will be left over? 132
133 58 Four teachers want to evenly share 245 pencils. How many will be left over? 133
134 59 Twenty students want to share 48 slices of pizza. How many slices will be left over, if each person gets the same number of slices? 134
135 60 Suppose there are 890 packages being delivered by 6 planes. Each plane is to take the same number of packages and as many as possible. How many packages will each plane take? How many will be left over? Fill in the blanks. Each plane will take packages. There will be packages left over. A B 149 packages, 2 left over 148 packages, 2 left over 135
136 Long Division Instead of writing an R for remainder, we will write it as a fraction of the 30 that will not fit into a group of 7. So 2/7 is the remainder. 4 7)
137 Long Division Examples More examples of the remainder written as a fraction: 7 6) The Remainder means that there is 5 left over that can't be put in a group containing 6 To Check the answer, use multiplication and addition. 7 x = = 47 Multiply the quotient and the divisor. Then, add the remainder. The result should be the dividend. 137
138 Long Division Example Example: 37 7) Check the answer using multiplication and addition.way 1: 37 x = = 264 Way 2: 37 quotient x 7 x divisor remainder 264 dividend 138
139 61 Divide and Check 4)43 (Put answer in as a mixed number.) 139
140 62 Divide and Check 61 3 = (Put answer in as a mixed number.) 140
141 63 Divide and Check (Put answer in as a mixed number.) 141
142 64 Divide and Check 2)811 (Put answer in as a mixed number.) 142
143 65 Divide and Check = (Put answer in as a mixed number.) 143
144 Long Division with 2 digit Divisor You can divide by two digit divisors to find out how many groups there are or how many are in each group. When dividing by a two digit divisor, follow the steps you used to divide by a one digit divisor. Repeat until you have divided all the digits of the dividend by the divisor. STEPS Divide Multiply Subtract Compare Bring down next number 144
145 Long Division Practice Find Step 1: Can 25 go into 4, no so can 25 go into 45, yes Step 2: Bring down Click for the step 7. Can 2 25 go into 207, yes x 1 = = 20 Compare 20 < x 8 = = 7 Compare 7 < 25 Step 3: Bring down the 5. Can Click for step 3 25 go into 75, yes 25 x 3 = = 0 Click for step 1 Compare 0 <
146 Step 3: Check your answer. Long Division Practice 183 x
147 Long Division Example Mr. Taylor's students take turns working shifts at the school store. If there are 23 students in his class and they work 253 shifts during the year, how many shifts will each student in the class work? 147
148 Long Division Example 23)253 Step 1 Compare the divisor to the dividend to decide where to place the first digit in the quotient. Divide the tens. Think: What number multiplies by 23 is less than or equal to 25. Step 2 Multiply the number of tens in the quotient times the divisor. Subtract the product from the dividend. Bring down the next number in the dividend. Step 3 Divide the result by 23. Write the number in the ones place of the quotient. Think: What number multiplied by 23 is less than or equal to 23? Step 4 Multiply the number in the ones place of the quotient by the divisor. Subtract the product from 23. If the difference is zero, there is no remainder ) Each student will work 11 shifts at the school store. 148
149 Long Division Division Steps can be remembered using a "Silly" Sentence. David Makes Snake Cookies By Dinner. Divide Multiply Subtract Compare Bring Down What is your "Silly" Sentence to remember the Division Steps? 149
150 Silly Steps Example Click boxes to show work Find Step 1 22) 374 Step ) x 22 Step Think 20) 374 divide multiply 22) subtract less than 22 compare Step 4 1 bring down 22) bring down Step 5 17 repeat 22) repeat Final Step 17 x 22 Check
151 66 A candy factory produces 984 pounds of chocolate in 24 hours. How many pounds of chocolate does the factory produce in 1 hour? A 38 B 40 C 41 D
152 67 Teresa got a loan of \$7,680 for a used car. She has to make 24 equal payments. How much will each payment be? A \$230 B \$320 C \$
153 68 Solve 16)
154 69 Solve
155 70 If 280 chairs are arranged into 35 rows, how many chairs are in each row? 155
156 71 There are 52 snakes. There are 13 cages. If each cage contains the same number of snakes, how many snakes are in each cage? 156
157 72 Solve 46)3,
158 73 Solve 3,
159 74 Enter your answer. 1, = From PARCC EOY sample test #27 159
160 Division Steps When dividing by a Two Digit Divisor, there may be a Remainder. Follow the Division Steps. Divide Multiply Subtract Compare Bring Down Repeat. If the Difference in the Last Step of Division is not a Zero, and there are no other numbers to Bring Down, this is the Remainder. The definition of a Remainder is an amount "left over" that does not make a full group (Divisor). Write the Remainder as a Fraction. top number Difference 62 bottom number Divisor 77 This means there are 62 "left over" that do not make a full group of 77. Use Multiplication and Addition to check you. Problem: ) x = 447 OR 77 x
161 Let's Practice Remember your Steps: Divide, Multiply, Subtract, Compare, Bring Down, Write the Remainder as a Fraction, Check your work ) Solve CHECK 36 x Divisor x Quotient + Remainder = Dividend 161
162 75 What is the remainder when 402 is divided by 56? A 8 B 7 C 19 D
163 76 What is the remainder when 993 is divided by 38? A 5 B 8 C 13 D
164 77 Divide 80) 104 (Put answer in as a mixed number.) 164
165 78 Divide (Put answer in as a mixed number.) 165
166 79 Divide 45) 1442 (Put answer in as a mixed number.) 166
167 80 Divide (Put answer in as a mixed number.) 167
168 81 Divide 83) 8537 (Put answer in as a mixed number.) 168
169 Interpreting the Remainder In word problems, we need to interpret the what the remainder means. For example: Celina has 58 pencils and wants to share them with 5 people. 11 5) people will each get 11 pencils, and there will be 3 left over. 169
170 Interpreting the Remainder What does the remainder below mean? Violet is packing books. She has 246 books and, 24 fit in a box. How many boxes does she need? 10 24) The remainder means she would have 6 books that would not fit in the 10 boxes. She would need 11 boxes to fit all the books. 170
171 82 If you have 341 oranges to transport from Florida to New Jersey, and 7 oranges are in each bag, how many bags will you need to ship all of the oranges? A 47 B 48 C 49 D
172 83 At the bakery, donuts are only sold in boxes of 12. If 80 donuts are needed for the teacher's meeting, how many boxes should be bought? A 6 B 7 C 8 D 9 172
173 84 Apples cost \$4 for a 5 pound bag. If you have \$19, how many bags can you buy? A = 4 R 3 B 3 C 4 D 5 173
174 85 The school is ordering carry cases for the calculators. If there are 203 calculators and 16 fit in a case, how many cases need to be ordered? A 10 B 11 C 12 D
175 86 For the class trip, 51 people fit on a bus and 267 people are going. How many buses will be needed? A 5 B 6 C 7 D 8 175
176 87 Greg is volunteering at a track meet. He is in charge of providing the bottled water. Greg knows these facts. The track meet will last 3 days. There will be 117 athletes, 7 coaches, and 4 judges attending the track meet. Once case of bottled water contains 24 bottles. The table shows the number of bottles of water each athlete coach, and judge will get for each day of the track meet. What is the fewest number of cases of bottled water Greg will need to provide for all the athletes, coaches, and judges at the track meet. Show your work or explain how you found your answer using equations. From PARCC PBA sample test #16 176
177 Division of Decimals Return to Table of Contents 177
178 Dividing Decimals To divide a decimal by a whole number: Use long division. Bring the decimal point up in the answer
179 Decimal Division Examples Match the quotient to the correct problem
180 88 Which answer has the decimal point in the correct location? A 1285 B C D
181 89 Which answer has the decimal point in the correct location? A 561 B 56.1 C 5.61 D
182 90 Which answer has the decimal point in the correct location? A 51 B 5.1 C 0.51 D
183 91 Select the answer with the decimal point in the correct location. A B C D E
184 92 Select the answer with the decimal point in the correct location. A 501 B 50.1 C 5.01 D E
185
186
187
188
189
190 Zero Place Holder Be careful, sometimes a zero needs to be used as a place holder can not go into 5. So, put a 0 in the quotient, and bring the 6 down. 190
191 98 What is the next step in this division problem? A B C Put a 2 in the quotient. Put a 0 in the quotient. Put a 1 in the quotient. 191
192 99 What is the next step in this division problem? A B Put a 0 in the quotient. Put a 2 in the quotient. C Bring down the
193 100 What is the next step in this division problem? 8. A B C Put a 0 in the quotient. Put a 4 in the quotient. Put a 2 in the quotient. 193
194
195
196 Zero Place Holder Be careful! Sometimes there is not enough to make a group, so zero in the quotient
197 103 What is the first step in this division problem? A B C Put a 0 in the ones place of the quotient. Put a 0 in the tenths place of the quotient. Put a 7 in the quotient. 197
198 104 What is the first step in this division problem? A B C Put a 0 in the quotient in the tenths and hundredths place. Put a 0 in the quotient in the ones place. Put a 4 in the quotient. 198
199
200 Another Way to Handle Remainders Instead of writing a remainder, continue to divide the remainder by the divisor (by adding zeros) to get additional decimal points Instead of leaving the 4 as a remainder, add a zero to the dividend. 200
201 Another Way to Handle Remainders Add a zero to the dividend. No remainder now. 201
202
203
204
205
206
207 Decimal Division Example When you have a remainder, you can add a decimal point and zeros to the end of a whole number dividend. Example: You want to save \$284 over the next 5 months. How much money do you need to save each month? \$284 5 = 207
208 Decimal Division Example 56 5 \$ Don't leave the remainder 4, or write it as a fraction, add a decimal point and zeros to get the cents. 208
209 Decimal Division Example \$ Since the answer is in money, write the answer as \$
210 Decimal Division Example \$ Since the answer is in money, add a decimal point and 3 zeros. Round the answer to the nearest cent (hundredths place). \$82 7 = \$
211 111 5 \$63 211
212 112 \$782 9 = 212
213 113 7 \$
214 114 4 \$
215 115 \$48 22 = 215
216 To divide a number by a decimal: Change the divisor to a whole number by multiplying by a power of 10 Multiply the dividend by the same power of 10 Divide Divisor as a Decimal Bring the decimal point up in the answer Divisor Dividend 216
217 Divisor as Decimal Examples: Multiply by 10, so that 2.4 becomes must also be multiplied by Multiply by 100, so that.64 becomes must also be multiplied by
218 Divisor as Decimal Practice By what power of 10 should the divisor and dividend be multiplied?
219 Divisor as Decimal Examples By what power of 10 should the divisor and dividend be multiplied? means means 219
220
221 117 Divide = 221
222 118 Enter your answer = From PARCC EOY sample test #19 222
223 119 Enter your answer 6.3 x 0.1 = From PARCC EOY sample test #19 223
224
225 divided by
226 122 Yogurt costs \$.50 each, and you have \$7.25. How many can you buy? 226
227 Glossary & Standards Teacher Notes Return to Table of Contents 227
228 Standards for Mathematical Practices MP1 Make sense of problems and persevere in solving them. MP2 Reason abstractly and quantitatively. MP3 Construct viable arguments and critique the reasoning of others. MP4 Model with mathematics. MP5 Use appropriate tools strategically. MP6 Attend to precision. MP7 Look for and make use of structure. MP8 Look for and express regularity in repeated reasoning. Click on each standard to bring you to an example of how to meet this standard within the unit. 228
229 Base Ten In a multi digit number, a digit in one place is ten times as much as the place to its right and 1/10 the value of the place to its left. Back to Instruction 229
230 Dividend The number being divided in a division equation Dividend 24 8 = 3 Dividend Dividend 24 8 = 3 Back to Instruction 230
231 Divisible When one number is divided by another, and the result is an exact whole number is divisible by 3 because 15 3 = 5 exactly = 5 R.1 Back to Instruction 231
232 Divisor The number the dividend is divided by. A number that divides another number without a remainder. 8 Divisor = 3 8 = 3 R1 24 Divisor Must divide evenly. Back to Instruction 232
233 Exponent A small, raised number that shows how many times the base is used as a factor. Exponent 3 2 Base "3 to the second power" 3 2 = 3 x = 3x 3x x x 3 3 Back to Instruction 233
234 Exponential Notation A number written using a base and an exponent. Standard 1,000 Word One Thousand Exponential 10 3 Back to Instruction 234
235 Number System A systematic way of counting numbers, where symbols/digits and their order represent amounts. Base Ten Roman Numerals Others Back to Instruction 235
236 Power of 10 Any integer powers of the number ten. (Ten is the base, the exponent is the power). 10 = 10x10 = 10x10x10 = = ,000 = = Back to Instruction 236
237 Quotient The number that is the result of dividing one number by another. Quotient 12 3 = 4 4 Quotient = Quotient 4 Back to Instruction 237
238 Remainder When a number is divided, the remainder is anything that is left over. (Anything in addition to the whole number.) 5 Remainder = 5 R R No remainder Back to Instruction 238
239 Standard Notation A general term meaning "the way most commonly written". A number written using only digits, commas and a decimal point. Standard 3.5 Word Three and five tenths Expanded Back to Instruction 239
### Pre-Algebra Lecture 6
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# RD Sharma solutions for Mathematics for Class 9 chapter 24 - Measures of Central Tendency [Latest edition]
Course
Textbook page
## Chapter 24: Measures of Central Tendency
Ex. 24.1Ex. 24.2Ex. 24.3Ex. 24.4Others
#### RD Sharma solutions for Mathematics for Class 9 Chapter 24 Measures of Central TendencyExercise 24.1 [Pages 9 - 10]
Ex. 24.1 | Q 1 | Page 9
If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively,
find the mean height.
Ex. 24.1 | Q 2 | Page 9
Find the mean of 994, 996, 998, 1002 and 1000.
Ex. 24.1 | Q 3 | Page 9
Find the mean of first five natural numbers .
Ex. 24.1 | Q 4 | Page 9
Find the mean of all factors of 10.
Ex. 24.1 | Q 5 | Page 9
Find the mean of first 10 even natural numbers.
Ex. 24.1 | Q 6 | Page 9
Find the mean of x, x + 2, x + 4, x +6, x + 8.
Ex. 24.1 | Q 7 | Page 9
Find the mean of first five multiples of 3.
Ex. 24.1 | Q 8 | Page 9
Following are the weights (in kg) of 10 new born babies in a hospital on a particular day:
3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. Find the mean x.
Ex. 24.1 | Q 9 | Page 9
The percentage of marks obtained by students of a class in mathematics are : 64, 36, 47, 23,
0, 19, 81, 93, 72, 35, 3, 1. Find their mean.
Ex. 24.1 | Q 10 | Page 9
The numbers of children in 10 families of a locality are:
2, 4, 3, 4, 2, 0, 3, 5, 1, 1, 5. Find the mean number of children per family.
Ex. 24.1 | Q 11 | Page 9
Explain, by taking a suitable example, how the arithmetic mean alters by
(i) adding a constant k to each term
(ii) subtracting a constant k from each them
(iii) multiplying each term by a constant k and
(iv) dividing each term by a non-zero constant k.
Ex. 24.1 | Q 12 | Page 9
The mean of marks scored by 100 students was found to be 40. Later on it was discovered
that a score of 53 was misread as 83. Find the correct mean.
Ex. 24.1 | Q 13 | Page 9
The traffic police recorded the speed (in kmlhr) of 10 motorists as 47, 53, 49, 60, 39, 42, 55,57, 52, 48. Later on an error in recording instrument was found. Find the correct overagespeed of the motorists if the instrument recorded 5 km/hr less in each case.
Ex. 24.1 | Q 14 | Page 9
The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the
excluded number.
Ex. 24.1 | Q 15 | Page 9
The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6
of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.
Ex. 24.1 | Q 16 | Page 10
The mean weight of 8 numbers is 15. If each number is multiplied by 2, what will be the new mean?
Ex. 24.1 | Q 17 | Page 10
The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
Ex. 24.1 | Q 18 | Page 10
The mean of 200 items was 50. Later on, it was discovered that the two items were misread
as 92 and 8 instead of 192 and 88. Find the correct mean.
Ex. 24.1 | Q 19 | Page 10
If M is the mean of x1 , x2 , x3 , x4 , x5 and x6, prove that
(x1 − M) + (x2 − M) + (x3 − M) + (x4 − M) + (x5 — M) + (x6 − M) = 0.
Ex. 24.1 | Q 20 | Page 10
Durations of sunshine (in hours) in Amritsar for first 10 days of August 1997 as reported by
the Meteorological Department are given below: 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9
(i) Find the mean 𝑋 ̅
(ii) Verify that = sum _ ( i = 1)^10(xi - x ) = 0
Ex. 24.1 | Q 21 | Page 10
Find the values of n and X in each of the following cases :
(i) sum _(i = 1)^n(xi - 12) = - 10 sum _(i = 1)^n(xi - 3) = 62
(ii) sum _(i = 1)^n (xi - 10) = 30 sum _(i = 6)^n (xi - 6) = 150 .
Ex. 24.1 | Q 22 | Page 10
The sums of the deviations of a set of n values 𝑥1, 𝑥2, … . 𝑥11 measured from 15 and −3 are − 90 and 54 respectively. Find the valùe of n and mean.
Ex. 24.1 | Q 23 | Page 10
Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean.
Ex. 24.1 | Q 24 | Page 10
If x is the mean of the ten natural numbers x_1, x_2 , x_3+....... + x_10 show that (x1 -x) + (x2 - x) +.........+ (x10 - x) = 0
#### RD Sharma solutions for Mathematics for Class 9 Chapter 24 Measures of Central TendencyExercise 24.2 [Pages 14 - 16]
Ex. 24.2 | Q 1 | Page 14
Calculate the mean for the following distribution:
x : 5 6 7 8 9 f : 4 8 14 11 3
Ex. 24.2 | Q 2 | Page 14
Find the mean of the following data:
x : 19 21 23 25 27 29 31 f : 13 15 16 18 16 15 13
Ex. 24.2 | Q 3 | Page 14
Find the mean of the following distribution:
x : 10 12 20 25 35 F : 3 10 15 7 5
Ex. 24.2 | Q 4 | Page 15
Five coins were simultaneously tossed 1000 times and at each toss the number of heads wereobserved. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.
No. of heads per toss No.of tosses 0 38 1 144 2 342 3 287 4 164 5 25 TOtal 1000
Ex. 24.2 | Q 5 | Page 15
The mean of the following data is 20.6. Find the value of p.
x : 10 15 p 25 35 f : 3 10 25 7 5
Ex. 24.2 | Q 6 | Page 15
If the mean of the following data is 15, find p.
x: 5 10 15 20 25 f : 6 p 6 10 5
Ex. 24.2 | Q 7 | Page 15
Find the value of p for the following distribution whose mean is 16.6
x: 8 12 15 p 20 25 30 f : 12 16 20 24 16 8 4
Ex. 24.2 | Q 8 | Page 15
Find the missing value of p for the following distribution whose mean is 12.58.
x 5 8 10 12 p 20 25 f 2 5 8 22 7 4 2
Ex. 24.2 | Q 9 | Page 15
Find the missing frequency (p) for the following distribution whose mean is 7.68.
x 3 5 7 9 11 13 f 6 8 15 p 8 4
Ex. 24.2 | Q 10 | Page 15
Find the value of p, if the mean of the following distribution is 20.
x: 15 17 19 20+p 23 f: 2 3 4 5p 6
Ex. 24.2 | Q 11 | Page 15
Candidates of four schools appear in a mathematics test. The data were as follows:
Schools No. of candidates Average score 1 60 75 2 48 80 3 N A 55 4 40 50
If the average score of the candidates of all the four schools is 66, find the number of
candidates that appeared from school 3.
Ex. 24.2 | Q 12 | Page 16
Find the missing frequencies in the following frequency distribution if its known that the mean of the distribution is 50.
x 10 30 50 70 90 f 17 f1 32 f2 19 Total =120
#### RD Sharma solutions for Mathematics for Class 9 Chapter 24 Measures of Central TendencyExercise 24.3 [Page 18]
Ex. 24.3 | Q 1 | Page 18
Find the median of the following data (1-8)
83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Ex. 24.3 | Q 2 | Page 18
Find the median of the following data (1-8)
133, 73, 89, 108, 94, 1O4, 94, 85, 100, 120
Ex. 24.3 | Q 3 | Page 18
Find the median of the following data (1-8)
31 , 38, 27, 28, 36, 25, 35, 40
Ex. 24.3 | Q 4 | Page 18
Find the median of the following data (1-8)
15, 6, 16, 8, 22, 21, 9, 18, 25
Ex. 24.3 | Q 5 | Page 18
Find the median of the following data (1-8)
41, 43, 127, 99, 71, 92, 71, 58, 57
Ex. 24.3 | Q 6 | Page 18
Find the median of the following data (1-8)
25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Ex. 24.3 | Q 7 | Page 18
Find the median of the following data (1-8)
12, 17, 3, 14, 5, 8, 7, 15
Ex. 24.3 | Q 8 | Page 18
Find the median of the following data (1-8)
92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Ex. 24.3 | Q 9 | Page 18
Numbers 50, 42, 35, 2x + 10, 2x − 8, 12, 11, 8 are written in descending order and their
median is 25, find x.
Ex. 24.3 | Q 10 | Page 18
Find the median of the following observations : 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Ex. 24.3 | Q 11 | Page 18
Find the median of the following data : 41, 43, 127, 99, 61, 92, 71, 58, 57 If 58 is replaced
by 85, what will be the new median.
Ex. 24.3 | Q 12 | Page 18
The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42,30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.
Ex. 24.3 | Q 13 | Page 18
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
#### RD Sharma solutions for Mathematics for Class 9 Chapter 24 Measures of Central TendencyExercise 24.4 [Page 20]
Ex. 24.4 | Q 1 | Page 20
Find out the mode of the following marks obtained by 15 students in a class:
Marks : 4, 6, 5, 7, 9, 8, 10, 4, 7, 6, 5, 9, 8, 7, 7.
Ex. 24.4 | Q 2 | Page 20
Find the mode from the following data:
125, 175, 225, 125, 225, 175, 325, 125, 375, 225, 125
Ex. 24.4 | Q 3 | Page 20
Find the mode for the following series :
7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7,7.5, 7.3, 7.2, 7.6, 7.2
Ex. 24.4 | Q 4.1 | Page 20
Find the mode of the following data :
14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
Ex. 24.4 | Q 4.2 | Page 20
Find the mode of the following data :
7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18, 7
Ex. 24.4 | Q 5 | Page 20
The demand of different shirt sizes, as obtained by a survey, is given below:
Size: 38 39 40 41 42 43 44 Total No of persons(wearing it) 26 39 20 15 13 7 5 125
Find the modal shirt sizes, as observed from the survey.
#### RD Sharma solutions for Mathematics for Class 9 Chapter 24 Measures of Central Tendency [Page 21]
Q 1 | Page 21
If the ratio of mode and median of a certain data is 6 : 5, then find the ratio of its mean and median.
Q 2 | Page 21
If the mean of x + 2, 2x + 3, 3x + 4, 4x + 5 is x + 2, find x.
Q 3 | Page 21
If the median of scores $\frac{x}{2}, \frac{x}{3}, \frac{x}{4}, \frac{x}{5}$ and $\frac{x}{6}$ (where x > 0) is 6, then find the value of $\frac{x}{6}$ .
Q 4 | Page 21
If the mean of 2, 4, 6, 8, xy is 5, then find the value of x + y.
Q 5 | Page 21
If the mode of scores 3, 4, 3, 5, 4, 6, 6, x is 4, find the value of x.
Q 6 | Page 21
If the median of 33, 28, 20, 25, 34, x is 29, find the maximum possible value of x.
Q 7 | Page 21
If the median of the scores 1, 2, x, 4, 5 (where 1 < 2 < x < 4 < 5) is 3, then find the mean of the scores.
Q 8 | Page 21
If the ratio of mean and median of a certain data is 2:3, then find the ratio of its mode and mean
Q 9 | Page 21
The arithmetic mean and mode of a data are 24 and 12 respectively, then find the median of the data.
Q 10 | Page 21
If the difference of mode and median of a data is 24, then find the difference of median and mean.
#### RD Sharma solutions for Mathematics for Class 9 Chapter 24 Measures of Central Tendency [Pages 21 - 22]
Q 1 | Page 21
Mark the correct alternative in each of the following:
Which one of the following is not a measure of central value?
• Mean
• Range
• Median
• Mode
Q 2 | Page 21
The mean of n observations is X. If k is added to each observation, then the new mean is
• X
• X + k
• X − k
• kX
Q 3 | Page 22
The mean of n observations is X. If each observation is multiplied by k, the mean of new observations is
• k bar(X)
• bar(X)/k
• bar(X) +k
• bar(X)- k`
Q 4 | Page 22
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is
• 98
• 99
• 100
• 101
Q 5 | Page 22
For which set of numbers do the mean, median and mode all have the same value?
• 2, 2, 2, 2, 4
• 1, 3, 3, 3, 5
• 1, 1, 2, 5, 6
• 1, 1, 1, 2, 5
Q 6 | Page 22
For the set of numbers 2, 2, 4, 5 and 12, which of the following statements is true?
• Mean = Median
• Mean > Mode
• Mean > Mode
• Mode = Median
Q 7 | Page 22
If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value x is
• 10
• 12
• 14
• 16
Q 8 | Page 22
If the mean of five observations x, x+2, x+4, x+6, x+8, is 11, then the mean of first three observations is
• 9
• 11
• 13
• none of these
Q 9 | Page 22
Mode is
• least frequent value
• middle most value
• most frequent value
• none of these
Q 10 | Page 22
The following is the data of wages per day : 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8
The mode of the data is
• 7
• 5
• 8
• 10
Q 11 | Page 22
The median of the following data : 0, 2, 2, 2, −3, 5, −1, 5, −3, 6, 6, 5, 6 is
• 0
• −1.5
• 2
• 3.5
Q 12 | Page 22
The algebraic sum of the deviations of a set of n values from their mean is
• 0
• n − 1
• n
• n + 1
Q 13 | Page 22
A, B, C are three sets of values of x
(a) A: 2, 3, 7, 1, 3, 2, 3
(b) 7, 5, 9, 12, 5, 3, 8
(c) 4, 4, 11, 7, 2, 3, 4
Which one of the following statements is correct?
• Mean of A = Mode of C
• Mean of C = Median of B
• Median of B = Mode of A
• Mean, Median and Mode of A are equal.
Q 14 | Page 22
The empirical relation between mean, mode and median is
• Mode = 3 Median − 2 Mean
• Mode = 2 Median − 3 Mean
• Median = 3 Mode − 2 Mean
• Mean = 3 Median − 2 Mode
Q 15 | Page 22
The mean of abcd and e is 28. If the mean of ac, and e is 24, What is the mean of band d?
• 31
• 32
• 33
• 34
## Chapter 24: Measures of Central Tendency
Ex. 24.1Ex. 24.2Ex. 24.3Ex. 24.4Others
## RD Sharma solutions for Mathematics for Class 9 chapter 24 - Measures of Central Tendency
RD Sharma solutions for Mathematics for Class 9 chapter 24 (Measures of Central Tendency) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics for Class 9 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Mathematics for Class 9 chapter 24 Measures of Central Tendency are Introduction of Statistics, Collection of Data, Presentation of Data, Graphical Representation of Data, Measures of Central Tendency.
Using RD Sharma Class 9 solutions Measures of Central Tendency exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 9 prefer RD Sharma Textbook Solutions to score more in exam.
Get the free view of chapter 24 Measures of Central Tendency Class 9 extra questions for Mathematics for Class 9 and can use Shaalaa.com to keep it handy for your exam preparation
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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Transcript
Ex 1.1,1 (v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x, y): x and y work at the same place} R = {(x, y): x and y work at the same place} Check reflexive Since x & x are the same person, they work at the same place So, (x, x) R R is reflexive. Check symmetric If x & y work at the same place then y & x also work at the same place If (x, y) R, (y, x) R. Hence, R is symmetric. Check transitive If x & y work at the same place and y & z work at the same place then x & z also work at the same place If (x, y) R and (y, z) R, (x, z) R R is transitive. Hence, R is reflexive, symmetric, and transitive Ex 1.1,1(v) (b) R = {(x, y): x and y live in the same locality} R = {(x, y): x and y live in the same locality} Check reflexive Since x & x are the same person, they live in the same locality So, (x, x) R R is reflexive. Check symmetric If x & y live in the same locality then y & x live in the same locality If (x, y) R, (y, x) R. Hence, R is symmetric. Check transitive If x & y live in the same locality and y & z live in the same locality then x & z also live in the same locality If (x, y) R and (y, z) R, (x, z) R R is transitive. Hence, R is reflexive, symmetric, and transitive Ex 1.1,1(v) (c) R = {(x, y): x is exactly 7 cm taller than y} R = {(x, y): x is exactly 7 cm taller than y} Check reflexive Since x & x are the same person, he cannot be taller than himself (x, x) R R is not reflexive. Check symmetric If x is exactly 7 cm taller than y. then, y is not taller than x. So, if (x, y) R , (y, x) R R is not symmetric. Check transitive If x is exactly 7 cm taller than y and y is exactly 7 cm taller than z. x is exactly 14 cm taller than z . So, x is not 7 cm taller than z So, If (x, y) R & (y, z) R, then(x, z) R R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. Ex 1.1,1(v) (d) R = {(x, y): x is wife of y} R = {(x, y): x is wife of y} Check reflexive Since x & x are the same person, she cannot be the wife of herself (x, x) R R is not reflexive. Check symmetric If x is the wife of y. then, y cannot be the wife of x. So, if (x, y) R , (y, x) R R is not symmetric. Check transitive If x is the wife of y then y cannot be the wife of anybody else So, if (x, y) R & (y, z) R, R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. Ex 1.1,1(v) (e) R = {(x, y): x is father of y} R = {(x, y): x is father of y} Check reflexive Since x & x are the same person, he cannot be the father of herself (x, x) R R is not reflexive. Check symmetric If x is the father of y. then, y cannot be the father of x. So, if (x, y) R , (y, x) R R is not symmetric. Check transitive If x is the father of y , & y is the father of z then x cannot be the father of z (he is the grandfather) So, if (x, y) R and (y, z) R. then (x, z) R R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.
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# 180 Days of Math for Fifth Grade Day 37 Answers Key
By accessing our 180 Days of Math for Fifth Grade Answers Key Day 37 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fifth Grade Answers Key Day 37
Directions Solve each problem.
Question 1.
Explanation:
we get the sum.
So, 52 + 18 = 70
Question 2.
10 × 12 _____
Explanation:
When we multiply the numbers 10 and 12,
we get the product = 120
Question 3.
Explanation:
Divided 54 by 6
54 ÷ 6 = 9
Question 4.
Arrange the numbers in ascending order.
956, 967, 942
_________________________________
942, 956, 967
Explanation:
When the numbers are arranged from smallest to the largest numbers are known as
Ascending order.
So, 942 is the smallest number in the given series.
Question 5.
Is $$\frac{1}{3}$$ greater than, less than, or equal to $$\frac{3}{6}$$?
______________
$$\frac{1}{3}$$ is less than $$\frac{3}{6}$$
Explanation:
$$\frac{1}{3}$$ = 0.3
$$\frac{3}{6}$$ = $$\frac{1}{2}$$ = 0.5
So, $$\frac{1}{3}$$ is less than $$\frac{3}{6}$$
Question 6.
(63 ÷ 7) – 10 = ___
– 1
Explanation:
The given equation below is a combined operation
(63 ÷ 7) – 10
= 9 – 10
= -1
Question 7.
9 + = 24
Explanation:
9 + = 24
Let the given unknown number be x
9 + x = 24
x = 24 – 9
x = 15
Question 8.
What is the area of the polygon?
______________
50m
Explanation:
The given polygon is rectangle.
The area of rectangle = length x width
length = 10m; width = 5m
A = 10 x 5 = 50m
Question 9.
Which 3-dimensional figure has 4 triangular faces?
_____________________
3-dimensional triangular pyramid.
Explanation:
A triangular pyramid has 4 faces,
one in the base and 3 lateral faces.
Question 10.
Favorite Foods
Tacos Spaghetti Pizza Hot Dogs 17 18 26 11
Which food was the favorite?
_____________________
Pizza
Explanation:
From the above table of favorite foods Pizza has more number of people,
when compared with other foods.
Question 11.
You have a bag of 12 marbles. Six are blue, two are green, three are yellow, and one is red. If you reach into the bag and pull out one marble, what is the probability that it will be yellow?
_____________________
0.25 yellow marbles
Explanation:
Bag contain 12 MARBLES
BLUE 6
GREEN 2
YELLOW 3
RED 1
TOTAL = 6 + 2 + 3 + 1 = 12
$$\frac{3}{12}$$ = 0.25
Question 12.
It took 10 minutes to set up a board game. It took 45 minutes to play the game. The game ended at 2:15. At what time did the game begin?
_____________________
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# Newton’s Method
6.006 Intro to Algorithms Recitation 12 October 26, 2011
## Newton’s Method
Find root of f ( x ) = 0 through successive approximation e.g., f ( x ) = x
2
− a x i x i+1 y = f(x)
Figure 1 :
Newton’s Method.
Tangent at ( x i
, f ( x i
)) is line y = f ( x i
) + f
0
( x i
)
·
( x
− x i
) where f
0
( x i
) is the derivative.
x i +1
= intercept on x-axis x i +1
= x i
− f ( x i
) f
0
( x i
)
### Square Roots
We want to find the square root of a number
Newton’s Method, so that our answer x a through successive approximation using will converge on a
1 / 2
. It looks like f ( x ) = a
1 / 2
− x has a root at x = a
1 / 2
. Let’s try to use it with Newton’s Method.
f ( x ) = f
0
( x ) =
χ i +1
=
=
=
= a
1 / 2
− x
1
χ i
− f ( χ i
) f
0
( χ i
) a
1 / 2
χ
χ i
χ i
+ a
1 / 2
1
χ i a
1 / 2
The choice of this function was unsuccessful, since we’re unable to compute
Let’s try squaring both sides of x = x i +1 using x i
.
a
1 / 2
. This time, it looks like we get something usable:
1
6.006 Intro to Algorithms Recitation 12 October 26, 2011 f ( x ) = x
2
− a
χ i +1
= χ i
( χ i
2
− a )
2 χ i
= a
χ i
+
χ i
2
### Error Analysis of Newton’s Method
Suppose
X n
=
√ a
·
(1 + n
) n may be + or -
Then,
X n +1
=
=
=
=
=
X n
+ a/X n
2 a (1 + n
) +
√ a a (1+ n
)
2 p
( a )
(1 + n
) +
1
(1+ n
) p p
(
( a a
)
)
2 + 2 n
2
+ n
2
2(1 + n
2 n
)
1 +
2(1 + n
)
Therefore, n +1
= n
2
2(1 + n
)
Quadratic convergence, as ] correct digits doubles each step.
### Complexity of Computing Square Roots
We apply a first level of Newton’s method to solve f ( x ) = x
2
− a
. Each iteration of this first level requires a division. If we set the precision to d digits right from the beginning, then convergence at the first level will require lg d iterations. This means the complexity of computing a square root will be Θ( d
α lg d ) if the complexity of multiplication is Θ( d
α
) , given that we have shown that the complexity of division is the same as the complexity of multiplication.
However, we can do better, if we recognize that the number of digits of precision we need at beginning of the first level of Newton’s method starts out small and then grows. If the complexity of a d -digit division is Θ( d
α
) , then a similar summation to the one above tells us that the complexity of computing square roots is Θ( d
α
) .
2
6.006 Intro to Algorithms Recitation 12 October 26, 2011
### Termination
χ i
+ b a/χ i c
Iteration:
χ i +1
= b c
2
Do floors hurt? Does program terminate? ( α and β are the fractional parts below.)
Iteration is
χ i +1
=
=
χ i
+
χ i
+ a
χ i
2 a
χ i
2
α
β
γ where γ =
α
2
+ β and 0
γ < 1
Since a + b
√ ab,
χ i
2 won’t stay stuck above if
+
2 i a
χ i
< 1
√ a
, so subtracting
γ
(good initial guess).
always leaves us
≥ b
√ a c
. This
### Cube Roots
Now that we’ve seen square roots, let’s calculate cube roots. This time, x a
1 / 3
, instead. We can use f ( x ) = a
− x
3
: will converge on f ( x ) = x
3 f
0
( x ) = 3 x
2
− a
χ i +1
=
=
=
χ i
− f ( χ i
) f
0
( χ i
)
χ i
1
3
χ i
− a
χ
2 i
2
3
χ i
+
1 a
3 χ
2 i
3
6.006 Intro to Algorithms Recitation 12 October 26, 2011
Suppose x i +1
Then,
= a
1 / 3
·
(1 + i
) i may be + or -
χ i +1
=
=
=
=
=
=
=
2
3
χ i
+
1 a
3 χ
2 i
2 a
3
1 / 3
(1 + i
) +
1 a
3 ( a
1 / 3
(1 + n
))
2 a
1 / 3
3
2(1 + i
) +
(1 +
1 i
)
2 a
1 / 3
1
2(1 + 3
3 (1 + i
)
2 a
1 / 3
3
3 + 6 i
+ 6
1 + 2 i
2 i
+ 2
+
2 i a
1 / 3
3 +
3
2 i
+ 2
3 i
3 1 + 2 i
+
2 i
3 i a
1 / 3
1 +
2 i
+
(1 +
2
3
3 i i
)
2 i
+ 3
2 i
+
3 i
) + 1
Notice that the cubic term gets dominated by the quadratic term in the numerator, and that the denominator is very close to 1. Therefore, we have quadratic convergence here, just like in the square root case. The running time analysis is very similar to the square root case as well: the only difference is one additional multiplication in the denominator, so the running time stays the same as the running time of a multiplication.
4
MIT OpenCourseWare http://ocw.mit.edu
Fall 201 1
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Lesson Video: Work Done by a Constant Force | Nagwa Lesson Video: Work Done by a Constant Force | Nagwa
# Lesson Video: Work Done by a Constant Force Mathematics
In this video, we will learn how to calculate the work done by a constant force acting on a particle.
14:04
### Video Transcript
In this video, we’ll learn how to calculate the work done by a constant force acting on a particle. As we’ll see, this work done can be positive, negative, or zero. This has to do with the direction a particle moves compared to the direction of the force acting on it.
To get started, let’s consider this situation of a box of some mass at rest on the ground next to a table. Say that we walk up to this box and then pick it up, applying a constant vertically upward force to do this. If we know the strength of that constant force, we’ll call it 𝐹, and the distance through which the box moved, we’ll call that 𝑑, then we can compute the work done on the box by our force 𝐹 according to this relationship.
In this equation, 𝑑 is technically a displacement rather than a distance. That’s important because it means the direction of 𝑑 is taken into account. In this case of us lifting up the box, the fact that 𝑑 and 𝐹 are in the same direction means that the overall work that we do on the box during this process is positive. We can represent that like this. When we apply a constant force 𝐹 to lift a box up, and that box does indeed go up, then the work done on the box by the force 𝐹 is positive. In terms of its magnitude, it would be the magnitude of the force 𝐹 multiplied by this distance 𝑑.
Next, let’s say that we place this box on top of the table. And let’s imagine further that this tabletop is smooth, that there’s essentially no friction between the box and the table. In that case, it would take literally no effort at all to push this box horizontally along the table. Here, the distance the box travels is to the right, while the force needed to create this motion is zero. This means that sliding our box across our smooth tabletop requires zero work.
But now let’s imagine a third motion for this box. Say we catch the box at the other side of the table. And then applying the same vertically upward force as before on the box, this time, we let it move downward at a constant speed under the force of gravity. Just like before then, the force we apply points up. But now the box moves down instead of up. And because the force and the box’s displacement are in opposite directions, this means the work that our applied force 𝐹 does on the box is actually negative.
As we’ve seen, the reason for this is that 𝐹 and 𝑑 are pointing in opposite directions. When it comes to work then, we do have this formula where we calculate work by multiplying 𝐹 and 𝑑. But we see that the outcome of this calculation depends on the relative directions of these two factors. And in fact, it’s only the parallel or antiparallel components of these factors that go into calculating work.
For example, say that we applied a force in this direction on our box. If the box moved only horizontally, as a result of this applied force, then we could say that it’s only the horizontal component of this applied force that did any work. The vertical component did no work. And we can see this because in this vertical dimension, the distance traveled 𝑑 is zero.
Now, it’s worth pointing out that even though we are talking about constant forces being applied to different objects, it’s still possible that this may accelerate the objects in question. For example, if we say that the floor, like our tabletop, is smooth so that this box slides frictionlessly across it, then that means there will be a nonzero net force in the horizontal direction. And this may remind us of Newton’s second law of motion, which says that the net force acting on an object in a given dimension equals that object’s mass times its acceleration in that dimension.
All this to say the force in our work equation may be an individual force or it may be a net force, or they may even be the same thing. This shows us that when we talk about work, it’s important to be specific about what force is doing that work.
Recall that when we were lowering our box from approximately here to here, the work we did by applying an upward force was negative because that force and the distance the box moved were in opposite directions. However, if we considered instead the work done by the force of gravity acting down on the box, in that case, we would have a downward-acting force with a downward-pointing distance. So the work done by gravity on the box would be positive.
Let’s get some practice with these ideas now through an example.
Calculate the work done by a force of 13 newtons acting on a body that moved 40 meters to the north if the force was acting towards the south. State your answer in joules.
Okay, so if we say that north points upward, like this, and we imagine we’re looking down on some body from above, then we’re told that this particular body moves 40 meters to the north and that all throughout that motion the body is subject to a 13-newton force acting southward. We want to calculate the work done by this 13-newton force.
We can recall that work is equal to force multiplied by displacement. Sometimes we hear this factor called a distance. But really, it’s important to keep in mind the direction of our object’s motion. A good example of that is our current situation. If we consider the northern direction to be positive, then we would say our body’s total displacement is positive 40 meters. This implies though that the southern direction is considered negative. And so we would say that the 13-newton force, we’ll call it 𝐹, is negative 13 newtons.
All this means that when we use this value of 𝐹 and 𝑑 in our equation for work, we have negative 13 newtons being multiplied by positive 40 meters. A newton times a meter is a joule. And so our answer comes out to negative 520 joules. This is the work done by the 13-newton force acting on this body.
Now let’s look at an example involving Newton’s second law of motion.
A force acted on a body of mass 400 grams which had been at rest, causing it to accelerate at 36 centimeters per second squared. If the work done by this force was 0.72 joules, find the distance the body moved.
Alright, so let’s say that this is our body, which starts out at rest, and then a force is applied to it. So it starts to accelerate. And we’ll call this acceleration 𝑎. We’re told the mass of our body, we’ll call that 𝑚, as well as the work done on it, we’ll call it 𝑊, by this force. Knowing that this work was done while the body moved across some distance, we’ll call it 𝑑, it’s that distance we want to solve for.
To start doing this, we can recall that work is equal to force multiplied by distance. Strictly speaking, this 𝑑 is a displacement. But in this case, we can treat it as a distance, a quantity with no direction associated with it. Along with this equation for work, we can recall Newton’s second law of motion. It tells us that the net force acting on an object is equal to that object’s mass multiplied by its acceleration.
In this exercise, we’re told that this force that acts on our body is what causes it to accelerate at this given rate. This means the force does represent the net force acting on the body. And so we can replace 𝐹 in our equation for work with 𝑚 times 𝑎 from Newton’s second law. In this equation, it’s the distance 𝑑 that we want to solve for.
If we divide both sides of the equation by 𝑚 times 𝑎, those factors cancel on the right. And we find that 𝑑 equals 𝑊 over 𝑚 times 𝑎. If we substitute in the given values of 𝑊, 𝑚, and 𝑎, we find that, because of the units involved, we’re not quite ready to calculate 𝑑. We’ll want to convert the mass of our body to the SI base unit of kilograms and our body’s acceleration to units of meters per second squared.
Recalling that 1,000 grams equals one kilogram and that 100 centimeters is one meter, we can write an equivalent expression in updated units like this. Our body’s mass is 0.400 kilograms, while its acceleration is 0.36 meters per second squared. At this point, all of our units agree. So we can calculate 𝑑. We find it to equal exactly five meters. This is the distance our body moved while accelerating under the given conditions.
Let’s look now at another example.
A construction worker of mass 100 kilograms is carrying some bricks up a ladder of height 15 meters. If the work done by the construction worker in climbing the ladder is 20,433 joules, find the mass of the bricks. Take 𝑔 to equal 9.8 meters per second squared.
Okay, so let’s say that this is our ladder, which we’re told has a height we’ll call ℎ of 15 meters, and that on this ladder there’s a construction worker with a mass we’ll call 𝑚 sub 𝑐, who has climbed all the way to the top carrying a load of bricks with a mass we’ll call 𝑚 sub 𝑏. Knowing that this construction worker did 20,433 joules of work in climbing this ladder with this load of bricks, we want to determine the mass of the bricks, 𝑚 sub 𝑏.
If we think about the combined mass involved, the mass of the bricks plus the mass of the construction worker, we know that this total mass multiplied by the acceleration due to gravity is equal to the weight force that acts on this worker. But then, over against this force, the construction worker is able to climb up a height of 15 meters. We can say then that the magnitude of the force the construction worker exerts, we’ll call it 𝐹 sub 𝑐, is equal to this total mass times 𝑔.
At this point, we can recall that the work done by a given force 𝐹 is equal to that force multiplied by the displacement of an object on which the force acts. In our case, our body’s displacement is the height of our ladder ℎ, while the force involved is the sum of the masses of the bricks and the construction worker times 𝑔. Therefore, we can write this expression for the work done by the construction worker. Recalling that it’s 𝑚 sub 𝑏 that we want to solve for, we divide both sides of the equation by 𝑔 times ℎ. Those factors then are canceled on the right. And following that, we subtract 𝑚 sub 𝑐, the mass of the construction worker, from both sides. That results in this equation for the mass of the bricks. When we plug in for the values of 𝑊, 𝑔, ℎ, and 𝑚 sub 𝑐, we obtain a result of 39 kilograms. This is the mass of the bricks the worker carried to the top of the ladder.
Let’s look now at one last example involving work.
A man of mass 94 kilograms ascended a plane of length 90 meters, which was inclined at an angle of 30 degrees to the horizontal. Determine the work done by his weight to the nearest joule. Take 𝑔 to equal 9.8 meters per second squared.
Okay, so let’s say that this is our plane at 30 degrees to the horizontal. And there’s a man of mass 94 kilograms climbing up the plane. We’ll call that mass 𝑚. And we’re told that the length of the plane is 90 meters, which we’ll call 𝑙. We want to calculate the work done by this man’s weight to the nearest joule.
We can say then that this man’s weight, his mass times the acceleration due to gravity, a downward-acting force, actually does work as this man climbs up the incline. That work is equal to the force involved, that is, the man’s weight force, multiplied by the vertical distance through which the man moves. That is, the distance we’ll use in our calculation won’t be the length 𝑙. But rather it will be this vertical distance 𝑑. That’s because this distance represents the only component of the length 𝑙, we could call it, that is parallel or antiparallel to the force involved.
Here’s where we can start then. We know that the force for which we want to calculate work done is the man’s weight force, 𝑚 times 𝑔. Going a step further, since this 30-degree slope is part of a right triangle, we can say that the distance 𝑑 is equal to 𝑙 times the sin of 30 degrees. This is so because the sine of this angle is equal to the ratio of 𝑑 to 𝑙. Rearranging that relationship, we get that 𝑑 equals 𝑙 sin 30.
Before we plug in the given values for 𝑚, 𝑔, and 𝑙, it’s important to set up a sign convention. Note that as this man climbs the incline, he moves upward vertically. This is in the opposite direction to his weight force, which acts down. To make sure that we combine quantities correctly, let’s say that one of these directions is positive, making the other negative. If we say that the upward direction is positive, then that means the value of 𝑙 times the sin of 30 degrees is positive. But then that means that 𝑚 times 𝑔, which is directed downward, must be negative. The point here is that the force involved, the man’s weight force, is acting in an opposite direction to the displacement 𝑑. The net effect of this is that the work done by this weight force will be negative.
Now that we’ve figured out the signs, we’re ready to substitute in for 𝑚, 𝑔, and 𝑙. The man’s mass is 94 kilograms, 𝑔 is 9.8 meters per second squared, and 𝑙 is 90 meters. Entering this expression on our calculator, we get a result, to the nearest joule, of negative 41,454 joules. This, to the nearest joule, is the work done by this man’s weight as he ascends this plane.
Let’s finish up now by summarizing a few key points. In this video, we’ve seen that the work done by a constant force 𝐹 on some object is equal to the product of that force and the displacement that the object undergoes. For both 𝐹 and 𝑑, the direction is very important. In general, the work done by a constant force can be positive, negative, or zero. Lastly, we saw that it’s only the parallel or antiparallel components of 𝐹 and 𝑑 that contribute to the total work done by the force 𝐹.
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Back to all chapters
# Derivatives
A derivative is simply a rate of change. Whether you're modeling the movement of a particle or a supply/demand model, this is a key instrument of Calculus.
# Discrete First Derivatives Warmup
$\begin{array} {c|c|c|c|c|c|c|c} n & 1 & 2 & 3 & 4 & 5 & 6 & \cdots \\ a_n = n^2 & \color{red}{1} & \color{red}{4} & \color{red}{9} & 16 & 25 & 36 & \cdots \\ \Delta(a_n) = a_{n+1} - a_n & \color{blue}{3} & \color{blue}{5} & ???\\ \end{array}$
The second row of the table above shows the first few terms of the sequence $$a_n = n^2.$$ Each entry in the third row is the difference between the next term in row 2 and the current term in row 2. For example, $$\color{blue}{3} = \color{red}{4} -\color{red}{1}$$ and $$\color{blue}{5} = \color{red}{9} - \color{red}{4}.$$ What number should go in the cell marked '???'?
$\begin{array} {c|c|c|c|c|c|c|c} n & 1 & 2 & 3 & 4 & 5 & 6 & \cdots \\ a_n = n^3 & 1 & 8 & 27 & 64 & 125 & 216 & \cdots \\ \Delta(a_n) = a_{n+1} - a_n & 7 & 19 & 37 & 61 & ???\\ \end{array}$
The second row of the table above shows the first few terms of the sequence $$a_n = n^3.$$ Each entry in the third row is the difference between the next term in row 2 and the current term in row 2. For example, $$7 = 8 -1$$ and $$19 = 27 -8.$$ What number should go in the cell marked '???'?
$\begin{array} {c|c|c|c|c|c|c|c} n & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & n & n +1 \\ a_n = n^3 & 1 & 8 & 27 & 64 & 125 & 216 & \cdots &n^3 & (n+1)^3 \\ \Delta(a_n) = a_{n+1} - a_n & 7 & 19 & 37 & 61 & 91 & & \cdots & ? \\ \end{array}$
Given $$a_n = n^3,$$ find $$\Delta(a_n)$$.
In other words, what is the general formula for the discrete derivative of $$\{n^3\}?$$
Given $$a_n = n^2 + n,$$ find $$\Delta(a_n),$$ the discrete derivative of $$\{a_n\}.$$
Which of these sequences has a constant discrete derivative?
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Transcript of 6.2 Properties of Logarithms - Sam Houston State University...
• 6.2 Properties of Logarithms 437
6.2 Properties of Logarithms
In Section 6.1, we introduced the logarithmic functions as inverses of exponential functions and discussed a few of their functional properties from that perspective. In this section, we explore the algebraic properties of logarithms. Historically, these have played a huge role in the scientific development of our society since, among other things, they were used to develop analog computing devices called slide rules which enabled scientists and engineers to perform accurate calculations leading to such things as space travel and the moon landing. As we shall see shortly, logs inherit analogs of all of the properties of exponents you learned in Elementary and Intermediate Algebra. We first extract two properties from Theorem 6.2 to remind us of the definition of a logarithm as the inverse of an exponential function.
Theorem 6.3. (Inverse Properties of Exponential and Log Functions) Let b > 0, b 6= 1.
• ba = c if and only if logb(c) = a
• logb (bx) = x for all x and blogb(x) = x for all x > 0
Next, we spell out what it means for exponential and logarithmic functions to be one-to-one.
Theorem 6.4. (One-to-one Properties of Exponential and Log Functions) Let f(x) = bx and g(x) = logb(x) where b > 0, b 6= 1. Then f and g are one-to-one. In other words:
• bu = bw if and only if u = w for all real numbers u and w.
• logb(u) = logb(w) if and only if u = w for all real numbers u > 0, w > 0.
We now state the algebraic properties of exponential functions which will serve as a basis for the properties of logarithms. While these properties may look identical to the ones you learned in Elementary and Intermediate Algebra, they apply to real number exponents, not just rational exponents. Note that in the theorem that follows, we are interested in the properties of exponential functions, so the base b is restricted to b > 0, b 6= 1. An added benefit of this restriction is that it eliminates the pathologies discussed in Section 5.3 when, for example, we simplified
( x2/3
)3/2 and
obtained |x| instead of what we had expected from the arithmetic in the exponents, x1 = x.
Theorem 6.5. (Algebraic Properties of Exponential Functions) Let f(x) = bx be an exponential function (b > 0, b 6= 1) and let u and w be real numbers.
• Product Rule: f(u+ w) = f(u)f(w). In other words, bu+w = bubw
• Quotient Rule: f(u− w) = f(u) f(w)
. In other words, bu−w = bu
bw
• Power Rule: (f(u))w = f(uw). In other words, (bu)w = buw
While the properties listed in Theorem 6.5 are certainly believable based on similar properties of integer and rational exponents, the full proofs require Calculus. To each of these properties of
http://en.wikipedia.org/wiki/Slide_rule http://www.redorbit.com/news/space/73297/nasa_marks_35th_anniversary_of_first_moon_landing/
• 438 Exponential and Logarithmic Functions
exponential functions corresponds an analogous property of logarithmic functions. We list these below in our next theorem.
Theorem 6.6. (Algebraic Properties of Logarithm Functions) Let g(x) = logb(x) be a logarithmic function (b > 0, b 6= 1) and let u > 0 and w > 0 be real numbers.
• Product Rule: g(uw) = g(u) + g(w). In other words, logb(uw) = logb(u) + logb(w)
• Quotient Rule: g ( u w
) = g(u)− g(w). In other words, logb
( u w
) = logb(u)− logb(w)
• Power Rule: g (uw) = wg(u). In other words, logb (uw) = w logb(u)
There are a couple of different ways to understand why Theorem 6.6 is true. Consider the product rule: logb(uw) = logb(u) + logb(w). Let a = logb(uw), c = logb(u), and d = logb(w). Then, by definition, ba = uw, bc = u and bd = w. Hence, ba = uw = bcbd = bc+d, so that ba = bc+d. By the one-to-one property of bx, we have a = c + d. In other words, logb(uw) = logb(u) + logb(w). The remaining properties are proved similarly. From a purely functional approach, we can see the properties in Theorem 6.6 as an example of how inverse functions interchange the roles of inputs in outputs. For instance, the Product Rule for exponential functions given in Theorem 6.5, f(u+w) = f(u)f(w), says that adding inputs results in multiplying outputs. Hence, whatever f−1
is, it must take the products of outputs from f and return them to the sum of their respective inputs. Since the outputs from f are the inputs to f−1 and vice-versa, we have that that f−1 must take products of its inputs to the sum of their respective outputs. This is precisely what the Product Rule for Logarithmic functions states in Theorem 6.6: g(uw) = g(u)+g(w). The reader is encouraged to view the remaining properties listed in Theorem 6.6 similarly. The following examples help build familiarity with these properties. In our first example, we are asked to ‘expand’ the logarithms. This means that we read the properties in Theorem 6.6 from left to right and rewrite products inside the log as sums outside the log, quotients inside the log as differences outside the log, and powers inside the log as factors outside the log.1
Example 6.2.1. Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers.
1. log2
( 8
x
) 2. log0.1
( 10x2
) 3. ln
( 3
ex
)2
4. log 3
√ 100x2
yz5 5. log117
( x2 − 4
) Solution.
1. To expand log2 (
8 x
) , we use the Quotient Rule identifying u = 8 and w = x and simplify.
1Interestingly enough, it is the exact opposite process (which we will practice later) that is most useful in Algebra, the utility of expanding logarithms becomes apparent in Calculus.
• 6.2 Properties of Logarithms 439
log2
( 8
x
) = log2(8)− log2(x) Quotient Rule
= 3− log2(x) Since 23 = 8
= − log2(x) + 3
2. In the expression log0.1 ( 10x2
) , we have a power (the x2) and a product. In order to use the
Product Rule, the entire quantity inside the logarithm must be raised to the same exponent. Since the exponent 2 applies only to the x, we first apply the Product Rule with u = 10 and w = x2. Once we get the x2 by itself inside the log, we may apply the Power Rule with u = x and w = 2 and simplify.
log0.1 ( 10x2
) = log0.1(10) + log0.1
( x2 )
Product Rule
= log0.1(10) + 2 log0.1(x) Power Rule
= −1 + 2 log0.1(x) Since (0.1)−1 = 10
= 2 log0.1(x)− 1
3. We have a power, quotient and product occurring in ln (
3 ex
)2 . Since the exponent 2 applies
to the entire quantity inside the logarithm, we begin with the Power Rule with u = 3ex and w = 2. Next, we see the Quotient Rule is applicable, with u = 3 and w = ex, so we replace ln (
3 ex
) with the quantity ln(3) − ln(ex). Since ln
( 3 ex
) is being multiplied by 2, the entire
quantity ln(3)− ln(ex) is multiplied by 2. Finally, we apply the Product Rule with u = e and w = x, and replace ln(ex) with the quantity ln(e) + ln(x), and simplify, keeping in mind that the natural log is log base e.
ln
( 3
ex
)2 = 2 ln
( 3
ex
) Power Rule
= 2 [ln(3)− ln(ex)] Quotient Rule
= 2 ln(3)− 2 ln(ex)
= 2 ln(3)− 2 [ln(e) + ln(x)] Product Rule
= 2 ln(3)− 2 ln(e)− 2 ln(x)
= 2 ln(3)− 2− 2 ln(x) Since e1 = e
= −2 ln(x) + 2 ln(3)− 2
4. In Theorem 6.6, there is no mention of how to deal with radicals. However, thinking back to Definition 5.5, we can rewrite the cube root as a 13 exponent. We begin by using the Power
• 440 Exponential and Logarithmic Functions
Rule2, and we keep in mind that the common log is log base 10.
log 3
√ 100x2
yz5 = log
( 100x2
yz5
)1/3 = 13 log
( 100x2
yz5
) Power Rule
= 13 [ log ( 100x2
) − log
( yz5 )]
Quotient Rule
= 13 log ( 100x2
) − 13 log
( yz5 )
= 13 [ log(100) + log
( x2 )] − 13
[ log(y) + log
( z5 )]
Product Rule
= 13 log(100) + 1 3 log
( x2 ) − 13 log(y)−
1 3 log
( z5 )
= 13 log(100) + 2 3 log(x)−
1 3 log(y)−
5 3 log(z) Power Rule
= 23 + 2 3 log(x)−
1 3 log(y)−
5 3 log(z) Since 10
2 = 100
= 23 log(x)− 1 3 log(y)−
5 3 log(z) +
2 3
5. At first it seems as if we have no means of simplifying log117 ( x2 − 4
) , since none of the
properties of logs addresses the issue of expanding a difference inside the logarithm. However, we may factor x2 − 4 = (x + 2)(x − 2) thereby introducing a product which gives us license to use the Product Rule.
log117 ( x2 − 4
) = log117 [(x+ 2)(x− 2)] Factor = log117(x+ 2) + log117(x− 2) Product Rule
A couple of remarks about Example 6.2.1 are in order. First, while not explicitly stated in the above example, a general rule of thumb to determine which log property to apply first to a complicated problem is ‘reverse order of operations.’ For example, if we were to substitute a number for x into the expression log0.1
( 10x2
) , we would first square the x, then multiply by 10. The last step is the
multiplication, which tells us the first log property to apply is the Product Rule. In a multi-step problem, this rule can giv
|
# A solid metallic right circular cone 20 cm high and whose vertical angle is $60^{o}$, is cut into two parts at the middle of its height by a plane parallel to it base.If the frustum so obtained drawn into a wire of diameter $\frac{1}{12}$cm, find the length of the wire.
Given: A solid metallic right circular cone 20 cm high and whose vertical angle is $60^{o}$, is cut into two parts at the middle of its height by a plane parallel to it base. If the frustum so obtained drawn into a wire of diameter $\frac{1}{12}$cm.
To do: To find the length of the wire.
Solution:
Let ACB be the cone whose vertical angle $\angle ACB = 60^{o}$.
Let $R$ and $x$ be the radii of the lower and upper end of the frustum.
Here, height of the cone, $OC = 20 cm=H$
Height $CP = h = 10\ cm$
Let us consider P as the mid-Point of OC
Cutting the cone into two parts through P.
OP =$\frac{20}{2}= 10\ cm$
Also,$\angle ACO$ and $\angle OCB =$\frac{1}{2} \times 60^{o} =30^{o} $After cutting cone CQS from cone CBA, the remaining solid obtained is a frustum. Now, in triangle CPQ$tan30^{o}=\frac{x}{10}\frac{1}{\sqrt{3}} =\frac{x}{10}\Rightarrow x=\frac{10}{\sqrt{3}}\ cm$In triangle COB$tan30^{o}=\frac{R}{20}\Rightarrow \frac{1}{\sqrt{3}} =\frac{R}{20}\Rightarrow R=\frac{20}{\sqrt{3}}$Volume of the frustum,$V=\frac{1}{3} \pi \left( R^{2} H-x^{2} h\right)\Rightarrow V=\frac{1}{3} \pi \left(\left(\frac{20}{\sqrt{3}}\right)^{2} .20-\left(\frac{10}{\sqrt{3}}\right)^{2} .10\right)\Rightarrow V=\frac{1}{3} \pi \left(\frac{400\times 20}{3} -\frac{100\times 10}{3}\right)\Rightarrow V=\frac{1}{3} \pi \left(\frac{8000-1000}{3}\right)\Rightarrow V=\frac{7000}{9} \pi \ cm^{3}$Let us assume the length of the wire l. Given diameter of the wire obtained from the frustum$=\frac{1}{12}\ cm$Radius of the wire,$r=\frac{1}{2} \times \frac{1}{12} =\frac{1}{24}\ cm$Volume of the wire$=\pi r^{2} l=\pi \left(\frac{1}{24}\right)^{2} l=\frac{\pi l}{576} cm^{3}$The volumes of the frustum and the wire formed are equal,$\frac{7000}{9} \pi =\frac{\pi l}{576}\Rightarrow \frac{7000}{9} =\frac{l}{576}\Rightarrow l=\frac{7000\times 576}{9}\Rightarrow l=448000\ cm\Rightarrow l=4480\ cm\$
Therefore, length of the wire is 480 cm.
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# How do you calculate the total resistance of a parallel circuit?
## How do you calculate the total resistance of a parallel circuit?
The Effective Resistance of Resistors Connected in Parallel
1. There are three important characteristics in a parallel circuit:
(a) The potential difference is the same across each resistor.
(b) The current that passes through each resistor is inversely proportional to the resistance of the resistor.
(c) The total current in the circuit equals to the sum of the currents passing through the resistors in its parallel branches.
2. When two or more resistances are connected between two common points so that the same potential difference is applied across each of them, they are said to be connected is parallel.
When such a combination of resistance is connected to a battery, all the resistances have the same potential difference across their ends.
3. Derivation of mathematical expression of parallel combination:
Let, V be the potential difference across the two common points A and B. Then, from Ohm’s law
Current passing through R1, I1 = V/R1 … (i)
Current passing through R2, I2 = V/R2 … (ii)
Current passing through R3, I3 = V/R3 … (iii)
4. If R is the equivalent resistance, then from Ohm’s law, the total current flowing through the circuit is given by,
I = V/R … (iv)
and I = I1 + I2 + I3 … (v)
5. Substituting the values of I, I1, I2 and I3 in Eq. (v),
$$\frac{\text{V}}{\text{R}}=\frac{\text{V}}{{{\text{R}}_{\text{1}}}}+\frac{\text{V}}{{{\text{R}}_{\text{2}}}}+\frac{\text{V}}{{{\text{R}}_{\text{3}}}}\text{ }……..\text{ (vi)}$$
6. Cancelling common V term, one gets
$$\frac{\text{1}}{\text{R}}=\frac{\text{1}}{{{\text{R}}_{\text{1}}}}+\frac{\text{1}}{{{\text{R}}_{2}}}+\frac{\text{1}}{{{\text{R}}_{3}}}$$
The equivalent resistance of a parallel combination of resistance is less than each of all the individual resistances.
7. The equivalent circuit is shown in Figure.
Important results about parallel combination:
1. Total current through the circuit is equal to the sum of the currents flowing through it.
2. In a parallel combination of resistors the voltage (or potential difference) across each resistor is the same and is equal to the applied voltage i.e. V1 = V2 = V3 = V.
3. Current flowing through each resistor is inversely proportional to its resistances, thus higher the resistance of a resistors, lower will be the current flowing through it.
|
# Calculating The Number Of Days In 30 Years | Understanding Leap Years
//
Thomas
Learn how to calculate the number of days in 30 years, including leap years. Understand the importance of knowing the total days for financial planning, project management, and historical analysis.
## Calculating the Number of Days in 30 Years
Calculating the number of days in 30 years may seem like a daunting task, but fear not! We’ll break it down step by step to make it easy for you to understand. So let’s dive in!
### Days in a Year
Before we can calculate the number of days in 30 years, let’s first understand how many days are in a single year. In most cases, a year consists of 365 days. However, there are exceptions to this rule, which we’ll explore in the next subsection.
### Leap Years
Leap years are the exception to the standard 365-day year. These special years occur every four years and have an extra day, making them 366 days long. The additional day is added to the month of February, and it’s called a leap day. This adjustment is necessary to keep our calendar in sync with the Earth’s orbit around the sun.
### Calculation Formula
Now that we know about leap years, let’s move on to the calculation formula for determining the number of days in a given number of years, including both regular and leap years. To calculate the days in 30 years, we’ll need to consider the following:
1. Regular years: Multiply the number of regular years (non-leap years) by 365.
2. Leap years: Multiply the number of leap years by 366.
3. Add the results of step 1 and step 2 together to get the total number of days.
### Example Calculation
To illustrate the calculation formula, let’s take an example. Suppose we want to calculate the number of days in 30 years.
First, we determine the number of regular years and the number of leap years within this period. Since a leap year occurs every four years, we have 7 leap years (30 divided by 4, rounded down) and 23 regular years.
Next, we calculate the days in these years:
– Regular years: 23 years multiplied by 365 days equals 8,395 days.
– Leap years: 7 years multiplied by 366 days equals 2,562 days.
Finally, we add the results together:
8,395 days (regular years) + 2,562 days (leap years) equals a total of 10,957 days in 30 years.
Congratulations! You now know how to calculate the number of days in 30 years. This knowledge can come in handy in various scenarios, such as financial planning, project management, and historical analysis. So let’s explore why it’s important to have this information in the next section.
## Understanding Leap Years
Leap years are an important concept to understand when calculating the number of days in a 30-year period. Let’s dive into the details of what leap years are, how they are calculated, and how many leap years occur in a 30-year span.
### Definition of a Leap Year
A leap year is a year that contains an extra day, February 29th, making it 366 days instead of the usual 365. This adjustment is necessary to keep our calendar in sync with the Earth’s orbit around the sun. Without leap years, our calendar would gradually fall out of alignment with the astronomical year.
### Calculation of Leap Years
The calculation of leap years follows a specific rule. A leap year occurs every four years, except for years that are divisible by 100. However, years that are divisible by 400 are still considered leap years. This rule ensures that the calendar remains accurate over long periods.
To determine if a year is a leap year, we use the following criteria:
1. If a year is divisible by 4, it is a leap year.
2. If a year is divisible by 100, it is not a leap year, unless…
3. If a year is divisible by 400, it is a leap year.
For example, the year 2020 is a leap year because it is divisible by 4. However, the year 1900 was not a leap year because it is divisible by 100 but not by 400.
### Leap Years in a 30-Year Period
Now that we understand how leap years are calculated, let’s determine how many occur in a 30-year span. To do this, we divide the number of years by 4 and subtract any that fall within the divisible-by-100 exception.
In a 30-year period, there are typically 7 leap years. However, it’s important to consider the exceptions. For example, if the 30-year period includes a year divisible by 100 but not by 400, we subtract one leap year from the total count.
Calculating the number of leap years in a 30-year span allows us to accurately determine the total number of days in that period, taking into account both regular years and .
Understanding leap years is crucial when performing various calculations related to time, such as financial planning, project management, and historical analysis. Now that we have a solid foundation on leap years, we can explore further aspects of calculating the number of days in a 30-year period.
## Calculation Method for Days in a Year
### Days in a Non-Leap Year
In order to calculate the number of days in a non-leap year, we need to understand the basic structure of a calendar year. A non-leap year consists of 365 days, which are divided into 12 months. Most months have 30 or 31 days, except for February, which has 28 days in a non-leap year.
To calculate the number of days in a non-leap year, we can simply add up the days in each month. Here is a breakdown of the days in each month:
• January: 31 days
• February: 28 days
• March: 31 days
• April: 30 days
• May: 31 days
• June: 30 days
• July: 31 days
• August: 31 days
• September: 30 days
• October: 31 days
• November: 30 days
• December: 31 days
By adding up the days in each month, we can determine that a non-leap year has a total of 365 days.
### Days in a Leap Year
A leap year, on the other hand, has one extra day compared to a non-leap year. This extra day is added to the month of February, making it 29 days instead of the usual 28 days.
To calculate the number of days in a leap year, we can use the same method as for a non-leap year, but with a modification for the month of February. Here is the breakdown of the days in each month in a leap year:
• January: 31 days
• February: 29 days
• March: 31 days
• April: 30 days
• May: 31 days
• June: 30 days
• July: 31 days
• August: 31 days
• September: 30 days
• October: 31 days
• November: 30 days
• December: 31 days
By adding up the days in each month, we can determine that a leap year has a total of 366 days.
Knowing the calculation method for days in a year is essential for various purposes, including financial planning, project management, and historical analysis. It allows us to accurately estimate timeframes, allocate resources, and analyze trends over extended periods.
## Calculation Formula for Days in 30 Years
### Calculation Steps
To calculate the number of days in 30 years, you need to follow a simple calculation formula. Here are the steps:
1. Start by determining the number of days in a year. This will depend on whether it’s a leap year or not, which we will explain in the next section.
2. Calculate the number of days in a non-leap year. Multiply the number of days in a non-leap year by the number of non-leap years in 30 years.
3. Calculate the number of days in a leap year. Multiply the number of days in a leap year by the number of leap years in 30 years.
4. Add the total number of days in non- and leap years together to get the final result.
### Example Calculation
Let’s take a moment to illustrate the calculation with an example:
1. Number of days in a non-leap year: 365 days
2. Number of days in a leap year: 366 days
Now, let’s assume we want to calculate the number of days in 30 years.
• Number of non-leap years in 30 years: 30 – 6 (leap years) = 24
• Number of in 30 years: 6
Calculating the days in non-: 365 days x 24 years = 8,760 days
Calculating the days in leap years: 366 days x 6 years = 2,196 days
Adding the total number of days in non-leap years and leap years: 8,760 days + 2,196 days = 10,956 days
Therefore, there are 10,956 days in 30 years.
By following these calculation steps and using the example above, you can easily determine the number of days in any given 30-year period. This knowledge can be useful in various scenarios, as we will explore in the following sections.
## Importance of Knowing the Number of Days in 30 Years
In our everyday lives, understanding the number of days in a 30-year period may not seem particularly important at first glance. However, this knowledge can have significant implications in various fields, ranging from financial planning to project management and historical analysis. Let’s explore how this seemingly simple calculation can provide valuable insights and aid decision-making in these areas.
### Financial Planning
When it comes to financial planning, having a clear understanding of the number of days in 30 years can be immensely beneficial. Consider scenarios where long-term investments or retirement planning are involved. By accurately calculating the number of days in three decades, financial professionals can make more informed decisions regarding investment durations, interest rates, and compounding periods. This knowledge allows for precise projections of returns, helping individuals and businesses plan for their future with greater confidence.
### Project Management
In the realm of project management, time is a critical factor. Knowing the exact number of days in a 30-year span enables project managers to create realistic timelines and set achievable goals. Whether it’s a construction project, software development, or event planning, having an accurate understanding of the time available allows for effective resource allocation, task scheduling, and milestone tracking. By incorporating the precise number of days in 30 years, project managers can enhance their ability to deliver projects on time and within budget.
### Historical Analysis
History holds a wealth of knowledge and lessons for us to learn from. Understanding the number of days in 30 years is essential for historians and researchers who analyze timelines and events over extended periods. By accurately calculating the number of days, historians can create more nuanced narratives, identify patterns, and gain deeper insights into historical phenomena. This knowledge enables a better understanding of long-term trends, societal changes, and the impact of historical events on subsequent generations.
In conclusion, while the number of days in 30 years may seem like a trivial detail, it holds great significance in various fields. From aiding financial planning and project management to enhancing historical analysis, this seemingly simple calculation provides valuable insights and facilitates informed decision-making. By recognizing the importance of knowing the number of days in 30 years, individuals and professionals can harness its power to make better choices and achieve their goals more effectively.
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# Vectors
Vectors. Some quantities can be described with only a number. These quantities have magnitude (amount) only and are referred to as scalar quantities. Scalar - a quantity which can be described with only a magnitude
## Vectors
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### Presentation Transcript
1. Vectors Some quantities can be described with only a number. These quantities have magnitude (amount) only and are referred to as scalar quantities. Scalar - a quantity which can be described with only a magnitude Quantities that are described with both a magnitude and a direction are called vector quantities. Vector - a quantity with both magnitude and direction
2. Vectors We can use a graphical method to help us picture vectors, and to allow us to understand how vectors are added and subtracted. We will represent vector quantities with arrows. The length of the arrow represents its magnitude, the arrow tip and the direction of the arrow represent the vector direction. tip tail The point of the arrow is referred to as the “Tip”. The non-pointed end is called the “Tail”
3. The length and direction gives all the info about a vector. Vectors A and B are completely equivalent. B = 12 m east A = 12 m east A = B Vectors A vector can be moved (translated) without changing it. As long as the length and direction do not change, the vector has not changed.
4. If vectors are to be worked with graphically, axes and an appropriate scale must be chosen. Suppose we are describing the motion of a bug walking on the ground. If the bug walks 12 m to the east, we call this the bug’s displacement. We would write this as A = 12 m east . Once we establish a coordinate system, we can represent this displacement vector with an arrow like the one below. N W E S Vectors A = 12 m east
5. Vector Addition The process of adding two or more vectors is called Vector Addition. The sum of two or more vectors is called the resultant vector.
6. N C B W E A S C = A + B Vector Addition “Triangle Method” of Vector Addition Steps for adding two vectors using the tip-to-tail method: 1) Draw the first vector (A) to scale, with its tail at the origin 2) Draw the second vector (B) also to scale, with its tail at the tip of the first vector 3) The resultant vector runs from the tail end of the first vector to the tip of the second vector.
7. N C B W E A S C = A + B “Parallelogram Method” of Vector Addition Vector Addition Steps for adding two vectors using the parallelogram method: 1) Draw the first vector (A) to scale, with its tail at the origin 2) Draw the second vector (B) also to scale, and also with its tail at the origin. 3) Starting at the tip of one vector, draw a dotted line parallel to the other vector. Repeat, starting from the tip of the second vector. 4) The resultant vector runs from the origin to the intersection of the two dotted lines.
8. Notice - whichever method you use, you get the same value for vector C ! N N C C B B W E W E A A S S C = A + B C = A + B Tip - to - Tail Method Parallelogram Method Vector Addition
9. N N B B W W E E C C A A S S A + B = C A + B = C Vector Addition The methods work no matter the direction of the original vectors. Tip - to - Tail Method Parallelogram Method
10. N N C D D B B W W E E C A A S S C + B + A = D A + B + C = D Vector Addition If you need to add more than two vectors, the tip-to-tail method is usually easiest. Note that the order the vectors are added does not affect the result.
11. A -A Vector Subtraction For any vector A, the vector -A is simply a vector equal in magnitude and opposite in direction to A.
12. Subtracting one vector from another is the same as adding a negative vector. A - B = A + (-B) addition: A + B A + B = C C B B -B -B A A A A subtraction: A - B A + (-B) = D D Vector Subtraction
13. N C B W E A S C = A + B Tip - to - Tail Method Vector addition: Mathematical Method To add two perpendicular vectors mathematically, make a sketch of the two vectors you are adding, and use either graphical method to sketch their resultant. To find the magnitude of the resultant, use the Pythagorean theorem. To find the direction of the resultant, use an inverse trig function, for example: tan-1 Tan = B/A = tan-1(B/A)
14. N CN C W E CE S Finding Vector Components: Mathematical Method To break a vector down into perpendicular components, first make a sketch of the vector. Sketch dotted lines from the tip of the vector and parallel to each axis. Draw in the components, from the origin to the dotted line. To calculate the magnitude of the components, use the sin and cos functions. Sin = CN/C C = CE + CN CN = C sin CE = C cos Cos = CE/C
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Composite Solids
Composite Solids
A composite solid is a solid that is composed, or made up of, two or more solids. The solids that it is made up of are generally prisms, pyramids, cones, cylinders, and spheres. In order to find the surface area and volume of a composite solid, you need to know how to find the surface area and volume of prisms, pyramids, cones, cylinders, and spheres.
Finding the Volume of a Parallelogram-Based Prism
Find the volume of the solid below.
This solid is a parallelogram-based prism with a cylinder cut out of the middle.
\begin{align*}V_{prism} &= (25 \cdot 25)30=18,750 \ cm^3\\ V_{cylinder} &= \pi (4)^2 (30)=480 \pi \ cm^3\end{align*}
The total volume is \begin{align*}18750 - 480 \pi \approx 17,242.04 \ cm^3\end{align*}.
Finding the Surface Area
Find the surface area of the following solid.
This solid is a cylinder with a hemisphere on top. Because it is one fluid solid, we would not include the bottom of the hemisphere or the top of the cylinder because they are no longer on the surface of the solid. Below, “\begin{align*}LA\end{align*}” stands for lateral area.
\begin{align*}SA&=LA_{cylinder}+LA_{hemisphere}+A_{base \ circle}\\ & = \pi rh+ \frac{1}{2} 4 \pi r^2+ \pi r^2\\ & = \pi (6)(13)+2 \pi 6^2+ \pi 6^2\\ & = 78 \pi +72 \pi +36 \pi \\ & = 186 \pi \ in^2\end{align*}
Finding the Volume
Find the volume of the following solid.
To find the volume of this solid, we need the volume of a cylinder and the volume of the hemisphere.
\begin{align*}V_{cylinder} & = \pi 6^2 (13)=78 \pi \\ V_{hemisphere}& = \frac{1}{2} \left ( \frac{4}{3} \pi 6^3 \right )=36 \pi\\ V_{total} & = 78 \pi +36 \pi =114 \pi \ in^3\end{align*}
Examples
Example 1
Find the volume of the composite solid. All bases are squares.
This is a square prism with a square pyramid on top. Find the volume of each separeatly and then add them together to find the total volume. First, we need to find the height of the pyramid portion. The slant height is 25 and the edge is 48. Using have of the edge, we have a right triangle and we can use the Pythagorean Theorem. \begin{align*}h=\sqrt{25^2-24^2}=7\end{align*}
\begin{align*}V_{prism}&=(48)(48)(18)=41472 \ cm^3\\ V_{pyramid}&=\frac{1}{3} (48^2)(7)=5376 \ cm^3\end{align*}
The total volume is \begin{align*}41472 + 5376 = 46,848 \ cm^3\end{align*}.
Example 2
Find the volume of the base prism. Round your answer to the nearest hundredth.
Use what you know about prisms.
\begin{align*}V_{prism}&=B \cdot h \\ V_{prism}&=(4\cdot 4)\cdot 5\\ V_{prism}&=80in^3\end{align*}
Example 3
Using your work from Example 2, find the volume of the pyramid and then of the entire solid.
Use what you know about pyramids.
\begin{align*}V_{pyramid}&=\frac{1}{3} B \cdot h \\ V_{pyramid}&=\frac{1}{3}(4 \cdot 4)(6)\\ V_{pyramid}&=32in^3\end{align*}
Now find the total volume by finding the sum of the volumes of each solid.
\begin{align*}V_{total}&=V_{prism}+V_{pyramid}\\ V_{total}&=112 in^3\end{align*}
Review
Find the volume of the composite solids below. Round your answers to the nearest hundredth.
1. The bases are squares. Find the volume of the green part.
2. A cylinder fits tightly inside a rectangular prism with dimensions in the ratio 5:5:7 and volume \begin{align*}1400 \ in^3\end{align*}. Find the volume of the space inside the prism that is not contained in the cylinder.
Find the surface area and volume of the following shapes. Leave your answers in terms of \begin{align*}\pi\end{align*}.
1. You may assume the bottom is open.
2. A sphere has a radius of 5 cm. A right cylinder has the same radius and volume. Find the height and total surface area of the cylinder.
Tennis balls with a 3 inch diameter are sold in cans of three. The can is a cylinder. Assume the balls touch the can on the sides, top and bottom.
1. What is the volume of one tennis ball?
2. What is the volume of the space not occupied by the tennis balls? Round your answer to the nearest hundredth.
One hot day at a fair you buy yourself a snow cone. The height of the cone shaped container is 5 in and its radius is 2 in. The shaved ice is perfectly rounded on top forming a hemisphere.
1. What is the volume of the ice in your frozen treat?
2. If the ice melts at a rate of \begin{align*}2 \ in^3\end{align*} per minute, how long do you have to eat your treat before it all melts? Give your answer to the nearest minute.
1. For exercise 12, answer the following:
1. What is the surface area of a cylinder?
2. Adjust your answer from part a for the case where \begin{align*}r = h\end{align*}.
3. What is the surface area of a sphere?
4. What is the relationship between your answers to parts b and c? Can you explain this?
Find the volume of the composite solids. Round your answer to the nearest hundredth.
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Color Highlighted Text Notes
Vocabulary Language: English
Pythagorean Theorem
The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.
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# What is (9 plus 15) plus divided by 3 plus 2?
Updated: 9/27/2023
Wiki User
6y ago
The rules given below transform a given number into a generally smaller number, while preserving divisibility by the divisor of interest. Therefore, unless otherwise noted, the resulting number should be evaluated for divisibility by the same divisor. In some cases the process can be iterated until the divisibility is obvious; for others (such as examining the last n digits) the result must be examined by other means.
For divisors with multiple rules, the rules are generally ordered first for those appropriate for numbers with many digits, then those useful for numbers with fewer digits.
Note: To test divisibility by any number that can be expressed as 2n or 5n, in which n is a positive integer, just examine the last n digits.
Divisor Divisibility condition Examples 1 Automatic. Any integer is divisible by 1. 2 The last digit is even (0, 2, 4, 6, or 8).[1][2]1,294: 4 is even. 3 Sum the digits.[1][3][4]405 → 4 + 0 + 5 = 9 and 636 → 6 + 3 + 6 = 15 which both are clearly divisible by 3.
16,499,205,854,376 → 1+6+4+9+9+2+0+5+8+5+4+3+7+6 sums to 69 → 6 + 9 = 15 → 1 + 5 = 6, which is clearly divisible by 3. Subtract the quantity of the digits 2, 5 and 8 in the number from the quantity of the digits 1, 4 and 7 in the number. Using the example above: 16,499,205,854,376 has four of the digits 1, 4 and 7; four of the digits 2, 5 and 8; ∴ Since 4 − 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3. 4 Examine the last two digits.[1][2]40832: 32 is divisible by 4. If the tens digit is even, and the ones digit is 0, 4, or 8.
If the tens digit is odd, and the ones digit is 2 or 6. 40832: 3 is odd, and the last digit is 2. Twice the tens digit, plus the ones digit. 40832: 2 × 3 + 2 = 8, which is divisible by 4. 5 The last digit is 0 or 5.[1][2]495: the last digit is 5. 6 It is divisible by 2 and by 3.[5]1,458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6. 7 Form the alternating sum of blocks of three from right to left.[6][4]1,369,851: 851 − 369 + 1 = 483 = 7 × 69 Subtract 2 times the last digit from the rest. (Works because 21 is divisible by 7.) 483: 48 − (3 × 2) = 42 = 7 × 6. Or, add 5 times the last digit to the rest. (Works because 49 is divisible by 7.) 483: 48 + (3 × 5) = 63 = 7 × 9. Or, add 3 times the first digit to the next. (This works because 10a + b − 7a = 3a + b − last number has the same remainder) 483: 4×3 + 8 = 20 remainder 6, 6×3 + 3 = 21. Multiply each digit (from right to left) by the digit in the corresponding position in this pattern (from left to right): 1, 3, 2, -1, -3, -2 (repeating for digits beyond the hundred-thousands place). Then sum the results. 483595: (4 × (-2)) + (8 × (-3)) + (3 × (-1)) + (5 × 2) + (9 × 3) + (5 × 1) = 7. 8 If the hundreds digit is even, examine the number formed by the last two digits. 624: 24. If the hundreds digit is odd, examine the number obtained by the last two digits plus 4. 352: 52 + 4 = 56. Add the last digit to twice the rest. 56: (5 × 2) + 6 = 16. Examine the last three digits[1][2]34152: Examine divisibility of just 152: 19 × 8 Add four times the hundreds digit to twice the tens digit to the ones digit. 34152: 4 × 1 + 5 × 2 + 2 = 16 9 Sum the digits.[1][3][4]2,880: 2 + 8 + 8 + 0 = 18: 1 + 8 = 9. 10 The last digit is 0.[2]130: the last digit is 0. 11 Form the alternating sum of the digits.[1][4]918,082: 9 − 1 + 8 − 0 + 8 − 2 = 22. Add the digits in blocks of two from right to left.[1]627: 6 + 27 = 33. Subtract the last digit from the rest. 627: 62 − 7 = 55. If the number of digits is even, add the first and subtract the last digit from the rest. 918,082: the number of digits is even (6) → 1808 + 9 − 2 = 1815: 81 + 1 − 5 = 77 = 7 × 11 If the number of digits is odd, subtract the first and last digit from the rest. 14,179: the number of digits is odd (5) → 417 − 1 − 9 = 407 = 37 × 11 12 It is divisible by 3 and by 4.[5]324: it is divisible by 3 and by 4. Subtract the last digit from twice the rest. 324: 32 × 2 − 4 = 60. 13 Form the alternating sum of blocks of three from right to left.[6]2,911,272: −2 + 911 − 272 = 637 Add 4 times the last digit to the rest. 637: 63 + 7 × 4 = 91, 9 + 1 × 4 = 13. Multiply each digit (from right to left) by the digit in the corresponding position in this pattern (from left to right): -3, -4, -1, 3, 4, 1 (repeating for digits beyond the hundred-thousands place). Then sum the results.[7]. 30,747,912: (2 × (-3)) + (1 × (-4)) + (9 × (-1)) + (7 × 3) + (4 × 4) + (7 × 1) + (0 × (-3)) + (3 × (-4)) = 13. 14 It is divisible by 2 and by 7.[5]224: it is divisible by 2 and by 7. Add the last two digits to twice the rest. The answer must be divisible by 14. 364: 3 × 2 + 64 = 70. 15 It is divisible by 3 and by 5.[5]390: it is divisible by 3 and by 5. 16 If the thousands digit is even, examine the number formed by the last three digits. 254,176: 176. If the thousands digit is odd, examine the number formed by the last three digits plus 8. 3,408: 408 + 8 = 416. Add the last two digits to four times the rest. 176: 1 × 4 + 76 = 80.
1168: 11 × 4 + 68 = 112.
Examine the last four digits.[1][2]157,648: 7,648 = 428 × 16. 17 Subtract 5 times the last digit from the rest. 221: 22 − 1 × 5 = 17. 18 It is divisible by 2 and by 9.[5]342: it is divisible by 2 and by 9. 19 Add twice the last digit to the rest. 437: 43 + 7 × 2 = 57. 20 It is divisible by 10, and the tens digit is even. 360: is divisible by 10, and 6 is even. If the number formed by the last two digits is divisible by 20. 480: 80 is divisible by 20. Step-by-step examplesDivisibility by 2
First, take any even number (for this example it will be 376) and note the last digit in the number, discarding the other digits. Then take that digit (6) while ignoring the rest of the number and determine if it is divisible by 2. If it is divisible by 2, then the original number is divisible by 2.
Example
1. 376 (The original number)
2. 37 6 (Take the last digit)
3. 6 ÷ 2 = 3 (Check to see if the last digit is divisible by 2)
4. 376 ÷ 2 = 188 (If the last digit is divisible by 2, then the whole number is divisible by 2)
Divisibility by 3
First, take any number (for this example it will be 492) and add together each digit in the number (4 + 9 + 2 = 15). Then take that sum (15) and determine if it is divisible by 3. The original number is divisible by 3 if and only if the final number is divisible by 3.
If a number is a multiplication of 3 consecutive numbers then that number is always divisible by 3. This is useful for when the number takes the form of (n × (n − 1) × (n + 1))
Ex.
1. 492 (The original number)
2. 4 + 9 + 2 = 15 (Add each individual digit together)
3. 15 is divisible by 3 at which point we can stop. Alternatively we can continue using the same method if the number is still too large:
4. 1 + 5 = 6 (Add each individual digit together)
5. 6 ÷ 3 = 2 (Check to see if the number received is divisible by 3)
6. 492 ÷ 3 = 164 (If the number obtained by using the rule is divisible by 3, then the whole number is divisible by 3)
Ex.
1. 336 (The original number)
2. 6 × 7 × 8 = 336
3. 336 ÷ 3 = 112
Divisibility by 4
The basic rule for divisibility by 4 is that if the number formed by the last two digits in a number is divisible by 4, the original number is divisible by 4;[1][2]this is because 100 is divisible by 4 and so adding hundreds, thousands, etc. is simply adding another number that is divisible by 4. If any number ends in a two digit number that you know is divisible by 4 (e.g. 24, 04, 08, etc.), then the whole number will be divisible by 4 regardless of what is before the last two digits.
Alternatively, one can simply divide the number by 2, and then check the result to find if it is divisible by 2. If it is, the original number is divisible by 4. In addition, the result of this test is the same as the original number divided by 4.
Ex.
General rule
1. 2092 (The original number)
2. 20 92 (Take the last two digits of the number, discarding any other digits)
3. 92 ÷ 4 = 23 (Check to see if the number is divisible by 4)
4. 2092 ÷ 4 = 523 (If the number that is obtained is divisible by 4, then the original number is divisible by 4)
Alternative example
1. 1720 (The original number)
2. 1720 ÷ 2 = 860 (Divide the original number by 2)
3. 860 ÷ 2 = 430 (Check to see if the result is divisible by 2)
4. 1720 ÷ 4 = 430 (If the result is divisible by 2, then the original number is divisible by 4)
Divisibility by 5
Divisibility by 5 is easily determined by checking the last digit in the number (475), and seeing if it is either 0 or 5. If the last number is either 0 or 5, the entire number is divisible by 5.[1][2]
If the last digit in the number is 0, then the result will be the remaining digits multiplied by 2. For example, the number 40 ends in a zero (0), so take the remaining digits (4) and multiply that by two (4 × 2 = 8). The result is the same as the result of 40 divided by 5(40/5 = 8).
If the last digit in the number is 5, then the result will be the remaining digits multiplied by two (2), plus one (1). For example, the number 125 ends in a 5, so take the remaining digits (12), multiply them by two (12 × 2 = 24), then add one (24 + 1 = 25). The result is the same as the result of 125 divided by 5 (125/5=25).
Ex.
If the last digit is 0
1. 110 (The original number)
2. 11 0 (Take the last digit of the number, and check if it is 0 or 5)
3. 11 0 (If it is 0, take the remaining digits, discarding the last)
4. 11 × 2 = 22 (Multiply the result by 2)
5. 110 ÷ 5 = 22 (The result is the same as the original number divided by 5)
If the last digit is 5
1. 85 (The original number)
2. 8 5 (Take the last digit of the number, and check if it is 0 or 5)
3. 8 5 (If it is 5, take the remaining digits, discarding the last)
4. 8 × 2 = 16 (Multiply the result by 2)
5. 16 + 1 = 17 (Add 1 to the result)
6. 85 ÷ 5 = 17 (The result is the same as the original number divided by 5)
Divisibility by 6
Divisibility by 6 is determined by checking the original number to see if it is both an even number (divisible by 2) and divisible by 3.[5]This is the best test to use.
Alternatively, one can check for divisibility by six by taking the number (246), dropping the last digit in the number (24 6, adding together the remaining number (24 becomes 2 + 4 = 6), multiplying that by four (6 × 4 = 24), and adding the last digit of the original number to that (24 + 6 = 30). If this number is divisible by six, the original number is divisible by 6.
If the number is divisible by six, take the original number (246) and divide it by two (246 ÷ 2 = 123). Then, take that result and divide it by three (123 ÷ 3 = 41). This result is the same as the original number divided by six (246 ÷ 6 = 41).
Ex.
General rule
1. 324 (The original number)
2. 324 ÷ 3 = 108 (Check to see if the original number is divisible by 3)
3. 324 ÷ 2 = 162 OR 108 ÷ 2 = 54 (Check to see if either the original number or the result of the previous equation is divisible by 2)
4. 324 ÷ 6 = 54 (If either of the tests in the last step are true, then the original number is divisible by 6. Also, the result of the second test returns the same result as the original number divided by 6)
Finding a remainder of a number when divided by 6
6 − (1, −2, −2, −2, −2, and −2 goes on for the rest) No period.
Minimum magnitude sequence
(1, 4, 4, 4, 4, and 4 goes on for the rest)
Positive sequence
Multiply the right most digit by the left most digit in the sequence and multiply the second right most digit by the second left most digit in the sequence and so on. Next, compute the sum of all the values and take the remainder on division by 6.
Example: What is the remainder when 1036125837 is divided by 6?
Multiplication of the rightmost digit = 1 × 7 = 7
Multiplication of the second rightmost digit = 3 × −2 = −6
Third rightmost digit = −16
Fourth rightmost digit = −10
Fifth rightmost digit = −4
Sixth rightmost digit = −2
Seventh rightmost digit = −12
Eighth rightmost digit = −6
Ninth rightmost digit = 0
Tenth rightmost digit = −2
Sum = −51
−51 modulo 6 = 3
Remainder = 3
Divisibility by 7‹ Whether to make the |reason= mandatory for the {{cleanup}} template is being discussed. See the request for comment to help reach a consensus.›
This section may require cleanup to meet Wikipedia's quality standards. (Consider using more specific cleanup instructions.) Please help improve this section if you can. The talk page may contain suggestions. (August 2010)
Divisibility by 7 can be tested by a recursive method. A number of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a small number (below 20 in absolute value) is obtained. The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7. For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10 = −7; thus, since −7 is divisible by 7, 371 is divisible by 7.
Another method is multiplication by 3. A number of the form 10x + y has the same remainder when divided by 7 as 3x + y. So get the leftmost digit of the original number, multiply by 3, add the next digit, get the remainder by 7, and continue from the beginning: multiply by 3, add the next digit, etc. For example, the number 371: 3×3 + 7 = 16 remainder 2, and 2×3 + 1 = 7. This method can be used to find the remainder of division by 7.
A more complicated algorithm for testing divisibility by 7 uses the fact that 100 ≡ 1, 101 ≡ 3, 102 ≡ 2, 103 ≡ 6, 104 ≡ 4, 105 ≡ 5, 106 ≡ 1, ... (mod 7). Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2= 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7 (hence 371 is divisible by 7 since 28 is).[8]
This method can be simplified by removing the need to multiply. All it would take with this simplification is to memorise the sequence above (132645...), and to add and subtract, but always working with one-digit numbers.
The simplification goes as follows:
• Take for instance the number 371
• Change all occurrences of a 7, 8 or 9 into 0, 1 or 2respectively. In this example, we get: 301. This second step may be skipped, except for the left most digit, but following it may facilitate calculations later on.
• Now convert the first digit (3) into the following digit in the sequence 13264513... In our example, 3 becomes 2.
• Add the result in the previous step (2) to the second digit of the number, and substitute the result for both digits, leaving all remaining digits unmodified: 2 + 0 = 2. So 301 becomes 21.
• Repeat the procedure until you have a recognisable multiple of 7, or to make sure, a number between 0 and 6. So, starting from 21 (which is anyway a recognisable multiple of 7) take the first digit (2) and convert it into the following in the sequence above: 2 becomes 6. Then add this to the second digit: 6 + 1 = 7.
• If at any point the first digit is an 8 or a 9, these should become 1, or 2 respectively. But if it is a 7 it should become 0, only if no other digits follow. Otherwise, it should simply be dropped. This is because that 7 would have become 0, and numbers with at least two digits before the decimal dot do not begin with 0, which is useless. According to this, our 7 becomes 0.
If through this procedure you obtain a 0 or any recognisable multiple of 7, then the original number is a multiple of 7. If you obtain any number from 1 to 6, that will indicate how much you should subtract from the original number to get a multiple of 7. In other words, you will find the remainder of dividing the number by 7. For example take the number 186:
• First, change the 8 into a 1: 116.
• Now, change 1 into the following digit in the sequence (3), add it to the second digit, and write the result instead of both: 3 + 1 = 4. So 116 becomes now 46.
• Repeat the procedure, since the number is greater than 7. Now, 4 becomes 5, which must be added to 6. That is 11.
• Repeat the procedure one more time: 1 becomes 3, which is added to the second digit (1): 3 + 1 = 4.
Now we have a number lower than 7, and this number (4) is the remainder of dividing 186/7. So 186 minus 4, which is 182, must be a multiple of 7.
Note: The reason why this works is that if we have: a+b=cand b is a multiple of any given number n, then a and c will necessarily produce the same remainder when divided by n. In other words, in 2 + 7 = 9, 7 is divisible by 7. So 2 and 9 must have the same reminder when divided by 7. The remainder is 2.
Therefore, if a number n is a multiple of 7 (i.e.: the remainder of n/7 is 0), then adding (or subtracting) multiples of 7 cannot possibly change that property.
What this procedure does, as explained above for most divisibility rules, is simply subtract little by little multiples of 7 from the original number until reaching a number that is small enough for us to remember whether it is a multiple of 7. If 1 becomes a 3 in the following decimal position, that is just the same as converting 10×10n into a 3×10n. And that is actually the same as subtracting 7×10n (clearly a multiple of 7) from 10×10n.
Similarly, when you turn a 3 into a 2 in the following decimal position, you are turning 30×10n into 2×10n, which is the same as subtracting 30×10n−28×10n, and this is again subtracting a multiple of 7. The same reason applies for all the remaining conversions:
• 20×10n − 6×10n=14×10n
• 60×10n − 4×10n=56×10n
• 40×10n − 5×10n=35×10n
• 50×10n − 1×10n=49×10n
First method example
1050 → 105 − 0=105 → 10 − 10 = 0. ANSWER: 1050 is divisible by 7.
Second method example
1050 → 0501 (reverse) → 0×1 + 5×3 + 0×2 + 1×6 = 0 + 15 + 0 + 6 = 21 (multiply and add). ANSWER: 1050 is divisible by 7.
Vedic method of divisibility by osculation
Divisibility by seven can be tested by multiplication by the Ekhādika. Convert the divisor seven to the nines family by multiplying by seven. 7×7=49. Add one, drop the units digit and, take the 5, the Ekhādika, as the multiplier. Start on the right. Multiply by 5, add the product to the next digit to the left. Set down that result on a line below that digit. Repeat that method of multiplying the units digit by five and adding that product to the number of tens. Add the result to the next digit to the left. Write down that result below the digit. Continue to the end. If the end result is zero or a multiple of seven, then yes, the number is divisible by seven. Otherwise, it is not. This follows the Vedic ideal, one-line notation.[9]
Vedic method example:
Is 438,722,025 divisible by seven? Multiplier = 5. 4 3 8 7 2 2 0 2 5 42 37 46 37 6 40 37 27 YES
Pohlman-Mass method of divisibility by 7
The Pohlman-Mass method provides a quick solution that can determine if most integers are divisible by seven in three steps or less. This method could be useful in a mathematics competition such as MATHCOUNTS, where time is a factor to determine the solution without a calculator in the Sprint Round.
Step A: If the integer is 1,000 or less, subtract twice the last digit from the number formed by the remaining digits. If the result is a multiple of seven, then so is the original number (and vice versa). For example:
112 -> 11 − (2×2) = 11 − 4 = 7 YES 98 -> 9 − (8×2) = 9 − 16 = −7 YES 634 -> 63 − (4×2) = 63 − 8 = 55 NO
Because 1,001 is divisible by seven, an interesting pattern develops for repeating sets of 1, 2, or 3 digits that form 6-digit numbers (leading zeros are allowed) in that all such numbers are divisible by seven. For example:
001 001 = 1,001 / 7 = 143 010 010 = 10,010 / 7 = 1,430 011 011 = 11,011 / 7 = 1,573 100 100 = 100,100 / 7 = 14,300 101 101 = 101,101 / 7 = 14,443 110 110 = 110,110 / 7 = 15,730 01 01 01 = 10,101 / 7 = 1,443 10 10 10 = 101,010 / 7 = 14,430 111,111 / 7 = 15,873 222,222 / 7 = 31,746 999,999 / 7 = 142,857 576,576 / 7 = 82,368
For all of the above examples, subtracting the first thee digits from the last three results in a multiple of seven. Notice that leading zeros are permitted to form a 6-digit pattern.
This phenomenon forms the basis for Steps B and C.
Step B: If the integer is between 1,001 and one million, find a repeating pattern of 1, 2, or 3 digits that forms a 6-digit number that is close to the integer (leading zeros are allowed and can help you visualize the pattern). If the positive difference is less than 1,000, apply Step A. This can be done by subtracting the first three digits from the last three digits. For example:
341,355 − 341,341 = 14 -> 1 − (4×2) = 1 − 8 = −7 YES 67,326 − 067,067 = 259 -> 25 − (9×2) = 25 − 18 = 7 YES
The fact that 999,999 is a multiple of 7 can be used for determining divisibility of integers larger than one million by reducing the integer to a 6-digit number that can be determined using Step B. This can be done easily by adding the digits left of the first six to the last six and follow with Step A.
Step C: If the integer is larger than one million, subtract the nearest multiple of 999,999 and then apply Step B. For even larger numbers, use larger sets such as 12-digits (999,999,999,999) and so on. Then, break the integer into a smaller number that can be solved using Step B. For example:
22,862,420 − (999,999 × 22) = 22,862,420 − 21,999,978 -> 862,420 + 22 = 862,442 862,442 -> 862 − 442 (Step B) = 420 -> 42 − (0×2) (Step A) = 42 YES
This allows adding and subtracting alternating sets of three digits to determine divisibility by seven. Understanding these patterns allows you to quickly calculate divisibility of seven as seen in the following examples:
Pohlman-Mass method of divisibility by 7, examples:
Is 98 divisible by seven? 98 -> 9 − (8×2) = 9 − 16 = −7 YES (Step A) Is 634 divisible by seven? 634 -> 63 − (4×2) = 63 − 8 = 55 NO (Step A) Is 355,341 divisible by seven? 355,341 − 341,341 = 14,000 (Step B) -> 014 − 000 (Step B) -> 14 = 1 − (4×2) (Step A) = 1 − 8 = −7 YES Is 42,341,530 divisible by seven? 42,341,530 -> 341,530 + 42 = 341,572 (Step C) 341,572 − 341,341 = 231 (Step B) 231 -> 23 − (1×2) = 23 − 2 = 21 YES (Step A) Using quick alternating additions and subtractions: 42,341,530 -> 530 − 341 = 189 + 42 = 231 -> 23 − (1×2) = 21 YES
Multiplication by 3 method of divisibility by 7, examples:
Is 98 divisible by seven? 98 -> 9 remainder 2 -> 2×3 + 8 = 14 YES Is 634 divisible by seven? 634 -> 6×3 + 3 = 21 -> remainder 0 -> 0×3 + 4 = 4 NO Is 355,341 divisible by seven? 3 * 3 + 5 = 14 -> remainder 0 -> 0×3 + 5 = 5 -> 5×3 + 3 = 18 -> remainder 4 -> 4×3 + 4 = 16 -> remainder 2 -> 2×3 + 1 = 7 YES Find remainder of 1036125837 divided by 7 1×3 + 0 = 3 3×3 + 3 = 12 remainder 5 5×3 + 6 = 21 remainder 0 0×3 + 1 = 1 1×3 + 2 = 5 5×3 + 5 = 20 remainder 6 6×3 + 8 = 26 remainder 5 5×3 + 3 = 18 remainder 4 4×3 + 7 = 19 remainder 5 Answer is 5
Finding remainder of a number when divided by 7
7 − (1, 3, 2, −1, −3, −2, cycle repeats for the next six digits) Period: 6 digits. Recurring numbers: 1, 3, 2, −1, −3, −2
Minimum magnitude sequence
(1, 3, 2, 6, 4, 5, cycle repeats for the next six digits) Period: 6 digits. Recurring numbers: 1, 3, 2, 6, 4, 5
Positive sequence
Multiply the right most digit by the left most digit in the sequence and multiply the second right most digit by the second left most digit in the sequence and so on and so for. Next, compute the sum of all the values and take the modulus of 7.
Example: What is the remainder when 1036125837 is divided by 7?
Multiplication of the rightmost digit = 1 × 7 = 7
Multiplication of the second rightmost digit = 3 × 3 = 9
Third rightmost digit = 8 × 2 = 16
Fourth rightmost digit = 5 × −1 = −5
Fifth rightmost digit = 2 × −3 = −6
Sixth rightmost digit = 1 × −2 = −2
Seventh rightmost digit = 6 × 1 = 6
Eighth rightmost digit = 3 × 3 = 9
Ninth rightmost digit = 0
Tenth rightmost digit = 1 × −1 = −1
Sum = 33
33 modulus 7 = 5
Remainder = 5
Digit pair method of divisibility by 7
This method uses 1, −3, 2 pattern on the digit pairs. That is, the divisibility of any number by seven can be tested by first separating the number into digit pairs, and then applying the algorithm on three digit pairs (six digits). When the number is smaller than six digits, then fill zero's to the right side until there are six digits. When the number is larger than six digits, then repeat the cycle on the next six digit group and then add the results. Repeat the algorithm until the result is a small number. The original number is divisible by seven if and only if the number obtained using this algorithm is divisible by seven. This method is especially suitable for large numbers.
Example 1:
The number to be tested is 157514. First we separate the number into three digit pairs: 15, 75 and 14.
Then we apply the algorithm: 1 × 15 − 3 × 75 + 2 × 14 = 182
Because the resulting 182 is less than six digits, we add zero's to the right side until it is six digits.
Then we apply our algorithm again: 1 × 18 − 3 × 20 + 2 × 0 = −42
The result −42 is divisible by seven, thus the original number 157514 is divisible by seven!
Example 2:
The number to be tested is 15751537186.
(1 × 15 − 3 × 75 + 2 × 15) + (1 × 37 − 3 × 18 + 2 × 60) = −180 + 103 = −77
The result −77 is divisible by seven, thus the original number 15751537186 is divisible by seven!
Divisibility by 13
Remainder Test 13 (1, −3, −4, −1, 3, 4, cycle goes on.) If you are not comfortable with negative numbers, then use this sequence. (1, 10, 9, 12, 3, 4)
Multiply the right most digit of the number with the left most number in the sequence shown above and the second right most digit to the second left most digit of the number in the sequence. The cycle goes on.
Example: What is the remainder when 321 is divided by 13?
Using the first sequence,
Ans: 1 × 1 + 2 × −3 + 3 × −4 = 9
Remainder = −17 mod 13 = 9
Example: What is the remainder when 1234567 is divided by 13?
Using the second sequence,
Answer: 7 × 1 + 6 × 10 + 5 × 9 + 4 × 12 + 3 × 3 + 2 × 4 + 1 × 1 = 178 mod 13 = 9
Remainder = 9
Beyond 20
Divisibility properties can be determined in two ways, depending on the type of the divisor.
Composite divisors
A number is divisible by a given divisor if it is divisible by the highest power of each of its prime factors. For example, to determine divisibility by 24, check divisibility by 8 and by 3.[5]Note that checking 4 and 6, or 2 and 12, would not be sufficient. A table of prime factors may be useful.
A composite divisor may also have a rule formed using the same procedure as for a prime divisor, given below, with the caveat that the manipulations involved may not introduce any factor which is present in the divisor. For instance, one can not make a rule for 14 that involves multiplying the equation by 7. This is not an issue for prime divisors because they have no smaller factors.
Prime divisors
The goal is to find an inverse to 10 modulo the prime (not 2 or 5) and use that as a multiplier to make the divisibility of the original number by that prime depend on the divisibility of the new (usually smaller) number by the same prime. Using 17 as an example, since 10 × (−5) = −50 = 1 mod 17, we get the rule for using y − 5x in the table above. In fact, this rule for prime divisors besides 2 and 5 is really a rule for divisibility by any integer relatively prime to 10 (including 21 and 27; see tables below). This is why the last divisibility condition in the tables above and below for any number relatively prime to 10 has the same kind of form (add or subtract some multiple of the last digit from the rest of the number).
Notable examples
The following table provides rules for a few more notable divisors:
Divisor Divisibility condition Examples 21 Subtract twice the last digit from the rest. 168: 16 − (8×2) = 0, 168 is divisible.
1050: 105 − (0×2) = 105, 10 − (5×2) = 0, 1050 is divisible. 23 Add 7 times the last digit to the rest. 3128: 312 + (8×7) = 368, 368 ÷ 23 = 16. 25 The number formed by the last two digits is divisible by 25.[2]134,250: 50 is divisible by 25. 27 Sum the digits in blocks of three from right to left. If the result is divisible by 27, then the number is divisible by 27. 2,644,272: 2 + 644 + 272 = 918 = 27×34. Subtract 8 times the last digit from the rest. 621: 62 − (1×8) = 54. 29 Add three times the last digit to the rest. 261: 1×3 = 3; 3 + 26 = 29 31 Subtract three times the last digit from the rest. 837: 83 − 3×7 = 62 32 The number formed by the last five digits is divisible by 32.[1][2]25,135,520: 35,520=1110×32 If the ten thousands digit is even, examine the number formed by the last four digits. 41,312: 1312. If the ten thousands digit is odd, examine the number formed by the last four digits plus 16. 254,176: 4176+16 = 4192. Add the last two digits to 4 times the rest. 1,312: (13×4) + 12 = 64. 33 Add 10 times the last digit to the rest; it has to be divisible by 3 and 11. 627: 62 + 7 × 10 = 132,
13 + 2 × 10 = 33. Add the digits in blocks of two from right to left. 2,145: 21 + 45 = 66. 35 Number must be divisible by 7 ending in 0 or 5. 37 Take the digits in blocks of three from right to left and add each block, just as for 27. 2,651,272: 2 + 651 + 272 = 925. 925 = 37×25. Subtract 11 times the last digit from the rest. 925: 92 − (5×11) = 37. 39 Add 4 times the last digit to the rest. 351: 1 × 4 = 4; 4 + 35 = 39 41 Subtract 4 times the last digit from the rest. 738: 73 − 8 × 4 = 41. 43 Add 13 times the last digit to the rest. 36,249: 3624 + 9 × 13 = 3741,
374 + 1 × 13 = 387,
38 + 7 × 13 = 129,
12 + 9 × 13 = 129 = 43 × 3. Subtract 30 times the last digit from the rest. 36,249: 3624 - 9 × 30 = 3354,
335 - 4 × 30 = 215 = 43 × 5. 45 The number must be divisible by 9 ending in 0 or 5.[5]495: 4 + 9 + 5 = 18, 1 + 8 = 9;
(495 is divisible by both 5 and 9.) 47 Subtract 14 times the last digit from the rest. 1,642,979: 164297 − 9 × 14 = 164171,
16417 − 14 = 16403,
1640 − 3 × 14 = 1598,
159 − 8 × 14 = 47. 49 Add 5 times the last digit to the rest. 1,127: 112+(7×5)=147.
147: 14 + (7×5) = 49 50 The last two digits are 00 or 50. 134,250: 50. 51 Subtract 5 times the last digit to the rest. 55 Number must be divisible by 11 ending in 0 or 5.[5]935: 93 − 5 = 88 or 9 + 35 = 44. 59 Add 6 times the last digit to the rest. 295: 5×6 = 30; 30 + 29 = 59 61 Subtract 6 times the last digit from the rest. 64 The number formed by the last six digits must be divisible by 64.[1][2]65 Number must be divisible by 13 ending in 0 or 5.[5]66 Number must be divisible by 6 and 11.[5]69 Add 7 times the last digit to the rest. 345: 5×7 = 35; 35 + 34 = 69 71 Subtract 7 times the last digit from the rest. 75 Number must be divisible by 3 ending in 00, 25, 50 or 75.[5]825: ends in 25 and is divisible by 3. 77 Form the alternating sum of blocks of three from right to left. 76,923: 923 - 76 = 847. 79 Add 8 times the last digit to the rest. 711: 1×8 = 8; 8 + 71 = 79 81 Subtract 8 times the last digit from the rest. 89 Add 9 times the last digit to the rest. 801: 1×9 = 9; 80 + 9 = 89 91 Subtract 9 times the last digit from the rest. Form the alternating sum of blocks of three from right to left. 5,274,997: 5 - 274 + 997 = 728 99 Add the digits in blocks of two from right to left. 144,837: 14 + 48 + 37 = 99. 101 Form the alternating sum of blocks of two from right to left. 40,299: 4 - 2 + 99 = 101. 111 Add the digits in blocks of three from right to left. 125 The number formed by the last three digits must be divisible by 125.[2]128 The number formed by the last seven digits must be divisible by 128.[1][2]143 Form the alternating sum of blocks of three from right to left. 1,774,487: 1 - 774 + 487 = -286 256 The number formed by the last eight digits must be divisible by 256.[1][2]333 Add the digits in blocks of three from right to left. 512 The number formed by the last nine digits must be divisible by 512.[1][2]989 Add the last three digits to eleven times the rest. 21758: 21 × 11 = 231; 758 + 231 = 989 999 Add the digits in blocks of three from right to left. Generalized divisibility rule
To test for divisibility by D, where D ends in 1, 3, 7, or 9, the following method can be used.[10]Find any multiple of D ending in 9. (If D ends respectively in 1, 3, 7, or 9, then multiply by 9, 3, 7, or 1.) Then add 1 and divide by 10, denoting the result as m. Then a number N = 10t + q is divisible by Dif and only if mq + t is divisible by D.
For example, to determine if 913 = 10×91 + 3 is divisible by 11, find that m = (11×9+1)÷10 = 10. Then mq+t = 10×3+91 = 121; this is divisible by 11 (with quotient 11), so 913 is also divisible by 11. As another example, to determine if 689 = 10×68 + 9 is divisible by 53, find that m = (53×3+1)÷10 = 16. Then mq+t = 16×9 + 68 = 212, which is divisible by 53 (with quotient 4); so 689 is also divisible by 53.
ProofsProof using basic algebra
Many of the simpler rules can be produced using only algebraic manipulation, creating binomials and rearranging them. By writing a number as the sum of each digit times a power of 10 each digit's power can be manipulated individually.
Case where all digits are summed
This method works for divisors that are factors of 10 − 1 = 9.
Using 3 as an example, 3 divides 9 = 10 − 1. That means (see modular arithmetic). The same for all the higher powers of 10: They are all congruent to 1 modulo 3. Since two things that are congruent modulo 3 are either both divisible by 3 or both not, we can interchange values that are congruent modulo 3. So, in a number such as the following, we can replace all the powers of 10 by 1:
which is exactly the sum of the digits.
Case where the alternating sum of digits is used
This method works for divisors that are factors of 10 + 1 = 11.
Using 11 as an example, 11 divides 11 = 10 + 1. That means . For the higher powers of 10, they are congruent to 1 for even powers and congruent to −1 for odd powers:
Like the previous case, we can substitute powers of 10 with congruent values:
which is also the difference between the sum of digits at odd positions and the sum of digits at even positions.
Case where only the last digit(s) matter
This applies to divisors that are a factor of a power of 10. This is because sufficiently high powers of the base are multiples of the divisor, and can be eliminated.
For example, in base 10, the factors of 101 include 2, 5, and 10. Therefore, divisibility by 2, 5, and 10 only depend on whether the last 1 digit is divisible by those divisors. The factors of 102 include 4 and 25, and divisibility by those only depend on the last 2 digits.
Case where only the last digit(s) are removed
Most numbers do not divide 9 or 10 evenly, but do divide a higher power of 10n or 10n − 1. In this case the number is still written in powers of 10, but not fully expanded.
For example, 7 does not divide 9 or 10, but does divide 98, which is close to 100. Thus, proceed from
where in this case a is any integer, and b can range from 0 to 99. Next,
and again expanding
and after eliminating the known multiple of 7, the result is
which is the rule "double the number formed by all but the last two digits, then add the last two digits".
Case where the last digit(s) is multiplied by a factor
The representation of the number may also be multiplied by any number relatively prime to the divisor without changing its divisibility. After observing that 7 divides 21, we can perform the following:
after multiplying by 2, this becomes
and then
Eliminating the 21 gives
and multiplying by −1 gives
Either of the last two rules may be used, depending on which is easier to perform. They correspond to the rule "subtract twice the last digit from the rest".
Proof using modular arithmetic
This section will illustrate the basic method; all the rules can be derived following the same procedure. The following requires a basic grounding in modular arithmetic; for divisibility other than by 2's and 5's the proofs rest on the basic fact that 10 mod m is invertible if 10 and m are relatively prime.
For 2n or 5n:
Only the last n digits need to be checked.
Representing x as
and the divisibility of x is the same as that of z.
For 7:
Since 10 × 5 ≡ 10 × (−2) ≡ 1 (mod 7) we can do the following:
Representing x as
so x is divisible by 7 if and only if y − 2zis divisible by 7
Velda Stanton
Lvl 10
2y ago
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# 5th Grade Fractions
In 5th Grade Fractions we will discuss about definition of fraction, concept of fractions and different types of examples on fractions.
A fraction is a number representing a part of a whole. The whole may be a single object or a group of objects.
Definition of Fraction: A number that compares part of an object or set with the whole, especially the quotient of two whole numbers, written in the form of $$\frac{x}{y}$$ is called a fraction.
The fraction $$\frac{2}{5}$$, which means 2 divided by 5, can be represented as 2 books out of a box of 5 books.
A fraction is a
(i) part of a whole
(ii) part of a collection
## Concept of 5th Grade Fractions:
A fraction is a number which represents/indicates a part or parts of a whole. Fractions can be represented in three ways:
(i) Fraction as a Part of a Whole:
In the given figure, the coloured parts represent $$\frac{5}{8}$$ of the whole,
i.e., $$\frac{5}{8}$$ 5 indicates 5 parts out of 8 equal parts of a whole.
So, $$\frac{5}{8}$$ is a fraction.
5 Numerator
8 → Denominator
Clearly, a fraction comprises two numbers separated by a horizontal line. The number above the horizontal line is called the numerator and the number below the horizontal line is called the denominator of the fraction.
(ii) Fraction as a Part of a Collection:
We can find the fractional part of a collection by dividing the collection into subgroups equal to the number representing the denominator of the fraction. Then, we take the number of subgroups equal to the number representing the numerator of the fraction.
Consider a collection of 9 balls. If we divide this collection into three equal parts, we get 3 balls in each of the three parts.
Thus, one-third of 9 is 3.
i.e., $$\frac{1}{3}$$ of 9 = 9 × $$\frac{1}{3}$$ = $$\frac{9}{3}$$ = 3
(iii) Fraction as Division:
A fraction can be expressed as a division. Conversely, division can be expressed as fraction.
If 42 pencils are distributed equally among 7 students then each student will get 42 ÷ 7 = 6 pencils.
But if 1 mango is to be distributed among 4 students, then how many mango will a student get?
Obviously, each student gets 1 ÷ 4 i.e., $$\frac{1}{4}$$ mango.
### Following are Some Examples of 5th Grade Fractions:
(i) Consider the fraction $$\frac{7}{12}$$. This fraction is read as ”seven-twelfth” which means that 7 parts out of 12 equal parts in which the whole is divided. In the fraction $$\frac{7}{12}$$, 7 is called the numerator and 12 is called the denominator.
(ii) The fraction $$\frac{5}{7}$$ is read as ”five-seventh” which means that 5 parts out of 7 equal parts in which the whole is divided. In the fraction $$\frac{5}{7}$$, 5 is called the numerator and 7 is called the denominator.
(iii) The fraction $$\frac{3}{10}$$ is read as ”three-tenth” which means that 3 parts out of 10 equal parts in which the whole is divided. In the fraction $$\frac{3}{10}$$, 3 is called the numerator and 10 is called the denominator.
(iv) The fraction $$\frac{1}{5}$$ is read as ”one-fifth” which means that 1 parts out of 5 equal parts in which the whole is divided. In the fraction $$\frac{1}{5}$$, 1 is called the numerator and 5 is called the denominator.
For example on 5th Grade Fractions:
1. Mrs. Brown has 24 apples. She ate $$\frac{1}{4}$$ of them.
(i) How many apples does she eat?
(ii) How many does she have left?
Solution:
(i) Here the fraction $$\frac{1}{4}$$ means take 1 part from 4 equal parts.
So, arrange 24 apples in four equal groups.
Clearly, each group will contain 24 ÷ 4 = 6 apples.
Thus, $$\frac{1}{4}$$ of 24 is 6.
Hence, Mrs. Brown ate 6 apples.
(ii) Number of left out apples = 24 – 6 = 18.
2. Andrea has a packet of 20 biscuits. She gives $$\frac{1}{2}$$ of them to Andy and $$\frac{1}{4}$$ of them to Sally. The rest she keeps.
(i) How many biscuits does Andy get?
(ii) How many biscuits does Sally get?
(iii) How many biscuits does Andrea keep?
Solution:
(i) Here, $$\frac{1}{2}$$ of 20 means take 1 part from two equal parts.
So, we arrange 20 biscuits in two equal parts.
Clearly, each part will contain 20 ÷ 2 = 10 biscuits.
Therefore, $$\frac{1}{2}$$ of 20 is 10.
Hence, Andy gets 10 biscuits.
(ii) $$\frac{1}{4}$$ of 20 means take 1 part from four equal parts.
So, we arrange 20 biscuits in four equal parts.
Clearly, each part will contain 20 ÷ 4 = 5 biscuits.
Therefore, $$\frac{1}{4}$$ of 20 is 5.
Hence, Sally gets 5 biscuits.
(iii) Clearly, left out biscuits are kept by Andrea.
Therefore, Andrea keeps 20 – 10 – 5 = 5 biscuits.
5th Grade Fractions
3. What fraction of a day is 8 hours?
Solution:
We have,
One day = 12 hours.
Therefore, 8 hours = $$\frac{8}{12}$$ of a day.
Hence, 8 hours is $$\frac{8}{12}$$ part of a day.
4. Determine $$\frac{2}{3}$$ of a collection of 9 balls.
Solution:
In order to find $$\frac{2}{3}$$ of a collection of 9 balls, we divide the collection of 9 balls in 3 equal parts and take 2 such parts. Clearly, each row has $$\frac{9}{3}$$ = 3 balls.
When, we take 2 rows out of 3 rows. It represents $$\frac{2}{3}$$ of 9 balls. There are 6 balls in 2 rows.
Hence, $$\frac{2}{3}$$ of 9 balls = 6 balls.
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• ### Equivalent Fractions | Fractions |Reduced to the Lowest Term |Examples
The fractions having the same value are called equivalent fractions. Their numerator and denominator can be different but, they represent the same part of a whole. We can see the shade portion with respect to the whole shape in the figures from (i) to (viii) In; (i) Shaded
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Representations of Fractions on a Number Line
Fraction as Division
Types of Fractions
Conversion of Mixed Fractions into Improper Fractions
Conversion of Improper Fractions into Mixed Fractions
Equivalent Fractions
Interesting Fact about Equivalent Fractions
Fractions in Lowest Terms
Like and Unlike Fractions
Comparing Like Fractions
Comparing Unlike Fractions
Addition and Subtraction of Like Fractions
Addition and Subtraction of Unlike Fractions
Inserting a Fraction between Two Given Fractions
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# ### Denominators
## What is a Denominator?
A denominator is the bottom number of a fraction. The word comes from the Latin root word “denominaire” and it means “that which gives a name to”. Ancient mathematicians chose this word because the denominator of a fraction names the size of a fractional part.
The size of a fractional part is determined by the number of parts in a whole. If there is a larger number of parts in the whole, the parts will be smaller because the whole is split into more pieces.
## Examples
The denominator of $$\frac{4}{5}$$ is 5.
The denominator of $$\frac{7}{12}$$ is 12.
The denominator of $$\frac{1}{2}$$ is 2.
## Fractions as Division
Fractions are usually interpreted as a parts of a whole. When most people see $$\frac{2}{3}$$, they think of one whole split into three parts with two of the parts shaded.
However, fractions can also be interpreted as division. For example, $$\frac{2}{3}$$ could be seen as two wholes that are divided equally among three people.
Either way, the amount represented by the fraction $$\frac{2}{3}$$ is the same. There are just two different ways of thinking about it. When fractions are interpreted as division, the denominator of the fraction is the divisor of the division problem.
## When Will I Use Denominators?
Any time you add or subtract fractions, you have to find common denominators because you can’t add fractions if the fraction parts are not the same size.
When you have a square root in the bottom of a fraction, you may be asked to rationalize the denominator before you submit your answer.
If you have variables like x or y in a fraction, that means you are dealing with a rational expression. You can follow these rules to simplify rational expressions or you can follow these steps to solve rational equations if there is an equal sign.
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What is meant by curve sketching?
What is meant by curve sketching?
In geometry, curve sketching (or curve tracing) are techniques for producing a rough idea of overall shape of a plane curve given its equation, without computing the large numbers of points required for a detailed plot. It is an application of the theory of curves to find their main features.
Is curve sketching important?
Curve sketching shows us how we can understand and predict the behavior of the function based on its first and second derivatives. Functions and their graphs are important not only in math but in other fields and applications as well.
How do you find the curve?
A curve is defined as a smoothly- flowing continuous line that has bent. It does not have any sharp turns. The way to identify the curve is that the line bends and changes its direction at least once.
What are the rules for curve tracing?
A) If Curve is symmetric about x-axis, We find asymptote parallel to y-axis by equating coefficient of highest degree term in y to zero. B) If Curve is symmetric about y-axis, We find asymptote parallel to x-axis by equating coefficient of highest degree term in x to zero.
What is a curve in mathematics?
curve, In mathematics, an abstract term used to describe the path of a continuously moving point (see continuity). Such a path is usually generated by an equation. The word can also apply to a straight line or to a series of line segments linked end to end.
What is a simple curve in mathematics?
Simple Curve: A simple curve changes direction but does not cross itself while changing direction. A simple curve can be open and closed both. 6. Non-simple curves: A curve that crosses its own path is called a non-simple curve.
How do you draw a graph step by step?
1. Step 1: Identify the variables.
2. Step 2: Determine the variable range.
3. Step 3: Determine the scale of the graph.
4. Step 4: Number and label each axis and title the graph.
5. Step 5: Determine the data points and plot on the graph.
6. Step 6: Draw the graph.
Who invented curve tracing?
5.6. Johann Bernoulli’s crawling curves. We shall now turn to a completely different type of curve tracing, devised by Johann Bernoulli.
What is a curve in math?
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Understanding And Calculating Combinations With 6 Numbers
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Thomas
Explore the world of combinations with 6 numbers. Understand the definition and importance of combinations, and learn how to calculate them using a step-by-step process. Discover real-life examples and strategies for generating combinations, and explore their applications in various fields such as probability, cryptography, and data analysis.
Understanding Combinations with 6 Numbers
Definition of Combinations
Combinations refer to the different ways in which a set of numbers can be arranged or selected without regard to the order. In the context of 6 numbers, combinations refer to the unique sets of numbers that can be formed by selecting 6 numbers out of a larger set, without repetition.
Importance of Combinations
Combinations play a crucial role in various fields and applications. They are particularly important in probability and statistics, cryptography and encryption, and data analysis. By understanding combinations, we can gain insights into the likelihood of events, secure sensitive information, and make informed decisions based on data samples.
Factors Affecting Combinations
Several factors can affect the number of possible combinations with 6 numbers. The main factors include:
– The total number of available numbers: The larger the pool of numbers to choose from, the more combinations can be formed.
– The presence of repetition: If repetition is allowed, the number of combinations increases significantly.
– The order of the numbers: Combinations do not consider the order of the selected numbers, so different orders of the same numbers are considered as one combination.
– The number of numbers to be selected: In this case, we are selecting 6 numbers, but the number of combinations would differ if we were selecting a different number of numbers.
Understanding these factors is essential for accurately calculating and analyzing combinations with 6 numbers. By considering these factors, we can explore the vast possibilities and potential outcomes that arise from combining numbers in different ways.
Calculating Combinations with 6 Numbers
Formula for Calculating Combinations
When it comes to understanding combinations with 6 numbers, it’s important to know the formula used for calculating them. The formula for calculating combinations is based on the concept of choosing a specific number of items from a larger set without considering the order in which they are selected.
The formula for calculating combinations is:
nCr = n! / (r!(n-r)!)
Here, “n” represents the total number of items in the set, and “r” represents the number of items being chosen. The exclamation mark denotes the factorial of a number, which means multiplying the number by all the positive integers less than it down to 1.
For example, if we want to calculate the number of combinations when choosing 6 numbers out of a set of 10, the formula would be:
10C6 = 10! / (6!(10-6)!) = 10! / (6!4!)
Step-by-Step Calculation Process
To calculate combinations with 6 numbers, we can follow a step-by-step process using the formula mentioned earlier. Let’s illustrate this process with an example:
Suppose we have a set of 8 numbers, and we want to calculate the number of combinations when choosing 6 numbers out of this set.
1. Identify the values of “n” and “r” in the formula. In this case, n = 8 (the total number of numbers in the set) and r = 6 (the number of numbers being chosen).
2. Calculate the factorials of n, r, and (n-r). In our example, we have:
n! = 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320
r!
= 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
(n-r)! = (8-6)! = 2! = 2 x 1 = 2
3. Substitute the values into the formula and perform the calculation. Using our example:
8C6 = 8! / (6!(8-6)!)
= 40,320 / (720 x 2)
= 40,320 / 1440
= 28
Therefore, there are 28 different combinations when choosing 6 numbers from a set of 8.
Common Mistakes to Avoid
Calculating combinations with 6 numbers can be straightforward if done correctly, but there are some common mistakes to avoid to ensure accurate results.
1. Mistake: Forgetting to calculate factorials correctly.
Solution: Double-check the factorial calculations to ensure accurate results. Mistakes in factorials can lead to incorrect combinations.
2. Mistake: Using the wrong formula.
Solution: Make sure to use the combination formula specifically when dealing with choosing items without considering their order. Using other formulas like permutations can lead to incorrect results.
3. Mistake: Misinterpreting the values of “n” and “r”.
Solution: Be clear about the total number of items in the set (n) and the number of items being chosen (r). Switching or misunderstanding these values can lead to incorrect combinations.
By being aware of these common mistakes and following the correct calculation process, you can confidently calculate combinations with 6 numbers and obtain accurate results.
Real-Life Examples of Combinations with 6 Numbers
Combinations in Lottery Games
Have you ever wondered how lottery games determine the winning numbers? It all comes down to combinations. In lottery games, a set of numbers is drawn from a larger pool of numbers, and players try to match the drawn numbers with their own selections. The winning combination is determined by the specific numbers drawn, and the order in which they are drawn doesn’t matter. This means that if you have chosen the numbers 1, 2, 3, 4, 5, and 6, and the winning numbers are 5, 3, 1, 4, 2, and 6, you would still be a winner! The concept of combinations is fundamental in lottery games, as it determines the chances of winning and helps ensure fairness in the selection process.
Combinations in Locks and Security Systems
Locks and security systems often rely on combinations to provide access control. For example, think of a combination lock with three rotating discs, each containing the numbers 0 to 9. To open the lock, you need to align the discs so that the correct combination of numbers is displayed. In this case, the combination consists of six numbers, and each number has ten possibilities. This gives us a total of 10^6 (1,000,000) potential combinations! The security of the lock depends on the uniqueness and secrecy of the combination. By utilizing a large number of possible combinations, lock manufacturers can create secure systems that are challenging to bypass.
We encounter combinations in our everyday lives when setting passwords or personal identification numbers (PINs) for various online accounts, mobile devices, or bank cards. A common requirement is to have a password or PIN that consists of six numbers. This seemingly simple requirement still offers a significant number of possible combinations, making it difficult for unauthorized individuals to guess or brute force their way into our accounts. By using a combination of numbers, we add an extra layer of security to our personal information and digital assets. Remember to choose a strong and unique combination for your passwords and PINs, and avoid using easily guessable patterns such as sequential numbers or repeating digits.
By exploring these real-life examples, we can see the practical applications of combinations with six numbers. Whether it’s determining winning numbers in lottery games, providing secure access control in locks and security systems, or safeguarding our personal information with strong passwords and PINs, combinations play a crucial role in various aspects of our daily lives.
Strategies for Generating Combinations with 6 Numbers
When it comes to generating combinations with 6 numbers, there are a few strategies you can employ to increase your chances of success. Let’s explore three popular approaches: random number generation, frequency analysis, and the systematic approach.
Random Number Generation
Random number generation is a common strategy used in various fields, including generating combinations with 6 numbers. The idea behind this approach is to rely on chance and ensure that each number in the combination has an equal probability of being selected.
One way to generate random numbers is by using a computer algorithm that is designed to produce unpredictable results. This method ensures that the numbers selected are truly random and not influenced by any external factors. Many online tools and software programs are available that can generate random numbers for you with just a click of a button.
Another approach to random number generation is through physical means, such as using a lottery ball machine or a deck of cards. These methods involve mixing and shuffling the numbers or cards to ensure randomness. While these physical methods may be more time-consuming, they can add an element of excitement and anticipation to the process.
Frequency Analysis
Frequency analysis is a strategy that involves analyzing the past occurrences of numbers in order to identify patterns and trends. By studying the frequency at which certain numbers appear, you can make more informed decisions when generating combinations with 6 numbers.
To perform frequency analysis, you can gather data from previous lottery draws or any other relevant sources. By organizing the numbers and their frequencies in a table or chart, you can visually identify which numbers have appeared more frequently and which have been less common. This information can help you make strategic choices when creating your combinations.
It’s important to note that frequency analysis does not guarantee a winning combination, as the occurrence of numbers is still subject to chance. However, it can provide valuable insights and increase your chances of selecting numbers that have been historically more likely to appear.
Systematic Approach
The systematic approach is a methodical strategy that involves creating combinations by following a specific set of rules or patterns. This approach allows you to cover a wide range of possible outcomes while ensuring that no combinations are overlooked.
One systematic approach is to create combinations based on different number patterns, such as selecting numbers that follow a specific sequence or have a specific mathematical relationship. For example, you could choose numbers that increase or decrease in increments, or numbers that have a common factor.
Another systematic approach is to divide the 6 numbers into smaller groups and create combinations within those groups. This method ensures that all possible combinations are accounted for and gives you a structured framework to work with.
By utilizing a systematic approach, you can explore different patterns and structures within the combinations, increasing the likelihood of finding a winning combination.
Applications of Combinations with 6 Numbers
Probability and Statistics
Probability and statistics play a crucial role in various fields, and combinations with 6 numbers are no exception. Understanding the principles of probability and statistics can help us analyze and predict outcomes, making informed decisions based on data. Here are some key applications of combinations in the context of probability and statistics:
Analyzing Probability of Winning
Combinations with 6 numbers are often utilized in lottery games, where players select a set of numbers in the hopes of matching the winning combination. By calculating the total number of possible combinations, we can determine the probability of winning. This information is valuable for both players and lottery organizers, as it helps set prize amounts and ensures fair gameplay.
Estimating Population Parameters
In data analysis and sampling, combinations with 6 numbers can be used to estimate population parameters. By randomly selecting samples and analyzing their characteristics, statisticians can make inferences about the larger population. Combinations provide a systematic way to generate these samples and ensure representativeness, allowing for accurate estimations.
Assessing Risk and Uncertainty
In various domains, such as finance and insurance, combinations with 6 numbers are employed to assess risk and uncertainty. By analyzing historical data and calculating the probabilities of different outcomes, professionals can make informed decisions on investments, insurance premiums, and risk management strategies. Combinations help quantify the range of possible outcomes and their associated probabilities, enabling better risk assessment.
Cryptography and Encryption
In the realm of cybersecurity, combinations with 6 numbers are utilized in cryptography and encryption algorithms to ensure the confidentiality and integrity of sensitive information. Here are some ways combinations are applied in this field:
Combinations with 6 numbers form the basis for creating secure passwords and PINs. By using a combination of numbers, users can create passwords that are difficult to guess or crack. Cryptographic algorithms often include combinations to enhance the strength of passwords and protect against unauthorized access.
Creating Encryption Keys
Encryption algorithms rely on combinations to generate encryption keys. These keys are used to encode and decode sensitive data, ensuring that only authorized individuals can access the information. Combinations provide a vast number of possibilities, making it extremely difficult for hackers to decrypt the data without the correct key.
Securing Digital Communication
Combinations are also employed in the secure exchange of information over digital channels. By using combinations as part of cryptographic protocols, data can be encrypted and decrypted securely, protecting it from interception and unauthorized access. This application of combinations is crucial in areas such as online banking, e-commerce, and confidential communication.
Data Analysis and Sampling
Combinations with 6 numbers have significant implications in the field of data analysis and sampling. They help researchers and analysts draw meaningful conclusions from data and make informed decisions. Here are some key applications:
Sampling Techniques
In data analysis, combinations are used to generate samples for statistical analysis. By selecting a combination of numbers, researchers can ensure a representative sample that accurately reflects the larger population. This approach is particularly useful when dealing with large datasets, as it allows for efficient analysis without examining every individual data point.
Statistical Inference
Combinations play a vital role in statistical inference, where researchers draw conclusions about a population based on a sample. By utilizing combinations, analysts can calculate probabilities, confidence intervals, and p-values, providing insights into the reliability and significance of their findings. This enables researchers to make data-driven decisions and draw accurate conclusions from limited information.
Data Visualization and Exploration
Combinations can also be leveraged for data visualization and exploration purposes. By organizing and grouping data into combinations, analysts can identify patterns, trends, and relationships within the dataset. This approach enhances data understanding and facilitates effective communication of findings to a broader audience.
In conclusion, combinations with 6 numbers find applications in various domains, including probability and statistics, cryptography and encryption, and data analysis and sampling. Whether it’s estimating probabilities, securing sensitive information, or drawing meaningful insights from data, understanding and utilizing combinations is essential in these fields. By harnessing the power of combinations, professionals can make informed decisions, protect data, and unlock valuable insights.
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## Cross Section Area Of A Wire
How can you calculate the surface area of a wire? What is the formula for cross-sectional area? This article will answer these questions and more. Keep reading to learn how to calculate the surface area of a wire. In addition, we’ll explain what the cross-sectional area of a wire is made up of. Once you understand these terms, you’ll be able to calculate the cross-sectional area of a wire.
## Cross section area of a wire
The cross section area of a wire is the area of a circular or elliptical plane cut through an object at a right angle to its length. A wire with a circular cross section will have an area equal to the diameter of the wire times its cross section angle. On the other hand, a rectangular block cut at an angle will have an area equal to its cross section angle. Therefore, when you measure the cross section area of a wire, you must first determine how thick the wire is.
A wire’s cross section area is calculated by dividing the length of one side by its width. Then, you multiply the two sides by their squares. For example, a wire with a thickness of 3/8 inch and a width of 4 inches is equal to 0.375 inches in cross section area. By converting these measurements, you get a measurement of the wire’s diameter in square inches. Then, you multiply this value by 4,000 to get the square area of the wire.
## What is the formula for cross-sectional area?
When measuring the cross-sectional area of a wire, you will need to know its diameter and length. Fortunately, the formula for cross-sectional area is fairly straightforward. A wire’s cross-sectional area is equal to the square of its diameter, measured in mils. It can be calculated by drawing a wire rod. You can also find the cross-sectional area of a wire by squaring its diameter in inches.
To calculate the cross-sectional area of a wire, you multiply its length by its diameter. However, the formula assumes a clean cut of 90 degrees. In fact, you can get a larger result if you make a 45-degree cut. The area of a wire in a circle is approximately 1000 mils per square inch. Multi-strand cables have a cross-sectional area that is 4,000 mils squared.
To calculate the cross-sectional area of a wire, first determine its average diameter. Divide this number by 10. Next, multiply the result by the number of cables. For example, a section of wire with a diameter of 16 mm needs 32 amperes of current. You can round down to four millimeters for ease of calculations. The formula for cross-sectional area of a wire falls within the tabular data category.
## What is area of cross section in electricity?
The area of a conductor’s cross section is measured in mm2. It is equal to the surface area of a circle divided by the radius of the object. In the case of electricity, the area of a wire’s cross section is equal to the surface area of a circle with the same radius. The area of a vein will always be round. When calculating resistance, the diameter of a vein must be larger than the cross section of the conductor.
The resistance of a wire is the difficulty of allowing current to flow through it. A long wire has a greater resistance than a short one because electrons collide with more ions as they pass through. A thin wire has fewer spaces for free electrons to pass through. The resistance and area of cross section are inversely related. But you’ll find this formula confusing when you need to consider multiple events.
## What is the surface area of a wire?
A wire’s surface area can be measured by calculating its cross-sectional area. The surface area of one strand of 0.20mm diameter copper wire is equivalent to the surface area of a circle of radius r. This measurement can be rounded to the nearest hundredth of a meter. Alternatively, you can find out the surface area of a cylinder made of one mm wire by using a known formula.
A large-diameter wire has a higher surface area than a small-diameter wire. Wires with smaller diameters are typically measured in millimeters. In a nutshell, wires can be measured in square units or in circular mils. The cross-sectional area of wire is most easily calculated in circular mils rather than in square units, as the measurement scale of wire size is inverse.
## What shape is the cross-sectional area of a wire?
The cross-sectional area of a wire is equal to the area of a circle of diameter d with radius r. As the diameter of the wire is larger than its thickness, the area will always be larger. This measurement is also useful to understand the differences between stranded wire and solid wire. The cross-sectional area of wires can also be used to determine the resistance.
To understand the formula, first we need to define the cross-section. A cross-section is the common region of a 3D object. For example, a long cylindrical tube will have a cross-section that is a concentric circle. A beam will be named based on the shape of its cross-section. Basically, the area of a cross-section is the same as its height, width, and thickness. A cross-section calculator will give you the cross-sectional area of a cylinder of diameter 10, height H, and thickness t1.
A wire can be circular or oval in cross-section. An ellipsoidal cross-section is also possible. Both wire shapes have their center waist narrower than the rest. The S-shaped wire has a much thinner wall and beam than a Z-shaped wire. This means that the S-shaped wire is easier to bend than a Z-shaped one. In general, the cross-sectional area of a wire will be greater than that of a square.
## Is cross-sectional area the same as diameter?
A cross-sectional area is the squared length of one side of a conductor. Then, multiply the square length of the other side. For example, take a rectangular conductor with a thickness of 3/8 inch and a width of four inches. The thickness is expressed as 0.375 inch. This is equivalent to 4,000 mils. Similarly, a width expressed in inches is equal to 375 mils. Then, multiply 375 mils by 4,000 mils to find the cross-sectional area.
A wire’s cross-sectional area is measured using the formula: A = 1/pd2, where p is the length in feet. Its diameter is the area of a circle with radius r. The cross-sectional area of an n-gauge wire is equal to the square of its diameter. Once you have found the cross-sectional area of the wire, you can calculate the average diameter of the wire.
## What is the difference between area and cross-section.
The size of a wire can be measured using the area and cross section of the conductor. The area of a wire refers to the space in which the copper wires can pass. It is important to note that the cross-section area and diameter are not the same. Likewise, stranded wire has a larger cross-section than solid wire. So, the size of a solid wire is more important than its cross-sectional area.
The cross-sectional area of a wire is smaller than its overall surface area. Generally speaking, large-diameter wires have greater cross-sectional area than small-diameter wires. The cross-sectional area of a wire can be expressed in either square units or in circular mils. The area of a wire can also be expressed in the gauge scale. The circular-mil measurement of a wire is more convenient to calculate, as it eliminates the “pi” and d/2 (radius) factors.
The cross-sectional area of a wire affects its resistance. A wider wire has lower resistance than a thinner one. Therefore, the wider the cross-section area, the lower the resistance of the wire. To further understand this, consider the example of a water pipe. The wider a pipe is, the more water it flows. Therefore, a wire with a wider cross-section area has less resistance to the flow of electric charge.
## What is area formula?
What is the area formula for cross section of a wire, and how do I find its area? Wire cross sections are shaped like a circle, but the surface area of each section varies. Wires are a mix of different materials, and one type is generally more dense than another. One type of wire is stranded, which is a single-core wire that has been twisted together.
The cross section of a wire is a two-dimensional representation of the object. When cutting a solid wire into multiple sections, the two-dimensional slices of the wire will be different. The cross-section area is known as the SS, and it is measured in mm2. A stranded wire will have a greater area than a solid wire. Both types of wire have a different resistance.
One method of calculating the cross-section area of a wire is to measure the diameter of the wire. This measurement is easy. Take a long piece of wire and wind it around a pencil until the “tails” fit tight together. Make sure to use full turns that fit tightly and have no gaps between them. You’ll need to divide the length of the segment by the number of turns to determine the diameter. For example, if the wire has 11 turns, the resulting diameter will be 7.5 mm. Then divide that number by 11 and get 0.68 mm.
## How to Find the Cross-Sectional Area of a Rectangle Duct
A rectangular duct can be divided into sixteen sections. Each section indicates the average velocity of the air flowing through the duct. The cross-sectional area of a duct can be calculated using a formula that represents the cross-sectional area and velocity of the air. The cross-sectional area of a rectangular duct is equal to the flow rate of one cubic meter of air per second, where V is the velocity of the air.
The area of a solid depends on its shape and the angle between its axis of symmetry and the plane in which it intersects. The area of a rectangular solid is equal to the base area x its height. The cross-section area of a rectangular duct is equal to the base area plus its height. If you want to calculate the cross-sectional area of a square, you would multiply the width of the cylinder by the height.
A duct’s cross-sectional area is measured in square inches, and this value can be calculated from the length and circumference of the cylinder. The resulting square area is then multiplied by the radius of the cylinder. If the duct is round, the area of the cylinder is equal to p*R2. This is true for both rectangular and oval shaped ducts, but rectangular ducts are more accurate.
## What is Meant by Cross-Sectional Area of Conductor?
The cross-sectional area of a conductor is the surface area that is the same lengthwise, no matter the configuration. Its area can be measured in square mils or in the actual cross-section of the conductor. The square mil is a unit of measurement, where one mil equals the area of a square with sides of 1 mil. A 3/8-inch conductor, for example, has a cross-sectional area of 3/8 of an inch, and it is 4 inches wide. Therefore, a 3/8-inch conductor is 4 inches wide and 3/8 inch thick. The area of a circular conductor is equal to 0.375 inch. A rectangular conductor will have an area of 9 square mils, so a 3/8-inch square will have a cross-sectional area of
A cable is a small pipe. The configuration determines its outline. For example, if you cut a round metal rod in half, the cross-section will be two circles of a specific thickness. The cross-sectional area of a conductor (SS) is measured in mm2, whereas the area of a vein is round. Using this formula, you can find the cross-sectional area of a conductor by multiplying the radius of the vein by its circumference, R.
Another application for cross-sectional area is in nuclear physics. The effective size of a nuclear atom is defined by the cross-section of the nucleus. The cross-sectional area of a nuclear atom is the area of a circle divided parallel to the base, and the probability of the neutron interacting with the target atom is expressed by its cross-section. Nuclear fission relies on this mechanism.
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# Find the cost of laying grass in a triangular field of sides $50m,65m$and $65m$at the rate of Rs 7 per ${m^2}$.
Last updated date: 13th Sep 2024
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Hint: We need to use the unitary method in this case. In the unitary method, if we know the price of a particular product we can find the price of the number of products by multiplication also we can find the price of a single product if we are given with price of the numbers of products with the division.
The area of the triangular field is found by using here’s formula i.e.
$\sqrt {S(s - a)(s - b)(s - c)}$ where a, b, c are sides of the triangle
S is semi perimeter.
Value of $s = \dfrac{{a + b + c}}{2}$
Therefore
Given sides of triangular grass field $50n,65m$and $65m$ respectively
Let $a = 50m$
$b = 65m$
and $c = 65m$
Now $S = \dfrac{{a + b + c}}{2} = \dfrac{{50 + 65 + 65}}{2} = 90m$
Area of triangular field is found by heron’s formula
i.e. Area $= \sqrt {S(s - a)(s - b)(s - c)}$
put $s = 90m,a = 50m,b = 65m$ and $c = 65m$
Area $= \left[ {{{10}^2} \times {3^2} \times {2^2} \times {{25}^2}} \right]$
$= \sqrt {90(40)(25)(25)}$
$= \sqrt {10 \times 910 \times 4 \times 25 \times 25}$ $\left[ {\because 90 = 9 \times 10\,\,40 = 4 \times 10} \right]$
$= \left[ {{{10}^2} \times {3^2} \times {2^2} \times {{25}^2}} \right]$ $\left[ {\because {3^2} = 9\,4 = {2^2}} \right]$
$= 10 \times 3 \times 2 \times 25$
Area$= 1500{m^2}$
BY unitary method,
Cost of laying $I{m^2}$ grass $= Rs.7$(Given)
Cost of laying $1500{m^2}$ grass
$= Rs\,1500 \times 7 = Rs\,10500$
Hence the cost of laying is Rs. $1050$.
Note: The question can also be solved by using the concept of isosceles triangle of triangle In, isosceles triangle two sides, are equal.
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 91/4 - 45/6 = 53/12 = 4 5/12 ≅ 4.4166667
Spelled result in words is fifty-three twelfths (or four and five twelfths).
### How do you solve fractions step by step?
1. Conversion a mixed number 9 1/4 to a improper fraction: 9 1/4 = 9 1/4 = 9 · 4 + 1/4 = 36 + 1/4 = 37/4
To find new numerator:
a) Multiply the whole number 9 by the denominator 4. Whole number 9 equally 9 * 4/4 = 36/4
b) Add the answer from previous step 36 to the numerator 1. New numerator is 36 + 1 = 37
c) Write a previous answer (new numerator 37) over the denominator 4.
Nine and one quarter is thirty-seven quarters
2. Conversion a mixed number 4 5/6 to a improper fraction: 4 5/6 = 4 5/6 = 4 · 6 + 5/6 = 24 + 5/6 = 29/6
To find new numerator:
a) Multiply the whole number 4 by the denominator 6. Whole number 4 equally 4 * 6/6 = 24/6
b) Add the answer from previous step 24 to the numerator 5. New numerator is 24 + 5 = 29
c) Write a previous answer (new numerator 29) over the denominator 6.
Four and five sixths is twenty-nine sixths
3. Subtract: 37/4 - 29/6 = 37 · 3/4 · 3 - 29 · 2/6 · 2 = 111/12 - 58/12 = 111 - 58/12 = 53/12
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(4, 6) = 12. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 6 = 24. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - thirty-seven quarters minus twenty-nine sixths = fifty-three twelfths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Bucket 4
The bucket has 5/8 gallons of water. The bucket tips over, and 7/12 gallon of water pours out. How much water is left in the bucket, in gallons, written as a fraction?
A basket contains three types of fruits weighing 87/4 kg in all. If 23/4 kilograms of these are oranges, 48/7 kg are mangoes, and the rest are apples. What is the weight of the apples in the basket?
• Leo hiked
Leo hiked 6/7 of a kilometer. Jericho hiked 2/3 kilometer. Who covered a longer distance? How much longer?
• Savings
Eva borrowed 1/3 of her savings to her brother, 1/2 of savings spent in the store and 7 euros left. How much did she save?
• Colored blocks
Tucker and his classmates placed colored blocks on a scale during a science lab. The brown block weighed 8.94 pounds, and the red block weighed 1.87 pounds. How much more did the brown block weigh than the red block?
• Visit to grandfather
Shane's family traveled 3/10 of the distance to his grandfather’s house on Saturday. They traveled 4/7 of the remaining distance on Sunday. What fraction of the total distance to his grandfather’s house was traveled on Sunday?
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Linear regression explained
# Linear regression explained
## Intro
But for now, let's focus on linear regression
## What is linear regression?
Let's say you measured a couple of points, and plotted them on a graph (x, y). These values can represent anything you want (The cost of a house as a function of its livable surface area, the weight of a tree as a function of its height, etc...).
For my example, these are the values that I'm going to use:
```(1,1)
(1.5, 2)
(2, 0.5)
(2.7, 3.3)
(3.6, 4.5)
(3.7, 4.7)
(4.5, 7.5)
(6.1, 6.5)
(6.4, 8.8)
(7.5, 9.6)
(8.5, 9.7)```
When plotted, this is what they look like:
Now, let's ask ourselves: If we added a point, knowing its x-value, could we predict its y-value? For example, what could be an estimation of the y-value for x=5?
To answer this, we'll need to find a relation between y and x. Since it's linear, we'll be able to write (as you might remember from school) something that looks like the equation of a line: y = ax + b
In linear regression, we write: y = b0 + b1*x
The whole idea behind linear regression is to find those b0 and b1 values.
Using our example, we could take x=5 and see what y-value would be found.
For the curious ones out there, this is what a good fit (although not the best) might look like:
Using this model, if we plug in x=5, we get y=1.25*5 = 6.25
## How does linear regression work?
To make linear regression work, we have to establish a relation between the points and the estimated line. This relation will be the "cost" (the error). With different existing cost functions, we'll choose one and try to minimize it. The smaller that value is, The best fit a line will be. We'll try to make it approach zero, but with real data, it's almost always impossible (except if the line is a perfect fit).
Here is an example of how different lines will yield different costs:
Notice that one of the lines fits better than all others, and this yields the lowest cost
To find the best fit, we'll have to simply find the best b_0 and b_1.
For the sake of simplicity, we'll use the y=ax+b notation and therefore call b_0 b and b_1 a.
The idea is to introduce a cost (cost(a, b)) function and try to minimize that function. We'll use the gradient descent method (which I'll soon write an article about) to change the values of a and b to minimize cost(a, b). We'll have to calculate the partial derivatives of the cost relative to a and b to be able to use the gradient descent method, so I'd recommend choosing a not-so-complicated cost function to start with.
This might be slightly confusing so let's actually get into an example:
For the cost function we'll choose:
Then we'll have to see in what way a and b affect the cost function. To do this we'll calculate both partial derivatives:
The result we got is actually the slope to increase the value of
cost(a, b) so we'll have to take its inverse, and add a learning rate to modify the values of a and b:
Now to minimize this function, we'll go over a certain number of epochs (iterations) repeating the same previous step. By the end, we should have optimized our a and b values to fit the points nicely.
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Wazeesupperclub.com General What is a kite shape?
What is a kite shape?
What is a kite shape?
Explanation: A kite is a four-sided shape with straight sides that has two pairs of sides. Each pair of adjacent sides are equal in length. A square is also considered a kite.
What are the properties of a kite in geometry?
Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Other important polygon properties to be familiar with include trapezoid properties, parallelogram properties, rhombus properties, and rectangle and square properties.
How do you draw a kite in geometry?
60 second suggested clip0:452:54Geometry 6.6c, Construct a Kite with a compass and straightedge – YouTubeYouTube
What is kite Class 8?
A kite is a quadrilateral that has 2 pairs of equal-length sides and these sides are adjacent to each other. Properties: The two angles are equal where the unequal sides meet.
What are the angles of kite?
The perimeter of a kite is equal to the sum of the length of all of its sides. The sum of the interior angles of a kite is equal to 360°.
How do you teach a kite in geometry?
What are the angles in a kite?
Angles in a kite A kite is symmetrical. So it has two opposite and equal angles.
What are the angles of a kite?
Angles in a kite
• y ∘ = 180 ∘ − ( 90 ∘ + 40 ∘ )
• y ∘ = 180 ∘ − 130 ∘
• y ∘ = 50 ∘
What is trapezium and kite?
It has two pairs of equal adjacent angles. Pairs of adjacent sides are equal for a Kite. One pair of opposite angles are equal for an Isosceles trapezoid. It has one pair of opposite angles that are equal.
What is kite Byjus?
Properties of kite The two angles are equal where the unequal sides meet. It can be viewed as a pair of congruent triangles with a common base. It has 2 diagonals that intersect each other at right angles. A kite is symmetrical about its main diagonal. The shorter diagonal divides the kite into 2 isosceles triangles.
What does kite mean In geometry?
In math what is kite? In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent.
What is a kite dealing with geometry?
– Line segments and their measures inches – Line segments and their measures cm – Segment Addition Postulate – Angles and their measures – Classifying angles – Naming angles – The Angle Addition Postulate – Angle pair relationships – Understanding geometric diagrams and notation
What are the characteristics of a geometric kite?
Two pairs of adjacent sides are equal.
• One pair of opposite angles are equal.
• The diagonals of a kite are perpendicular to each other.
• The longer diagonal of the kite bisects the shorter diagonal.
• The area of a kite is equal to half of the product of the length of its diagonals.
• What are the basic properties of a kite in geometry?
P1 : A four sided shape.
• P2 : All sides are of equal length.
• P3 : It has four angles.
• P4 : The opposite angles are of equal angle.
• P5 : If the adjacent angles are same then the angles are 90 ∘.
• P6 : If P5 happens,then the Rhombus is a Square.
• P7 : It has two diagonals.
• P8 : The diagonals bisects each other at 90 ∘.
|
# Solving Quadratic Equations by Factors
The factorised form of a quadratic equation is $(ax+b)(cx+d)=0$. Using the Null Factor law, we can solve this equation algebraically to find $x$
\begin{align} \displaystyle (ax+b)(cx+d) &= 0 \\ ax+b &= 0 \text{ or } cx+d =0 \\ x &= -\dfrac{b}{a} \text{ or } x = -\dfrac{d}{c} \end{align}
### Example 1
Solve $(x-1)(x+2)=0$.
\begin{align} \displaystyle (x-1)(x+2) &= 0 \\ x-1 &= 0 \text{ or } x+2 = 0 \\ \therefore x &= 1 \text{ or } x = -2 \end{align}
### Example 2
Solve $x(x-3)=0$.
\begin{align} \displaystyle x(x-3) &= 0 \\ x &= 0 \text{ or } x-3 = 0 \\ \therefore x &= 0 \text{ or } x = 3 \end{align}
### Example 3
Solve $(2x-1)(3x+2)=0$.
\begin{align} \displaystyle (2x-1)(3x+2) &= 0 \\ 2x-1 &= 0 \text{ or } 3x+2 = 0 \\ \therefore x &= \dfrac{1}{2} \text{ or } x = -\dfrac{2}{3} \\ \end{align}
## High School Math for Life: Making Sense of Earnings
Salary Salary refers to the fixed amount of money that an employer pays an employee at regular intervals, typically on a monthly or biweekly basis,…
## Mastering Integration by Parts: The Ultimate Guide
Welcome to the ultimate guide on mastering integration by parts. If you’re a student of calculus, you’ve likely encountered integration problems that seem insurmountable. That’s…
## Division using Exponents (Indices)
If we are given $a^8 \div a^3$, we can also write this as $\dfrac{a^8}{a^3}$, which means \$\dfrac{a \times a \times a \times a \times a…
## Probability Pro: Mastering Two-Way Tables with Ease
Welcome to a comprehensive guide on mastering probability through the lens of two-way tables. If you’ve ever found probability challenging, fear not. We’ll break it…
## Conversion of Parametric Equations to Cartesian Equations
Welcome to the world of parametric equations and the art of converting them into Cartesian form. If you’re a student wrestling with parametric equations or…
|
# Complements
In the previous lesson, we learned about number systems – binary, decimal, octal and hexadecimal number systems in detail. Complements are very helpful in performing subtraction and save computing time and lead to simple circuits.
Before we move to subtraction, let’s understand what complement is all about. In simple words, if there is group with have R items, and you took S items from the group, then you need the complement R-S items to complete the group.
These R-S items are called the complements.
In a number system, there is a fixed range of numbers and when you pick any number from the system, you have a complement for that number in the number system.
## Types of Complements
There are two types of complements.
1. Diminished Radix Complement or r-1’s complement.
2. Radix Complement or r’s complement.
### Diminished Radix Complement [r-1’s complement]
Suppose a number N is given
N = Number
r = base of the number
n = Number of digits in the number.
The r-1’s complement of the number is given by the formula.
(r^n - 1) - N
### Radix Complement [r’s complement]
To find radix complement of a number N, where
N = Number
r. = base of the number
n = Number of digits in the number.
The r’s complement of the number is given as
r^n - N
or
r-1’s \hspace{3px}complement + 1 => [( r^n -1) – N] + 1
## Examples – Diminished Radix Complements
Here are some example problems for diminished radix complements.
### Problem1:
Find 9’s complement of 258.
### Solution:
\begin{aligned}
&N = 258\\
&r = 10\\
&n=3\\\\
&We \hspace{3px} use \hspace{3px}the \hspace{3px}formula\hspace{3px} for \hspace{3px}r-1’s \hspace{3px}complement.\\\\
&(r^n – 1) – N\\
&(10^3 – 1) – 258\\
&999 – 258\\
&The \hspace{3px} 9’s \hspace{3px}complement \hspace{3px}of \hspace{3px}258 \hspace{3px}is \hspace{3px}741.
\end{aligned}
### Problem2:
Find 1’s complement for 1101.
### Solution:
In the given problem,
\begin{aligned}
&N = 1101\\
&r = 2\\
&n = 4\\\\
&The \hspace{3px}formula \hspace{3px}for \hspace{3px}r-1’s \hspace{3px}complement \hspace{3px}where r = 2\hspace{3px} is \hspace{3px}(r^n - 1) – N.\\\\
&(2^4 - 1) - 1101\\\\
&Note:\hspace{3px} Binary \hspace{3px}equivalent for \hspace{3px}2^4 = 10000.\\\\
&(10000 - 1)1101\\
&1111 - 1101\\
&The \hspace{3px} 1’s \hspace{3px}complement \hspace{3px}for \hspace{3px}1101 \hspace{3px}is \hspace{3px}0010.
\end{aligned}
### Method 2
The second method to find the r-1’s complement is to subtract each digit in the number by r-1.
In our case, r = 2 so r – 1 = 1
Given that the 1’s complement of binary number is
\begin{aligned}
&1 – 1 = 0\\
&1 – 1 = 0\\
&1 – 0 = 1\\
&1 – 1 = 0\\\\
\end{aligned}
### Problem3:
Find 7’s complement of octal number 234.
### Solution:
The range of digits for octal number is 0-to-7. Subtract each digit with 7 will give r-1’s complement for octal.
\begin{aligned}
&7 - 2 = 5\\
&7 - 3 = 4\\
&7 - 4 = 3\\\\
&The \hspace{3px} r-1's \hspace{3px} complement \hspace{3px}is \hspace{3px} 543.
\end{aligned}
## Examples – Radix Complements
There are two methods to find the r’s complement of a number.
\begin{aligned}
&Method \hspace{3px}1: r^n – N\\
&Method \hspace{3px}2: r-1’s \hspace{3px}complement + 1
\end{aligned}
### Problem4:
Find 10’s complement for 432.
### Solution:
Given that
\begin{aligned}
&N = 432\\
&r = 10\\
&n = 3\\
&
\end{aligned}
Method 1:
\begin{aligned}
&First \hspace{3px}we \hspace{3px}will \hspace{3px}find \hspace{3px}the \hspace{3px} 10’s \hspace{3px}complement \hspace{3px}using \hspace{3px}r^n – N.\\
&r^n – N = 1000 – 432 = 568\\
&Therefore,\\
&10’s \hspace{3px}complement \hspace{3px}of\hspace{3px} 432 \hspace{3px}is \hspace{3px}568.
\end{aligned}
Method 2:
By using method 2, you need to find the 9’s complement and then add 1 to the result to get 10’s complement.
Step1: Subtract each digit by 9 to get 9’s complement.
\begin{aligned}
&9 - 4 = 5\\
&9 - 3 = 6\\
&9 - 2 = 7\\
\end{aligned}
Step2: Add 1 to the result.
\begin{aligned}
&567 + 1\\
&=568\\
&568 \hspace{3px}is \hspace{3px}the \hspace{3px}10’s \hspace{3px}complement \hspace{3px}of \hspace{3px}432.
\end{aligned}
### Problem5:
Find 2’s complement of the binary number 1001.
### Solution:
Method 1:
We can obtain the solution in using two methods.
\begin{aligned}
&To \hspace{3px} find \hspace{3px} 2’s \hspace{3px}complement \hspace{3px}using \hspace{3px}method 1 \hspace{3px}use \hspace{3px}the \hspace{3px}formula r^n – N.\\
&Given \hspace{3px} that\\
&N = 1001\\
&r = 2\\
&n = 4\\
&r^n - N = 10000 – 1001\\\\
&\hspace{8px}\sout10000\\
&-1001\\
&\hspace{8px}----\\
&\hspace{11px}0111\\\\
&The \hspace{3px}2’s \hspace{3px}complement \hspace{3px}for \hspace{3px}1001 \hspace{3px}is \hspace{3px}0111.
\end{aligned}
Method 2:
To use second method, find the 1’s complement for 1001 and then add 1 to the result.
Step1: find the 1’s complement of 1001.
\begin{aligned}
1001 -> 0110 (1’s complement)
\end{aligned}
Step2: Add 1 to result.
\begin{aligned}
&0110 + 1 => 0111\\
&Therefore,\\
&0111 \hspace{3px}is \hspace{3px} the \hspace{3px}2’s \hspace{3px}complement \hspace{3px}of \hspace{3px}0110.
\end{aligned}
## References
1. John.F.Wakerly. 2008. Digital Design: Principles And Practices, 4/E. Pearson Education, India.
2. Mano, M. Morris. 1984. Digital Design. Pearson.
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Q20.
Expert-verified
Found in: Page 581
### Algebra 2
Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903
# Find the next four terms of each arithmetic sequence.$\frac{18}{5},\frac{16}{5},\frac{14}{5},...$
The next four terms of the arithmetic sequence are $\frac{12}{5},2,\frac{8}{5},\frac{6}{5}$.
See the step by step solution
## Step 1. Given Information.
The given arithmetic sequence is $\frac{18}{5},\frac{16}{5},\frac{14}{5},...$
## Step 2. Calculation.
A sequence is said to be arithmetic when the difference between the consecutive terms is always the same.
Here the difference between the first two terms is $\frac{16}{5}-\frac{18}{5}=-\frac{2}{5}$
The difference between the next two terms is $\frac{14}{5}-\frac{16}{5}=-\frac{2}{5}$
So, the common difference is $-\frac{2}{5}$
The next term in the arithmetic sequence is $\frac{14}{5}-\frac{2}{5}=\frac{12}{5}$
The next term in the arithmetic sequence is $\frac{12}{5}-\frac{2}{5}=\frac{10}{5}=2$
The next term in the arithmetic sequence is $2-\frac{2}{5}=\frac{8}{5}$.
The next term in the arithmetic sequence is $\frac{8}{5}-\frac{2}{5}=\frac{6}{5}$.
## Step 3. Conclusion.
Hence, the next four terms of the arithmetic sequence are $\frac{12}{5},2,\frac{8}{5},\frac{6}{5}$.
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5
Q:
# The expression (11.98 * 11.98 + 11.98 * (x) + 0.02 * 0.02) will be a perfect square for x equal to :
A) 0.01 B) 0.02 C) 0.03 D) 0.04
Explanation:
Given = (11.98)^2 + (0.02)^2 + 11.98 * x.
For the given expression to be a perfecr square, we must have
11.98 * x = 2 * 11.98 * 0.02 => x = 0.04
Q:
1/8 as a decimal?
A) 0.125 B) 0.225 C) 0.325 D) 0.425
Explanation:
1/8 to convert into decimal,
Find an equivalent fraction with a denominator of a power of 10.
In this case, we will use 1000.
This can now be written as a decimal with 3 decimal places (because we have thousandths.
i.e, 0.125
Hence,
2 133
Q:
What is $\frac{1}{5}$ as a decimal?
A) 0.333 B) 0.666 C) 0.200 D) 0.452
Explanation:
Here I converted the original fraction to a fraction with the denominator being 10, because decimals can be written in the tenths place.
2 194
Q:
125 over 1000 in Simplest Form?
A) 2/9 B) 1/8 C) 3/8 D) 4/7
Explanation:
125 over 1000 in Simplest Form means $\frac{\mathbf{125}}{\mathbf{1000}}$ in its simple fraction form.
Now, to get the simplest form of 125/1000, find the HCF or GCD of both numerator and denominator i.e, 125 and 1000.
HCF of 125, 1000 = 125
Then, divide both numerator and denominator by 125
i.e,
Hence, $\frac{\mathbf{1}}{\mathbf{8}}$ is the simplest form of 125 over 1000.
3 131
Q:
$\frac{\mathbf{1}}{\mathbf{8}}$ as a decimal?
A) 0.375 B) 0.247 C) 0.214 D) 0.125
Explanation:
Here $\frac{\mathbf{1}}{\mathbf{8}}$ in decimal means 8th part in 1.
We know that $\frac{\mathbf{1}}{\mathbf{10}}$ = 0.1
Then,
=>
3 198
Q:
Express 0.875 and 0.375 as a Fraction.
A) 3/11 & 7/11 B) 7/9 & 9/13 C) 7/8 & 3/8 D) 7/9 & 9/11
Explanation:
Given 0.875 and 0.375
0.875 ::
After decimal point there are 3 digits 875
Now, mutiply and divide 0.875 with 1000
Similarly,
0.375 ::
After decimal point there are 3 digits 375
Now, mutiply and divide 0.375 with 1000
4 388
Q:
Express 3/8 and 1/10 as a Decimals?
A) 1.375 & 1 B) 0.375 & 1 C) 13.75 & 1.1 D) 0.375 & 0.1
Explanation:
3/8 as a Decimal :
In this we should convert the denominator into 10, 100, 1000, 10000, 100000,... to make it simple.
Now to make 8 into 1000, we multiply numerator and denominator with 125
=>
1/10 as a Decimal :
1 divided into 10 parts of each 0.1.
Therefore, 3/8 = 0.375 and 1/10 = 0.1 in decimals.
2 234
Q:
5/8 as a Decimal Equals To
A) 0.625 B) 0.541 C) 0.258 D) 0.147
Explanation:
5/8 is Nothing but to divide 5 into 8 parts.
As 5 is smaller than 8, the quotient starts with a decimal point to make 5 as 50 to be divisible by 8. (0.)
Now 50 goes for 6 times in 8 leaving a remainder 2. (0.6)
Again 2 is smaller than 8, As it already has decimal point in the quotient, now 2 becomes 20.
Now 20 goes for 2 times in 8 leaving a remainder 4. (0.62)
Again 4 is smaller than 8, As it already has decimal point in the quotient, now 4 becomes 40.
Now 40 goes for 5 times in 8 leaving a remainder 0. (0.625)
Therefore, the decimal value of $\frac{\mathbf{5}}{\mathbf{8}}$ is 0.625.
Now find a decimal value of $\left(\frac{\mathbf{11}}{\mathbf{17}}\right)$ and Discuss your at Discuss.
6 210
Q:
If two fractions, each of which has a value between 0 and 1, are multiplied together, the product will be :
A) always greater than either of the original fractions B) always less than either of the original fractions C) sometimes greater and sometimes less than either of the original fractions D) remains the same
Answer & Explanation Answer: B) always less than either of the original fractions
Explanation:
We can easily Answer this by taking some values.
The question states that the two fractions, each have values between 0 and 1.
Let us say one of the fraction is 1/2 and the other fraction is 1/3 .
The product of the two fractions is 1/2 x 1/3 = 1/6 .
is lesser than both 1/2 and 1/3 .
So, the correct answer is that the product is always less than either of the original fractions
|
# What is 8 of 15000?
#### Understand the Problem
The question is asking us to calculate the value of 8 percent of 15,000. This is a basic percentage calculation.
$1200$
The value of 8 percent of 15,000 is $1200$.
#### Steps to Solve
1. Identify the percentage and the whole amount
We need to find 8 percent of 15,000. Here, 8 percent can be represented as a fraction, which is $\frac{8}{100}$.
1. Convert percentage to decimal
Convert the percentage into a decimal by dividing by 100: $$8% = \frac{8}{100} = 0.08$$
1. Multiply by the whole amount
Now we can find 8 percent of 15,000 by multiplying the decimal by the whole amount: $$0.08 \times 15000$$
1. Calculate the final value
Now, let's do the multiplication: $$0.08 \times 15000 = 1200$$
The value of 8 percent of 15,000 is $1200$.
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Circumference and Area of a Circle: A Middle School Math Hands-On Lesson
Janine is a published author in Only Trollops Shave Above the Knees, appears on The Huffington Post and at Confessions of A Mommyaholic.
In Middle School Math, yet again another topic that comes to mind that middle schoolers need to learn and will be tested on is circles, specifically circumference and area. These two concepts can be downright boring if taught by the old chalk and talk method.
But lo and behold, I continually tried to find new and creative ways to teach some of the most mundane and boring math topics. Even before getting to actual activity at hand, I was lucky enough to teach alongside some really fabulous teachers and one can me this idea for how to introduce the two concepts. When thinking of circles, students are first and foremost introduced to a few basic principles.
So what are the words that kids must learn the definitions to before they can even begin to work with circles? Well look no further here they are.
Circle Definitions:
The radius of a circle is the distance from the center of the circle to the outside edge. In the picture to the right, the radius is labeled and is the the yellow line from the edge of the circle to the midpoint.
Diameter
The diameter of a circle is longest distance across a circle. (The diameter cuts through the center of the circle. This is what makes it the longest distance.) In the picture to the right, the diameter of the circle is clearly labeled and the yellow line that goes from one end of the circle to the other cutting directly through the middle of the circle.
Circumference
The definition of the circumference of a circle is quite simply the perimeter or the distance around the outer edge of the circle. Looking at the picture to the right, the circumference is the bright yellow line on the outside of the circle.
So the formula for circumference is C = π d, where d= the diameter of the circle and π = 3.141592...
Area
Area is defined as the amount of space inside the boundary of a flat or 2-dimensional object, such as a triangle or circle. In the circle picture to the right, the area is the inside of the circle that is shaded in purple.
The formula for area of a circle is A = π r2, where r = the radius of the circle and π = 3.141592...
So How Can We Remember the Actual Circle Formulas?
Once I briefly introduce these definitions, then I talk a bit about why in real life we would need to find area and circumference of a circle. I model on the smart board a google search about Real Life uses and show the top 5 according to Yahoo. They are as follows:
1. Car makers can measure car wheels to make sure they fit.
2. Race car engineers can use it to find out what size tire gives them the most performance.
3. Bakers can use it to make pies and other circular stuff.
4. Military engineers can use them to balance helicopter blades.
5. Aircraft engineer can use them for propeller efficiency.
Scroll to Continue
Bakers and a Mnemonic Device to Learn the Circumference and Area Definitions:
The real life example that I stop on is Bakers and how they use this with making pies. I bring in two fresh pies to illustrate my point. The reason for this is that I have a cute little mnemonic device to remember the actual formulas for circumference and area. For circumference, I show the class a cherry pie and teach them that "Cherry Pies Delicious" or C = π D. And for area, I then show them an apple pie and teach them that "Apple Pies Are Too" or A = π r2.
Now, we will measure the radius and the diameter of each pie and then will find out the area and circumference of both pies from finding both of these out and plugging them into both the formulas we just learned.
1. Apple Pie:
The apple pie was baked in a 9 inch pie pan. So we know from this bit of information that the diameter is 9 inches. Well, what is the radius? It will be half of the diameter and be 4.5 inches. So now let us plug into our formula to find both the circumference and area too!
So from earlier we know that for circumference, C = π d: C = π 9, (diameter = 9), so C =28.2743338. So if we round to the nearest tenth, the c = 28.3 inches.
Now for the area, we know that the formula is A = π r2. So A = π (4.5)2 = π (20.25) =63.61725123519331. Again, let's round and we get the area to the nearest tenth of the circle to be 63.6 inches.
2. Cherry Pie:
The cherry pie was baked in a 8 inch pie pan. So we know from this bit of information that the diameter is 8 inches. Well, what is the radius? It will be half of the diameter and be 4 inches. So now let us plug into our formula to find both the circumference and area too!
So from earlier we know that for circumference, C = π d: C = π 8, (diameter = 9), so C = 25.132741228718345. So if we round to the nearest tenth, the c = 25.1 inches.
Now for the area, we know that the formula is A = π r2. So A = π (4)2 = π (16) = 50.26548245743669. Again, let's round and we get the area to the nearest tenth of the circle to be 50.3 inches.
3. The Difference of the Circumference and Area of the Apple (9 Inch Pan) and Cherry Pie (8 Inch Pan):
Circumference Difference:
28.3 inches (Apple Pie Circumference) - 25.1 inches(Cherry Pie Circumference) = 3.2 inches.
Area Difference:
63.6 inches (Apple Pie Area) - 50.3 inches (Cherry Pie Area) = 13.3 inches.
What we have learned is the at even changing the diameter an inch can change both the circumference and area of the circle ever so slightly.
And now once we are done with the actual lesson, I usually offer a piece of either of the pies to anyone who wants to try them. So a good lesson was learned and a tasty reward to boot!!
Summing This Lesson Up..
I love this lesson, because it is another hands-on lesson using the two different types of pie something that yet again most middle school students are not only aware of, but interested in. Now, when they hear their parents or someone else speaking of making pies maybe they will remember a bit about the circle definitions and formulas learned even after the topic and test are long over and behind them. And as a teacher that is truly something you hope for that the student takes away something from your lesson and doesn't just forget it once the test is long gone! Anyone who has read any of my other math teaching articles previously will know from them that I am a strong believer in using stuff that interests middle school students to help them learn many of the basic concepts that are a requirement. I truly enjoy engaging my students and showing them how we can use math in everyday life and believe this lesson is another one that does just that.
Jennifer on September 18, 2018:
Brilliant! Ty!!!!
Janine Huldie (author) from New York, New York on September 17, 2012:
Thundermama, I wish I could still be a teacher and it would be my honor and privilege if I were to be able to teach your daughters. Thank you so much though for all your kindness here and for saying such lovely things about my lesson and teaching too :)
Catherine Taylor from Canada on September 17, 2012:
Janine, how I wish someone like you was teaching my daughters math. It is a shame you are not currently a teacher, but fantastic that you are willing to share your gifts for teaching here.
Janine Huldie (author) from New York, New York on August 29, 2012:
No problem about missing Joseph, believe me I have been there before too. Just happy you did find my article and are here now :) I too had to memorize a lot in school and remember having some really lame excuses for teachers who phoned it in. So that is why I vowed not to be like that myself as a teacher. Thank you for voting, sharing and tweeting too!!
Joseph De Cross from New York on August 29, 2012:
I did miss this one Janine. Totally awesome. There is no better way. Your care and lay out pops in our face. Sadly it was too late for me, by 6th grade I had to memorize these formulas by using them. But You did perfect. Voting this up and sharing and tweeting.
Janine Huldie (author) from New York, New York on August 29, 2012:
Eddy, your words here mean the world to mean. I tried my best when I was teaching to be the best for my students that I could be. I miss teaching greatly and do hope that someday soon in the future that I can return to the profession that I do have a passion for. Thank you seriously for your kindness here and fro your votes/support!! Hope you are enjoying your day too...Janine :)
Eddy Jones from Wales. on August 29, 2012:
Without a doubt you are anamazing teacher and the way in which you do this is wonderful.
A brilliant hub awesome/useful/interesting .
Eddy.
Janine Huldie (author) from New York, New York on August 28, 2012:
Dianna, thank you so much for the compliment. I have to tell you I really loved the challenge of coming up with a great hands on math lesson and even now have been going through my old lessons trying to retweek hoping that I will et to use these again in the future. Thank you also for the votes and share too!!
Dianna Mendez on August 28, 2012:
Once again you have posted an amazing hub on math. I enjoy the read and the application. Voted way up and shared too!
Janine Huldie (author) from New York, New York on August 28, 2012:
Any food usually would pull in young kids, lol!! Seriously, kids do love food so when I tried to think f motivation for certain topics this was truly was a winner. Thank you so much Glimmer Twin Fan for your lovely comments and very much do appreciate your continued support on Hubpages :)
Claudia Porter on August 28, 2012:
Another great math lesson. Loved using the pies as a teaching device. That must have pulled a lot of students in!
Janine Huldie (author) from New York, New York on August 28, 2012:
Oh Keith, thank you also for being so kind and saying that. Seems you are not alone in your dread of math. I only wish that I could find another full time job to teach math again. I loved it and miss the chance to make that difference. Seriously, thank you for your lovely words, it mean the world to me :)
KDuBarry03 on August 28, 2012:
Now this is a great way to teach math and circles! I always dreaded math not because of the subject matter, but the teacher made it so boring! I can see that you are a fun teacher to have and hope your students feel the same way :)
Janine Huldie (author) from New York, New York on August 28, 2012:
Myz, thank you for saying that. I have to tell you the positive feedback really makes me smile, because I know math is not of favorite of many, but did definitely try with my students to infuse the subject with interesting and innovative techniques. Thank you!!
DragonBallSuper on August 28, 2012:
whew!! very creative Janine!! even me, who hates math really love this hub
Janine Huldie (author) from New York, New York on August 28, 2012:
Richard, you are not alone with math not being your favorite subject. I think I have heard this from others time and time again, including my husband. As always, loved your bit of humor with a square pie being a cobbler, lol :) Seriously thank you for your comment and for all your support!!
Janine Huldie (author) from New York, New York on August 28, 2012:
Rema, thank you so much for saying math would have been your favorite subject if I had been your teacher. Thant truly is very much appreciated. Seriously thank you for all your kindness on here, it is so very much appreciated. God bless you too :)
Rich from Kentucky on August 27, 2012:
Math has never been a favorite of mine. You seem to make it much easier than when I struggled, but I have yet to see a square pie. In my day, they were called cobblers. Still, an enjoyable lesson, even though I think I gained five pounds in class today! Great Job!
Rema T V from Chennai, India on August 27, 2012:
Oh My! Janine, I am sure Math would have been my most favorite subject at school if I had a teacher like you or better still if You had been my teacher. Your hubs are great especially the Math lessons (I'm sure to repeat this sentence on your recipe hubs too lol- you seem to be master of everything) and I just loved the Cherry Pies Delicious and Apple Pies R2. Simply wonderful. Very sweet of you to offer a real pie to your students. They are truly blessed. God Bless You. Cheers, Rema.
Janine Huldie (author) from New York, New York on August 27, 2012:
Michelle, I totally love trying new and creative ways to teach math concepts and this one is so fun and tasty too. Seriously thank you for your kind comment, support and for sharing too!!
Michelle Liew from Singapore on August 27, 2012:
Janine, Area of Circle is one of my fave maths concepts. And you've a delicious, creative way of teaching it!! From a fellow teacher, I'm sharing!
Janine Huldie (author) from New York, New York on August 27, 2012:
Nell, very cute math joke indeed. Seriously, you sound like my husband. He too hated math, but would have loved to have become an architect (he had the art skills for it), but it was the math that kept him from pursuing that career. He always jokes that he hopes our girls get my love and knack for math. Seriously, thank you for stopping by and for your comment too. By the way, the pie would have gotten me too if I didn't love math already lol :)
Nell Rose from England on August 27, 2012:
Why couldn't you be my teacher? lol! I hated maths with a vengeance! I think it was because I missed a lot of school when I was young through illness and the teachers didn't give a jot. All the tt = rrr etc totally threw me! I always wanted to be a physicist! believe it or not, but just couldn't get the hang of maths, but pies? yep that will do me, better than Pi! haha! I made a math joke! lol!
Janine Huldie (author) from New York, New York on August 27, 2012:
Same here with math being on of my best subjects in school, Jackie. I very much about trying to make learning easier using short cuts and things to make kids remember math topics easier. I am a firm believer in using all this and more to bring these topics home. Thank you so much again for kind words here, so very appreciated!
Jackie Lynnley from the beautiful south on August 27, 2012:
Math was one of my best subjects and although I never had any problems I know many students did and I tutored in my sophomore year and taught the short cuts I did in my head before tackling a problem. I think anything that makes learning easier or more fun is great. Looks like you have a winner here.
Janine Huldie (author) from New York, New York on August 27, 2012:
Yes if you were in my class, you could definitely eat the pie Terrye lol :)
Terrye Toombs from Somewhere between Heaven and Hell without a road map. on August 27, 2012:
So, after the lesson, we get to eat the pies? :)
Janine Huldie (author) from New York, New York on August 27, 2012:
Coming for you, this compliment means the world to me. I would have so enjoyed teaching alongside you too and that would have been totally an awesome experience for me. Thank you seriously for your wonderful comment here as always so very appreciated!!
Bill Holland from Olympia, WA on August 27, 2012:
So let me get this straight....you tried to find new and creative ways to teach mundane subjects.....my goodness, Janine, you mean you cared enough about the students to do extra??? LOL You are priceless and I would have been proud to have taught alongside of you. Great lesson!
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Edit Article
# wikiHow to Construct Multiple Squares of Odd Order
In a Magic Square there are i) equal number of rows and columns. ii) We have to fill up consecutive numbers, generally beginning with 1 and going up to nxn i.e, n square, such that the total of each row, each column, and the two main diagonals is same. So for a 5x5 squares we will have numbers from 1 to 25 and the total required would be 65 for each row, column and diagonal. For the odd order square let us first look at what is known as THE HINDU RULE.
## Steps
1. 1
The right hand side 7x7 square illustrates the rule using alphabets. Write the first number Aa in the center of the topmost row. Next, write Ab in the lowest space of the vertical column next adjacent to the right. After, inscribe the remaining numbers in their natural order in the squares diagonally upwards towards the right that, upon reaching the right-hand margin, the inscription shall be continued from the left-hand margin in the row just above, and again, on reaching the upper margin, shall be continued from the lower margin in the column next adjacent to the right, noting that whenever we are arrested in our progress by a square already occupied we are to fill out the square next beneath the one we have filled. In this manner, for example, the 7×7 square given below has been formed:
• Eb Fd Gf Aa Bc Ce Dg
• Fc Ge Ag Bb Cd Df Ea
• Gd Af Ba Cc De Eg Fb
• Ae Bg Cb Dd Ef Fa Gc
• Bf Ca Dc Ee Fg Gb Ad
• Cg Db Ed Ff Ga Ac Be
• Da Ec Fe Gg Ab Bd Cf
• Please note that here Aa means A+a, Bb means B+b, and so on.
2. 2
Give specific values to A’s and a’s. For illustrating the method, let A=0, B=7, C=14, D=21, E=28, F=35, G=42, and a=1,b=2,c=3,d=4,e=5,f=6, and g=7, to get the square given below:
• 39 48 01 10 19 28
• 47 07 09 18 27 29
• 06 08 17 26 35 37
• 14 16 25 34 36 45
• 15 24 33 42 44 04
• 23 32 41 43 03 12
• 31 40 49 02 11 20
3. 3
To put it differently, you can start with the center square in the last column and get this 7x7 square:
• 12 04 45 37 29 28
• 03 44 36 35 27 19
• 43 42 34 26 18 10
• 41 33 25 17 09 01
• 32 24 16 08 07 48
• 23 15 14 06 47 39
• 21 13 05 46 38 30
• This method is neat and quick. De La Loubere, Envoy of Louis X1V to Siam learnt of this method here.
4. 4
To get multiple squares, start with the central cell of the bottom row, central cell of the first column or the last column as preferred. Giving A’s one of the values from 0,7,14,28,35,42, and a’s one of the values from 1,2,3,4,5,6,7 – with the provision that ‘D’ has to be given the value 21, you will be able to cover all numbers from 1 to 49 and get the magic sum of 175. It gives (7!x6!)/4 clear solutions.
5. 5
Alternatively, give A’s one of the values – 1,2,3,4,5,6,7 and to a’s – 0,7,14,28,35,42 – with ‘D’ being given the value 4. Of course, no value should be repeated; this automatically ensures that all the numbers are used. You will then be able to generate (7!x6!)/4 squares at one go by this method too. (Please note 7! Stands for 1x2x3x4x5x6x7 and 6! Is 1x2x3x4x5x6)
6. 6
To apply another method which will give an even larger number of squares -- as there is no restriction on the value to be given to the central number, take the 11x11 square by this method. The method is very general. The next line in this square starts with A’s from the 3rd column and a’s from the 4th column in the earlier line. One gets the basic square shown below:
• Aa Bb Cc Dd Ee Ff Gg Hh Ii Jj Kk
• Cd De Ef Fg Gh Hi Ij Jk Ka Ab Bc
• Eg Fh Gi Hj Ik Ja Kb Ac Bd Ce Df
• Gj Hk Ia Jb Kc Ad Be Cf Dg Eh Fi
• Ib Jc Kd Ae Bf Cg Dh Ei Fj Gk Ha
• Ke Af Bg Ch Di Ej Fk Ga Hb Ic Jd
• Bh Ci Dj Ek Fa Gb Hc Id Je Kf Ag
• Dk Ea Fb Gc Hd Ie Jf Kg Ah Bi Cj
• Fc Gd He If Jg Kh Ai Bj Ck Da Eb
• Hf Ig Jh Ki Aj Bk Ca Db Ec Fd Ge
• Ji Kj Ak Ba Cb Dc Ed Fe Gf Hg Ih
• Of course Aa is to be read as A+a, and so on. A’s are to be chosen from 1 to 11 and a’s from 0, 11, 22,33,44,55,66,77,88,99,110, (i.e. multiples of 11). You thus automatically get all numbers from 1 to 121 and the number of solutions is (11!x11!)/4, with a total of 671 for each row, column and the 2 diagonals.
7. 7
Select the values that the diagonals/rows/columns total come correctly, in cases when some of the alphabets will repeat. In case of 9x9 square it will be seen that A,D,G repeat in one diagonal and a,d,g; b,e,h and c,f,i repeat in the columns. Therefore ensure that A,D,G are so selected that they total 15, and a,d,g; b,e,h: c,f,i so that they each total 108, i.e 1/3rd of 324, which is the total of 0,9,18,27,36,45,54,63,72..
8. 8
Look at one of the 7x7 square by this method, with numbers from 1 to 49, with a total of 175. A=1. B=2, C=3.D=4, E=5, F=6, G=7, and a=0, b=7, c=14, d=21, e=28, f=35, g=42.
• 09 17 25 33 41 49
• 32 40 48 07 08 16
• 06 14 15 23 31 39
• 22 30 38 46 05 13
• 45 04 12 20 28 29
• 19 27 35 36 44 03
• 42 43 02 10 18 26
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# Power Rule
Now that we have some counting from Counting, Permutations and Combinations, the Binomial Theorem, and have seen what a limit and derivative are in Dividing by zero: Imploding the Universe., we are now ready to look at the power rule for derivatives. We will therefore begin by going through an example, followed by a proof that the rule we come up with will work in general.
Example
In a calculus class it is often helpful not to dive directly into a proof of something. Instead, I tend to ease into these proofs by working on an example or two so that we can outline the process. To this end, I would go through finding the derivatives of a constant k, and the function x and x2 with the students. The nice things about these examples is that we can check to see if we are correct, because the functions themselves are the tangent lines, and finding the slope is something that has been done before.
I won’t run through these here, however, I do want to show how I would find the derivative of f(x)=x3 with my students. To do this, I would remind them that
I’m using the same definition we found in Dividing by zero: Imploding the Universe., however, I have substituted an h for Δx since it is a little easier to write. In order to calculate the derivative of f(x)=x3, I break the process up into four smaller steps. I first learned this process from Dr. Karl Byleen when teaching a business calculus course with his book Calculus for Business. The process really just explicitly states what you would naturally do when trying to complete a complicated calculation. However, by walking your students through this and making them follow the steps, there are significantly fewer mistakes made on homework and tests. The process is,
1. First find f(x+h) and simplify. Note that instead of multiplying out (x+h)3, I would instead remind them of the binomial theorem because we will be using this for general case.
2. We would then find and simplify f(x+h)-f(x).
3. Then, find and simplify the entire inside.
4. Finally, find the limit noting that in the limit h is not 0, so we can substitute what we found.
We have now shown that that derivative of x3 is 3x2. Even at this point, when combined with the fact that the derivative of x2 is 2x, the students start to see the pattern. If no one can see it, I will show the derivative x4 is 4x3. Otherwise, I tell my students that I’m happy they’ve seen the pattern.
However, I also take this time to point out that we’ve applied argument due to inductive reasoning. That is, we saw some examples and tried to expand this. Is this guaranteed to work? If they say, yes, I will provide examples of misleading patterns, for example you’re walking and you go up steps. You’ve moved up 1, 2, 3, 4 steps, what will happen next? Whatever they choose, pick something different. That is, if they say you will have moved up 5 steps, state the stair case stopped and you stay at height 4. If they say 4, say it kept going up. If they say both, say the staircase ended and you fell back to level 0.
The goal here is to show that while we’ve made an educated guess, the pattern may change for some unseen reason. Therefore, we need to provide a deductive argument to show that the pattern will always hold. I would then continue to give the proof of the power rule.
General Case
We are now ready to show the power rule for f(x)=xn for all natural numbers n>1. I would now point out that our only option is to use the limit definition of derivative.
1. We would first use the binomial theorem to findIf you hadn’t previously explained the binomial theorem, make sure to let your students know what you mean by this notation. In particular, explain why the first two coefficients are 1 and n, and explain that the rest are just some number.
2. We then subtract f(x) and getNote that since the sum ends at i=n-2, there is still at least one h in each of the terms of the sum.
3. If we now divide by h, we get
4. Finally, taking the limit, we find Seeing that each term in the sum goes to 0 depends on seeing that there are still hs in each term, and you therefore get a number multiplied by 0.
We have now proven that the power rule that they saw earlier,
is true for all n>1. That is, not only do we think it will work because we saw a pattern, but we know that it will always work. I really try to emphasize this to my students so that they will understand the distinction between the guessing and knowing. I want them to realize that the equations and properties they’ve been using in mathematics aren’t true just because someone claimed them to be true, but rather someone took the time to prove, based on supplied assumptions, that the formulas used are correct.
I also point out, that our proof doesn’t work for n=0,1, but we can work with these situations with a few adjustments. In particular, the above formula would give that the derivative of x is, 1x1-1=x0. However, the derivative of x is 1. While these agree for all x except 0, we do note that 00 is undefined. We have a similar situation if we write 1=x0. Then the derivative is 0x-1. Again, there is a distinction between these at x=0, so I state they can use the power rule for these cases with the caveat that they need to be careful at this one point.
Conclusion
The importance of understanding why something is true and that we don’t just make claims without justification in mathematics is something that I really try to get across to my Calculus students. I feel like this is normally the first time students are introduced to underlying principles of mathematics. I hope that I can get a few of them to understand what mathematics will become if they continue with their courses. Realizing that mathematics isn’t just a box of tools that you pull out, either excites or terrifies students. Either way, it is good for them to know at this point instead of later in their studies.
Next time in this series, we will continue to work with the power rule. While it may seem we have finished, we have only shown the power rule works for all natural numbers. However, we will also need to look at the case where the powers are negative or rational.
If you are enjoying the posts, let me know by liking the posts here, on Facebook or Twitter. I would also appreciate it if you shared the page with anyone you know that may find the topics enjoyable or useful. Thank you.
Since it is October, if you are looking for a Calculus related horror book, I’d suggest reading the above Zombies and Calculus. It’s a fun story that explains how you can always manage to stay just ahead of the zombies that are trying to get at you.
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# What is the unit vector that is normal to the plane containing ( i +k) and ( i + 7 j + 4 k) ?
Aug 3, 2016
$\hat{v} = \frac{1}{\sqrt{107}} \cdot \left(\begin{matrix}7 \\ 3 \\ - 7\end{matrix}\right)$
#### Explanation:
first, you need to find the vector (cross) product vector, $\vec{v}$, of those 2 co-planar vectors, as $\vec{v}$ will be at right angles to both of these by definition:
$\vec{a} \times \vec{b} = \left\mid \vec{a} \right\mid \left\mid \vec{b} \right\mid \setminus \sin \theta \setminus {\hat{n}}_{\textcolor{red}{a b}}$
computationally, that vector is the determinant of this matrix, ie
$\vec{v} = \det \left(\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & 7 & 4\end{matrix}\right)$
$= \hat{i} \left(- 7\right) - \hat{j} \left(3\right) + \hat{k} \left(7\right)$
$= \left(\begin{matrix}- 7 \\ - 3 \\ 7\end{matrix}\right)$ or as we are only interested in direction
$\vec{v} = \left(\begin{matrix}7 \\ 3 \\ - 7\end{matrix}\right)$
for the unit vector we have
$\hat{v} = \frac{\vec{v}}{\left\mid \vec{v} \right\mid} = \frac{1}{\sqrt{{7}^{2} + {3}^{3} + {\left(- 7\right)}^{2}}} \cdot \left(\begin{matrix}7 \\ 3 \\ - 7\end{matrix}\right)$
$= \frac{1}{\sqrt{107}} \cdot \left(\begin{matrix}7 \\ 3 \\ - 7\end{matrix}\right)$
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DIVISION OF MONOMIALS BY MONOMIALS WORKSHEET
About "Division of monomials by monomials worksheet"
Division of monomials by monomials worksheet :
Here we are going to see some practice questions on division of monomial by another monomials.
Division of monomials by monomials worksheet - practice questions
(1) Simplify 5a2b2c2 ÷ 15abc
(2) Simplify 16x÷ 32x
(3) Simplify -42y÷ 7y2
(4) Simplify 30 a3b3c3 ÷ 45 abc
(5) Simplify 72 l4m5n8 ÷ (-8l2m2n3)
(6) Simplify 25x3y2 ÷ 15x2y
(7) Simplify (7m- 6m) ÷ m
Division of monomials by monomials worksheet - Solution
Question 1 :
Simplify 5a2b2c2 ÷ 15abc
Solution :
5a2b2c2 ÷ 15abc = 5a2b2c2 /15abc
By dividing 5 and 15, we get 3 in the denominator.
By dividing a2 by a, we get a in the numerator.
By dividing b2 by b, we get b in the numerator.
By dividing c2 by c, we get c in the numerator.
Hence the answer is abc/3
Question 2 :
Simplify 16x÷ 32x
Solution :
16x÷ 32x = 16x4 /32x
By dividing 16 and 32, we get 2 in the denominator.
By dividing x4 by x, we get x3 in the numerator.
Hence the answer is x3/2
Question 3 :
Simplify -42y3 ÷ 7y2
Solution :
-42y÷ 7y2 = -42y/ 7y2
By dividing 42 by 7, we get 6 in the numerator.
By dividing y3 by y2, we get y in the numerator.
Hence the answer is -6y
Question 4 :
Simplify 30 a3b3c3 ÷ 45 abc
Solution :
30 a3b3c3 ÷ 45 abc = 30 a3b3c3 /45 abc
by simplifying 30 and 45, we get 2 in the numerator and 3 in the denominator.
By dividing a3 by a, we get a2 in the numerator.
By dividing b3 by b, we get b2 in the numerator.
By dividing c3 by c, we get c2 in the numerator.
Hence the answer is (2/3) a b2 c2
Question 5 :
Simplify 72 l4m5n8 ÷ (-8l2m2n3)
Solution :
72 l4m5n8 ÷ (-8l2m2n3) = 72 l4m5n8 /(-8l2m2n3)
by simplifying 72 and -8, we get 9 in the numerator.
By dividing l4 by l2, we get l2 in the numerator.
By dividing m5 by m2, we get m3 in the numerator.
By dividing nby n3, we get n5 in the numerator.
Hence the answer is 9 l m2 n2
Question 6 :
Simplify 25x3y2 ÷ 15x2y
Solution :
25x3y2 ÷ 15x2y = 25x3y2 /15x2y
by simplifying 25 and 15, we get 5 in the numerator and 3 in the denominator.
By dividing x3 by x2, we get x in the numerator.
By dividing yby y, we get y in the numerator.
Hence the answer is (5/3)xy
Question 7 :
Simplify (7m2 - 6m) ÷ m
Solution :
(7m- 6m) ÷ m = (7m- 6m)/m
= (7m2/m) - (6m/m)
= 7m - 6
Hence the answer is 7m - 6.
After having gone through the stuff given above, we hope that the students would have understood "Division of a monomial by a monomial".
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# 6.5: Subtracting Mixed Numbers by Renaming
Difficulty Level: At Grade Created by: CK-12
## Introduction
Measuring Differences
After measuring the window, Travis worked the rest of the day helping his Uncle Larry and Mr. Wilson cut the hole for the window.
When they finished, they examined their work and felt good about what they had accomplished.
The total height of the wall is 8 ft. Travis is curious about the distance from the top of the window space to the edge where the wall meets the ceiling. He knows that there is going to be crown molding that goes around the edge of the room when finished, and he hopes that he can help select the style of crown molding.
Travis knows that the window is on a wall that is 8 ft high, and he knows that the distance from the floor to the top of the window space is .
Given this information, what is the distance from the top of the window space to the edge where the wall meets the ceiling?
Travis is stuck on how to work through this problem. He knows that he will need to convert the 8 foot wall to inches and then subtract, but he can’t remember exactly how to do the subtraction.
This is where you come in. To accomplish this task, Travis will need to know how to subtract mixed numbers with renaming. This lesson will help you learn all that you need to know!!
What You Will Learn
In this lesson, you will learn how to complete the following:
• Subtract mixed numbers with renaming.
• Evaluate numerical expressions involving differences of mixed numbers requiring renaming.
• Solve real-world problems involving differences of mixed numbers requiring renaming.
Teaching Time
I. Subtract Mixed Numbers with Renaming
We have already learned about renaming fractions when we have two different denominators. To rename a fraction means that we are going to take the fractions and write an equivalent fraction that has a common denominator.
This is an example of renaming one-third to be three-ninths. The fractions are equivalent or equal, but the second fraction has a denominator of 9.
Sometimes when we subtract mixed numbers, we must rename the mixed numbers in a different way.
What does this mean? It means that when we are subtracting a mixed number from a whole number, we must rename to subtract correctly.
Let’s look at an example.
Example
In this example, we are trying to take a fraction from a whole number. We have to RENAME this whole number to do this.
How can we rename 6 to have a whole number part and a fraction part?
Think back. Remember when we learned that any fraction where the numerator and the denominator are the same that the fraction is equal to one?
or or
Do you remember this? Well, if we can take one and rename it as a fraction, then we can use that to help us in our subtraction.
In the example above, we need to take 6 and make it a mixed number so that we can subtract. To do this, we borrow a one from the six and make it five. Then we take that one and make it into a fraction that has sixths as a denominator since that is the denominator of the fraction we are subtracting.
Next, we rewrite the problem.
Example
Sometimes, we also have to rename a mixed number if the fraction we are subtracting is larger than the first fraction.
Example
At first glance, this problem looks simple. We have two mixed numbers. This is unlike the first example where we had a whole number and a mixed number. But watch out!! This one is tricky. Four-ninths is larger than one-ninth. We cannot subtract four-ninths from one-ninth.
To make this work, we have to rename the top mixed number!
How do we do this?
We can do this by changing the whole number six into five and nine-ninths-then we add that to the one-ninth to make larger fraction.
Now we can rewrite the problem and subtract.
Notice that we simplified, so this is our final answer.
Try a few of these on your own.
1. Rename 8 as an equivalent mixed number.
Take a few minutes to check your work with a peer.
II. Evaluate Numerical Expressions Involving Differences of Mixed Numbers Requiring Renaming
Sometimes, you will find numerical expressions that have multiple operations in them, but will that will still require renaming.
Example
To work on this problem, we need to think of it as two separate problems. Working in order from left to right, we complete the subtraction problem first and then add the final mixed number to the difference.
To work on this problem, we first need to rename 5. We rename it to a mixed number equivalent of 5 with a fraction in eighths.
Now we can subtract easily.
Next, we add this mixed number with the last mixed number in the original expression.
Notice that we ended up with an extra whole at the end.
This is our answer in simplest form.
Here are a few problems for practice.
That second one is tricky! Check your work step by step with your neighbor.
## Real Life Example Completed
Measuring Differences
Now that you have learned all about renaming mixed numbers, you are ready to work with Travis. Here is the problem once again.
After measuring the window, Travis worked the rest of the day helping his Uncle Larry and Mr. Wilson cut the hole for the window.
When they finished, they examined their work and felt good about what they had accomplished.
The total height of the wall is 8 ft. Travis is curious about the distance from the top of the window space to the edge where the wall meets the ceiling.
Travis knows that the window is on a wall that is 8 ft high, and he knows that the distance from the floor to the top of the window space is 64 .
Given this information, what is the distance from the top of the window space to the edge where the wall meets the ceiling?
Travis is stuck on how to work through this problem. He knows that he will need to convert the 8 foot wall to inches and then subtract, but he can’t remember exactly how to do the subtraction.
First, go back and underline any important information.
Let’s convert the 8 ft wall measurement into inches since our window measurement is in inches.
There are 12 inches in 1 foot, so 12 8 = 96 inches. The wall is 96 inches high.
Next, we subtract the total from the floor to the top of the window space from the height of the wall.
To do this, we are going to need to rename 96 in terms of eighths.
Now we can subtract.
From the top of the window space to the edge where the wall meets the ceiling is .
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Rename
to write an equivalent form of a whole number or a fraction.
Equivalent
equal
## Time to Practice
Directions: Rename each whole number as a mixed number with a fraction terms of sixths.
1. 4
2. 5
3. 6
4. 10
5. 9
6. 12
Directions: Find each difference. Rename mixed numbers as needed and be sure that your answer is in simplest form.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
### Notes/Highlights Having trouble? Report an issue.
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# SAT Math : How to find consecutive integers
## Example Questions
### Example Question #1 : How To Find Consecutive Integers
The sum of three consecutive even integers is 108. What is the largest number?
36
42
40
38
34
38
Explanation:
Three consecutive even integers can be represented by x, x+2, x+4. The sum is 3x+6, which is equal to 108. Thus, 3x+6=108. Solving for x yields x=34. However, the question asks for the largest number, which is x+4 or 38. Please make sure to answer what the question asks for!
You could have also plugged in the answer choices. If you plugged in 38 as the largest number, then the previous even integer would be 36 and the next previous even integer 34. The sum of 34, 36, and 38 yields 108.
### Example Question #1 : How To Find Consecutive Integers
The sum of three consecutive even integers equals 72. What is the product of these integers?
13800
17472
13728
12144
10560
13728
Explanation:
Let us call x the smallest integer. Because the next two numbers are consecutive even integers, we can call represent them as x + 2 and x + 4. We are told the sum of x, x+2, and x+4 is equal to 72.
x + (x + 2) + (x + 4) = 72
3x + 6 = 72
3x = 66
x = 22.
This means that the integers are 22, 24, and 26. The question asks us for the product of these numbers, which is 22(24)(26) = 13728.
The answer is 13728.
### Example Question #1 : How To Find Consecutive Integers
Four consecutive integers have a mean of 9.5. What is the largest of these integers?
8
13
12
9
11
11
Explanation:
Four consecutive integers could be represented as n, n+1, n+2, n+3
Therefore, by saying that they have a mean of 9.5, we mean to say:
(n + n+1 + n+2 + n+ 3)/4 = 9.5
(4n + 6)/4 = 9.5 → 4n + 6 = 38 → 4n = 32 → n = 8
Therefore, the largest value is n + 3, or 11.
### Example Question #4 : How To Find Consecutive Integers
The sum of four consecutive odd integers is equal to 96. How many of the integers are prime?
3
1
4
0
2
1
Explanation:
Let x be the smallest of the four integers. We are told that the integers are consecutive odd integers. Because odd integers are separated by two, each consecutive odd integer is two larger than the one before it. Thus, we can let + 2 represent the second integer, + 4 represent the third, and + 6 represent the fourth. The sum of the four integers equals 96, so we can write the following equation:
x + (+ 2) + (+ 4) + (+ 6) = 96
Combine x terms.
4x + 2 + 4 + 6 = 96
Combine constants on the left side.
4x + 12 = 96
Subtract 12 from both sides.
4x = 84
Divide both sides by 4.
x = 21
This means the smallest integer is 21. The other integers are therefore 23, 25, and 27.
The question asks us how many of the four integers are prime. A prime number is divisible only by itself and one. Among the four integers, only 23 is prime. The number 21 is divisible by 3 and 7; the number 25 is divisible by 5; and 27 is divisible by 3 and 9. Thus, 23 is the only number from the integers that is prime. There is only one prime integer.
The answer is 1.
### Example Question #1 : How To Find Consecutive Integers
The sum of three consecutive integers is 60. Find the smallest of these three integers.
Explanation:
Assume the three consecutive integers equal , , and . The sum of these three integers is 60. Thus,
Explanation:
### Example Question #7 : How To Find Consecutive Integers
In the repeating pattern 9,5,6,2,1,9,5,6,2,1......What is the 457th number in the sequence?
1
2
5
9
1
5
Explanation:
There are 5 numbers in the sequnce.
How many numbers are left over if you divide 5 into 457?
There would be 2 numbers!
The second number in the sequence is 9,5,6,2,1
### Example Question #8 : How To Find Consecutive Integers
If are consecutive, non-negative integers, how many different values of are there such that is a prime number?
Explanation:
Since are consecutive integers, we know that at least 2 of them will be even. Since we have 2 that are going to be even, we know that when we divide the product by 2 we will still have an even number. Since 2 is the only prime that is even, we must have:
What we notice, however, is that for , we have the product is 0. For , we have the product is 24. We will then never have a product of 4, meaning that is never going to be a prime number.
### Example Question #1 : How To Find Consecutive Integers
Four consecutive odd integers have a sum of 32. What are the integers?
Explanation:
Consecutive odd integers can be represented as x, x+2, x+4, and x+6.
We know that the sum of these integers is 32. We can add the terms together and set it equal to 32:
x + (x+2) + (x+4) + (x+6) = 32
4x + 12 = 32
4x = 20
x = 5; x+2=7; x+4 = 9; x+6 = 11
Our integers are 5, 7, 9, and 11.
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# How do you graph 2x - 3y = 9 using x- and y- intercepts?
Mar 19, 2018
See a solution process below:
#### Explanation:
First, find the y-intercept: Set $x = 0$ and solve for $y$:
$\left(2 \cdot 0\right) - 3 y = 9$
$0 - 3 y = 9$
$- 3 y = 9$
$\frac{- 3 y}{\textcolor{red}{- 3}} = \frac{9}{\textcolor{red}{- 3}}$
$y = - 3$ or $\left(0 , - 3\right)$
Next, find the x-intercept: Set $y = 0$ and solve for $x$:
$2 x - \left(3 \cdot 0\right) = 9$
$2 x - 0 = 9$
$2 x = 9$
$\frac{2 x}{\textcolor{red}{2}} = \frac{9}{\textcolor{red}{2}}$
$x = \frac{9}{2}$ or $\left(\frac{9}{2} , 0\right)$
We can next plot the two points on the coordinate plane:
graph{(x^2+(y+3)^2-0.075)((x-(9/2))^2+y^2-0.075)=0 [-20, 20, -10, 10]}
Now, we can draw a straight line through the two points to graph the line:
graph{(2x - 3y - 9)(x^2+(y+3)^2-0.075)((x-(9/2))^2+y^2-0.075)=0 [-20, 20, -10, 10]}
|
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# RS Aggarwal Class 12 Chapter 12 Solutions (Indefinite Integral)
RS Aggarwal Solutions for Class 12 Chapter 12 are created with the viewpoint of bringing you at ease with the complex topics of the chapter. Indefinite Integral is a fundamental chapter that will help you understand the chapters ahead. In this chapter, you will be introduced to the concepts of anti-derivatives and area function. You will also get to understand the most important theorems of integral which are the first fundamental theorem of Calculus and the second fundamental theorem of Calculus that have a broad range of applications in maths and other subjects.
There is only one exercise that has 32 questions and an additional exercise of objective 41 questions covering a wide variety of functions under each topic. If you are able to grasp the methodology of how to solve these questions, you will be able to approach the questions in exams correctly and with confidence. These questions accompanied by your textbook questions are enough to prepare you for your school exam of CBSE, UP board etc. Solving these questions using Instasolv as your source of reference will make you adhere to the guidelines of the board.
The Maths experts of Instasolv have prepared these answers after thorough research and massive experience with students. Thus, they are well aware of your vulnerable point where you might end up losing marks. You will find the RS Aggarwal solutions accurate and detailed.
## Summary of Important Topics Covered in RS Aggarwal Solutions for Class 12 Chapter 12 – Indefinite Integrals
Introduction:
Integration is a method of adding things up. It joins slices and makes them whole.
Integration is the inverse of Differentiation. To find the derivative of f(x), we use:
d/dx(f(x))=g(x)
So to calculate the antiderivative, we use –
∫ g(x) dx = f(x) + c
Here, c is the constant of integration.
Area Function:
By definition, ∫ab f(x) dx is the area of the region bounded by the curve y = f(x), the x-axis and the coordinates ‘x = a’ and ‘x = b’. If ‘x’ is a point in [a, b], then ∫ax f(x) dx shows the area under the curve y.
First Fundamental Theorem of Calculus:
If ‘f’ is a continuous function on the closed interval [a, b] and A (x) is the area function. Then A′(x) = f (x), for all x ∈ [a, b].
Second Fundamental Theorem of Calculus:
If ‘f’ is a continuous function defined on the closed interval [a, b] and F is an anti-derivative of ‘f’. Then
ab f(x) dx = [F(x)]ab = F(b) – F(a)
The area under the curve:
For now, we wanted to calculate the area of the curve y=f(x), which will be
∫ f(x) dx = h(x) + c
Formulae for Indefinite Integrals:
Some of the basic indefinite integrals formulae are:
• xn dx = xn+1 + c
________ (n + 1)
• ∫ sin x dx = – cos x + c
• ∫ cos x dx = sin x + c
• 1 dx = ln x + c
x
• ∫ ex dx = ex + c
Properties of Indefinite Integrals:
• Theorem 1
The method of differentiation and integration are the opposite of each other.
Proof: If F be the anti-derivative of f, i.e., d⁄dx F(x) = f(x)
Then ∫ f(x) dx = F(x) + C
Hence, d⁄dx ∫ f(x) dx = d⁄dx [F(x) + C] = d ⁄ dx F(x) = f(x).
Similarly, f ′(x) = d ⁄ dx f(x) and hence ∫ f ′(x) dx = f(x) + C.
Where C is the constant of integration.
• Theorem 2
The integration of the sum of two integrands is the sum of integrations of two integrands.
∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
Proof: Using Theorem 1, we have
d⁄dx [∫ [f(x) + g(x)] dx] = f(x) + g(x) … (1)
d⁄dx [∫ f(x) dx + ∫ g(x) dx] = d⁄dx ∫ f(x) dx + d⁄dx ∫ g(x) dx = f(x) + g(x) … (2)
From (1) and (2), we have, ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
• Theorem 3
Let k be any real number. We can say that, ∫ k f(x) dx = k ∫ f(x) dx.
Proof: Using theorem 1, d⁄dx ∫ k f(x) dx = k f(x) … (1)
Also, d⁄dx [∫ k f(x) dx] = k d⁄dx ∫ f(x) dx = k f(x) … (2)
From (1) and (2), we have, ∫ k f(x) dx = k ∫ f(x) dx.
The above result can be generalized to:
ʃ k1 [f1(x) + k2 f2(x) +…+ kn fn(x)] dx
= k1 ∫ f1(x) dx + k2 ∫ f2 (x) dx + … + kn ∫ fn(x) dx.
### Exercise-Wise Discussion of Questions in RS Aggarwal Solutions for Class 12 Chapter 12 – Indefinite Integrals
1. In the only exercise of RS Aggarwal Chapter 12, you will find a diverse set of questions covering all the topics in the chapter which makes these solutions crucial from the exam perspective.
2. The questions are also of objective or short-answer type besides the classic long answers.
3. These exercise solutions will give a very apt idea about the kind of questions that are asked in CBSE exams.
4. The questions are arranged in an orderly manner so as to pace with your level of understanding as you advance towards more complex topics.
5. It is advised that you do the final polishing of your concepts of Calculus before your exam using Instasolv for an interactive study session.
## Benefits of RS Aggarwal Solutions for Class 12 Maths Chapter 12 by Instasolv
1. The RS Aggarwal class 12 maths Solutions chapter 12 are prepared in easy to understand language by the experts at Instasolv.
2. We, at Instasolv, believe that efficient utilisation of study hours is mandatory to stand out among the crowd, which you will witness once you will refer to our solutions. You can complete the whole chapter quickly without any distraction.
3. The Instasolv experts have covered all the questions in detail without compromising the quality of the exams.
More Chapters from Class 12
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# 3rd Class Mathematics Numbers
Numbers
Category : 3rd Class
Numbers
• The number system contains ten digits 0,1,2,3,4,5,6,7,8,9.
Largest and smallest numbers:
• The smallest one digit number is 1.
• The largest one digit number is 9.
• The smallest two digit number is 10.
• The largest two digit number is 99.
• The smallest three digit number is 100.
• The largest three digit number is 999.
• 10 Ones = 1 Ten
• 10 Tens = 1 Hundred
• 10 Hundreds = 1 Thousand
Place Value of each digit in a number:
In a four digit number, there are four places named:
Ones place. Tens place. Hundreds place and Thousands place.
Ex:
Ascending Order of numbers:
Arranging numbers from the least number to the largest number is called Ascending Order.
Ex: 13, 67, 132, 168 are in ascending order
Descending Order of numbers:
Arranging numbers from the largest number to the least number is called Descending Order.
Ex: 168, 132, 67, 13 are in descending order.
Symbols used for comparison of two numbers:
This is the "equal to” sign. When we use it between two numbers, it means that the numbers on both sides of the sign have the same value.
Ex: 53 = 53
This is the "greater than" sign. It is placed between two numbers in such a way that the
open side is towards the bigger number.
Ex: 43 > 34
This is the “less than" sign. It is placed between two numbers in such a way that the closed side is towards the smaller number.
Ex: 57 < 69
Rules for comparison of numbers:
Rule 1:
A numeral containing more digits is greater.
Ex: 163 > 35
Rule 2:
If two numerals contain the same number of digits, the numeral having greater digit at the left most place will be greater.
Rule 3:
If the left most digits are also the same, we go to next digit from left and compare.
Successor of a numeral:
Successor of a particular numeral comes just after that numeral. So, we can find out the successor of a numeral by adding 1 to the given numeral.
Ex : The successor of 99 is 99 + 1 or 100.
Predecessor of a numeral:
Predecessor of a particular numeral comes just before that numeral. So, we can find out the
predecessor of a numeral by subtracting 1 from the number.
Ex : The predecessor of 100 is 100 - 1 or 99.
Even numbers:
The numbers that have 2, 4, 6, 8 and 0 in the ones place are called even numbers.
Ex: 4,32, 168, 490
Odd numbers:
The numbers that have 1, 3, 5, 7 and 9 in the ones place are called odd numbers.
#### Other Topics
##### Notes - Numbers
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## Intermediate Algebra (12th Edition)
$\left( \dfrac{3}{2},\infty \right]$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $\dfrac{6x+3}{-4} \lt -3 ,$ use the properties of inequality. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Multiplying both sides by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} -4\left(\dfrac{6x+3}{-4}\right) \gt -4(-3) \\\\ 6x+3 \gt 12 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 6x \gt 12-3 \\\\ 6x \gt 9 \\\\ x \gt \dfrac{9}{6} \\\\ x \gt \dfrac{3}{2} .\end{array} In interval notation, the solution set is $\left( \dfrac{3}{2},\infty \right] .$
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Created By : Awnish kumar
Reviewed By : Rajashekhar Valipishetty
Last Updated : Apr 28, 2023
Reciprocal of a Fraction Calculator: Reciprocal of a fraction calculator is a free online tool that helps you to find reciprocals of the given values easily. Follow the guidelines to calculate the reciprocal of the fraction calculator using a calculator.
### What is meant by Reciprocal?
In maths, the term reciprocal is nothing but multiplicative inverse. Taking the inverse of the given number is known as the reciprocal.
### Reciprocal of a Fraction - Definition
In order to find the reciprocal of fractions, we have to interchange the numerator and denominator. Simply we can say that x/1 becomes x and a/b becomes b/a.
### How to use the Reciprocal of a Fraction calculator?
The process to apply the reciprocal of fractions using a calculator is shown below.
• Enter the values of numerator and denominator in the required input fields.
• After that click on the solve button.
• Now the multiplicative inverse or reciprocal of the given fraction will be displayed on your screen.
Examples on finding Reciprocal of a Fraction
Question 1: Find the reciprocal of the fraction 3/5?
Solution:
Given Fraction is 3/5
To find the reciprocal of a fraction simply interchange the denominator and numerator.
Thus, the reciprocal of the fraction is 5/3.
Question 2: Find the reciprocal of the fraction 1/2?
Solution:
Given fraction is 1/2
To find the reciprocal of a fraction simply interchange the denominator and numerator.
Thus, the reciprocal of the fraction is 2/1
Question 3: Calculate the reciprocal of fraction 5/105?
Solution:
The given fraction is 5/105
A reciprocal is one of a pair of numbers that when multiplied with another number equals the number 1.
So, the reciprocal or multiplicative inverse of 5/105 is 105/5.
### FAQs on Reciprocal of a Fraction
1. How do you find the reciprocal of a fraction on a calculator?
You can find the Reciprocal of a fraction on a calculator by putting the given values in the required field and then clicking on the solve button. The output will be displayed on your screen.
2. What is the reciprocal of 5 as a fraction?
The reciprocal of a fraction 5/1 is 1/5
3. What are reciprocal fractions?
The reciprocal fractions are the fractions that obtain by switching the top number and the bottom number.
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# The line joining two points A (2, 0) and B (3, 1) is rotated about A in anti-clockwise direction through an angle of ${{15}^{\circ }}$ . The equation of the line in the new position, is:A. $\sqrt{3}x-y-2\sqrt{3}=0$B. $x-3\sqrt{y}-2=0$C. $\sqrt{3}x+y-2\sqrt{3}=0$D. $x+\sqrt{3}y-2=0$
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Hint: The formula to calculate the slope of a line between two points is given as follows
$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Another formula to calculate the slope of a line is given as follows
$m=\tan \theta$
(Where $\theta$ is the angle measured in the anti-clockwise direction made by the line with the x-axis of which the slope is being calculated)
As the line is rotated anticlockwise keeping the point A as fixed, so the slope of the new line is increased by ${{15}^{\circ }}$.
Hence, in this question, we will first find the slope of the line that joins the two points and then write the slope in terms of tan function. Then we can add the needed rotation and then get the new slope as the rotation is done keeping the point A as fixed and also in the anti-clockwise direction.
As mentioned in the question, we have to find the equation of the new line that is formed when the original line is rotated anticlockwise keeping the point A fixed.
Now, we know that the slope of a line ‘m’ passing through two given points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by the formula:
$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Thus, we can find the slope of our line using this formula.
Putting the value of points A and B in the formula for slope, we get our slope as:
$m=\dfrac{1-0}{3-2}=1$
Thus, the slope of the line is 1.
We also know that the slope of a line is given by $\tan \theta$ where $\theta$ is the angle made by the line with the x-axis measured in the anticlockwise direction.
Hence, the angle made by this line with the x-axis is given as follows
\begin{align} & \tan \theta =1 \\ & \Rightarrow \theta ={{\tan }^{-1}}\left( 1 \right) \\ & \Rightarrow \theta ={{45}^{\circ }} \\ \end{align}
Now, the line is rotated by ${{15}^{\circ }}$, hence, the angle made by the rotated line will be ${{15}^{\circ }}$ more than the angle made by the line when it was not rotated. Let this angle be $\alpha$.
Thus, the angle $\alpha$ made by the line with the x-axis is given as:
\begin{align} & \alpha ={{45}^{\circ }}+{{15}^{\circ }} \\ & \Rightarrow \alpha ={{60}^{\circ }} \\ \end{align}
Thus, the slope of the new line obtained is given by:
\begin{align} & m=\tan \alpha \\ & m=\tan {{60}^{\circ }} \\ & m=\sqrt{3} \\ \end{align}
Now, we know that when a line has a fixed slope ‘m’ and it passes through a fixed point (h,k), then its equation is given as:
$y-k=m\left( x-h \right)$
Now, we know that our required line has a slope of $\sqrt{3}$ and it passes through the fixed point A given by (2,0).
Hence, the equation of the new line can be written as follows:
\begin{align} & y-0=\sqrt{3}\left( x-2 \right) \\ & \Rightarrow y=\sqrt{3}x-2\sqrt{3} \\ & \Rightarrow \sqrt{3}x-y-2\sqrt{3}=0 \\ \end{align}
Therefore, the equation of the new line is given as:
$\sqrt{3}x-y-2\sqrt{3}=0$
So, the correct answer is “Option A”.
Note: Remember while measuring the angle made by the line with the x-axis, it is always measured in the anticlockwise direction. If we measure it in the clockwise direction, we should add a minus (-) sign in front of it otherwise it will result in an error in the solution. We also could have used a different general equation to find our required equation given by:
$y=mx+c$
Where, ‘m’ is the slope of the line and ‘c’ is the y-intercept. Here, $m=\sqrt{3}$
Thus, putting this value in the general equation we get:
$y=\sqrt{3}x+c$
But we know that this line also passes through the point A(2,0). Thus, by putting this point in the equation, we will get the value of ‘c’.
It is given as:
\begin{align} & 0=\sqrt{3}.2+c \\ & \Rightarrow c=-2\sqrt{3} \\ \end{align}
Thus, our equation will become:
\begin{align} & y=\sqrt{3}x-2\sqrt{3} \\ & \Rightarrow \sqrt{3}x-y-2\sqrt{3}=0 \\ \end{align}
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### Prompt Cards
These two group activities use mathematical reasoning - one is numerical, one geometric.
### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
# Money Bags
##### Stage: 2 Challenge Level:
We had some very good submissions showing us the ways that different pupils approached the activity and the way that they presented their results.
the following pdf from Angus at Canberra Grammar School in Australia. Here you see just a snippet so why not download the rest?
See the rest of this exceptionally good presentation here.pdf
Laolu from the same school went to great lengths, as did Angus, to produce another presentation:
The full pdf can be found here.pdf
Bethany from Christ Church sent in the following explanation:
Step 1. Bag 1 has to have 1 penny in it to make 1 pence.
Step 2. To make 2 pence, bag two has to either have 1 penny or 2 pennies in it. However, if it has 1 penny in it, you can't make 3 pence. This would mean we would need yet another 1 penny to make 3 pence, so Bag 2 has to have two pennies in it.
Step 3. You can now make 1, 2 and 3 pence, but not four. This means that bag 3 has to have either 1, 2 or 4 pence inside it. Again, using a 1 pence or a two pence would mean you wouldn't be able to make higher values, so bag 3 has to have 4 pennies in it.
Step 4. You can now make all values up to 7 pence, so the next bag needs to hold a 1 pence, 2 pennies, 3 pennies, 4 pennies, 6 pennies or 7 pennies. however, (yet again!) this choice becomes easier when you think of the other values up to 15 pence. The only number of pennies you can have in this bag to ensure you could make all other digits is 8 pennies.
Solution: Bag 1 has 1 penny, bag 2 has 2 pennies, bag 3 has 4 pennies and bag 4 has 8 pennies.
Another way of proving this is using base two.
As you can see, we can use just the numbers in base two as the numbers of pennies, as I am going to prove below:
You can continue this chart forever, and it shows how every value can be made.
From Neve and Saba recorded their work in this way:
Number of bags
1
2
3
4
5
6
Amount of money
1p
2p
4p
8p
16p
32p
Total money you can get
1p
3p
7p
15p
31p
63p
It would go on with the amount of money in the bags doubling (which is the pattern, we found out) forever!
We get our totals like this:
Number of bags
1
2
Amount of money.
1p
2p
4p
Total money you can get
1p
3p
7p
We added up the amounts of money in each of the bags to get the amount of money in the total column. 1p+2p+4p=7p
Rik from Cheongna Dalton School in South Korea, who was the youngest to send in their work, sent in this:
My mind put 1 penny in the first money bag. Then 2 pennies for the second money bag. But not 3 pennies for the third money bag. I put 4 pennies since I couldn't make four.
Then 4+1=5.
Then 4+2=6.
Then 4+2+1=7.
So I put 8 pennies in the last money bag.
Then 8+1=9.
Then 8+2=10.
Then 8+2+1=11.
Then 8+4=12.
Then 8+4+1=13.
Then 8+4+2=14.
Then if you add all of them together it equals 15.
Year 4 at Abbey Primary School sent in their solution as a picture that shows all the combinations, well done!
Thank you to all who sent in their solutions. It's good to see the different ways of working and presenting.
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# Division with Remainders using the Area Model
Examples, videos, and solutions to help Grade 4 students learn how to solve division problems with remainders using the area model.
Common Core Standards: 4.NBT.6, 4.OA.3
New York State Common Core Math Grade 4, Module 3, Lesson 21
NYS Math Module 3, Grade 4, Lesson 21 Problem Set
Problem 1: Solve 37 ÷ 2 using an area model. Use long division and distributive property to record your work.
Problem 2. Solve 76 ÷ 3 using an area model. Use long division and the distributive property to record your work.
1. Carolina solved the following division problem by drawing an area model.
a. What division problem did she solve?
b. Show how Carolina’s model can be represented using the distributive property.
NYS Math Module 3 Grade 4 Lesson 21 Problem Set
Problem 1 - Problem 3
Solve the following problems using the area model. Support the area model with long division or the distributive property.
1. 48 ÷ 3
Solve the following problems using the area model. Support the area model with long division or the distributive property.
5. 49 ÷ 3
6. 56 ÷ 4
7. 58 ÷ 4
8. 66 ÷ 5
9. 79 ÷ 3
10. Seventy-three students are divided into groups of 6 students each. How many groups of 6 students are there? How many students will not be in a group of 6?
NYS Math Module 3 Grade 4 Lesson 21 Homework
1. Solve 35 ÷ 2 using an area model. Use long division and the distributive property to record your work.
2. Solve 79 ÷ 3 using an area model. Use long division and the distributive property to record your work.
3. Paulina solved the following division problem by drawing an area model.
a. What division problem did she solve?
b. Show how Paulina’s model can be represented using the distributive property.
Solve the following problems using the area model. Support the area model with long division or the distributive property.
4. 42 ÷ 3
5. 43 ÷ 3
6. 52 ÷ 4
7. 54 ÷ 4
8. 61 ÷ 5
9. 73 ÷ 3
10. Ninety-seven lunch trays were placed equally in 4 stacks. How many lunch trays were in each stack? How many lunch trays will be leftover?
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Lesson 6- Comparing Decimals
Objectives
At the end of this lesson, students should be able to:
• Identify which decimal is larger or smaller than the other.
• Identify decimals or decimal fractions that are equal.
• Use the signs >, < or = to compare decimals.
Comparing decimals involves identifying which decimal is greater than (>), less than (<) or equal to (=) the other decimal to which it is being compared.
With the knowledge of decimal place value, decimals can be compared easily.
Example 1
Put in (>, < or =)
a) 24 _____ 3.89
b) 821 _____ 24.102
c) 2 ______ 98.75
d) 2 ______ 8.4739
Solution
a) 24˃3.89
Looking at the whole numbers, 8 is larger than 3, therefore 8.24 is greater than 3.89.
b) 821 ˂ 24.102
Comparing whole numbers, 24 is greater than 16, therefore, 16.821 is less than 24.102
c) 2˃98.75
865 is greater than 98, therefore, 865.2 is greater.
d) 2˃ 8.4739
In this example, one may wonder why the number with a greater number of digits isn’t the greater decimal. However, the number of digits in a decimal does not particularly play any role in comparison if the whole numbers are different. Therefore 9.2 is greater than 8.4739.
Example 2
Compare these:
a) 2 _____ 4.9
b) 25 _____ 16.09
c) 182 ______ 6.190
d) 241 ______ 5.247
Solution
Example 3
Put in >,< or =
Solution
Unit 10 Lesson 6: Exercise 1
Compare the following decimals using >, < or =
1) 8.6 _______ 8.9 2) 2.65 ________ 3.65
3) 7.953 ________ 7.928 4) 2.976 _______ 9.5
5) 5.148 ________ 5.142 6) 19.11 _______ 19.28
7) 0.87 ________ 0.09 8) 0.47 ________ 4.07
9) 240.8 ________ 248.0 10) 0.4 ________ 0.41
Unit 10 Lesson 6: Exercise 2
Unit >, < or = to make each statement true.
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# SSC CGL solved question paper Tier-I exam 2016: Quantitative Aptitude
May 9, 2018 14:44 IST
In this article, we have solved 25 questions of Quantitative Aptitude asked in SSC CGL Tier-1 exam 2016. This exam was organized from 27thAugust to 11thSeptember, 2016. read full question paper here-
SSC CGL solved paper
In this article, we have solved 25 questions of Quantitative Aptitude asked in SSC CGL Tier-1 exam 2016. This exam was organized from 27thAugust to 11thSeptember, 2016. These solved questions will give a gist of the type of questions, their difficulty level, and expected time spent on each question.
In this section, Most of the questions were from number system, geometry, and data interpretation. It is surprising that SSC did not put the questions from mixture & allegations, partnership, algebra, and time & work. However, we would suggest you not to leave these topics during preparation. The difficulty level of questions is moderate and it is very hard to tackle them with less practice. Let us go through all of them-
Question1. A clock gains 15 minutes per day. If it is set right at 12 noon, the time it shows at 4 AM is
a. 4.20 AM
b. 4.30 AM
c. 4.02 AM
d. 4.10 AM
Ans. d.
Explanation:
The difference between 12:00 noon and 4 AM is 16 hours.
if a clock gains 15 minutes in 24 hours, then it will gain= 15*16/24= 10 minutes in 16 hours.
Hence, the clock will show time = 4:10 AM at 4 AM.
Question2. A home theatre set is Rs. 4950. If two successive discounts of 20% and 15% are given, then its selling price is
a. Rs.3366
b. Rs.6633
c. Rs.3636
d. Rs.6363
Ans. a.
Explanation: After first discount, the selling price of home theatre = 4950*80%= Rs. 3960.
After second discount, the selling price= 3960*85%= Rs. 3366.
Question3. If 15% of x is three times of 10% of y, then x:y =
a.1:2
b. 2:1
c. 3:2
d. 2:3
Ans. b.
Explanation: Given: x*15% = 3*y*10%;
x: y= 30: 15= 2: 1;
SSC CGL 2018 Quantitative Aptitude Preparation Strategy: Detailed Chapterwise and Yearwise Analysis
Question4. A bookseller bought 500 text books for 20,000. He wanted to sell them at a profit so that he get 50 books free. At what profit percent should he sell them?
a.10
b.20
c.15
d. 10.5
Ans. a.
Explanation: The price of 500 text books = Rs. 20000 (given).
If bookseller wants to sell 450 books after saving 50 textbooks, then the price of 450 textbooks = 20000*450/500 = Rs. 18000.
The profit to be earned = Rs. 20000 – Rs. 18000 = Rs. 2000.
Hence, the profit in %= 2000*100/20000= 10%.
Question5. 20% of a man's salary is paid as rent, 60% are his living expenses and 10% are his savings. If he spends remaining 30 on the education of his children, find his salary
a. 300
b. 900
c. 3000
d. 9000
Ans. a.
Explanation: The salary in percentage spent on eduction of children= 100-(20 + 60 + 10) = 10%.
Let the salary is Rs. x.
x*10% = 30; = > x= Rs. 300.
Question6. A gun is fired at a distance of 6.64 km away from Ram. He hears the sound 20 seconds later. Then the speed of sound is
a. 664 m/s
b. 664 km/s
c. 332 m/s
d. 332 km/s
Ans. c.
Explanation: The speed of sound = Distance/time;
= (6.64 *1000 m)/20 seconds= 332 m/s.
Question7. The simple interest on Rs. 2000 for 2 years at Rs. 75 per thousand per annum will be
a. Rs.150
b. Rs.300
c. Rs.600
d. Rs.400
Ans. b.
Explanation: Interest rate= Rs. 75 per thousand= 7.5 per cent.
Hence, the simple interest= 2000*2*7.5/100= 300.
Question8. If x = 1 + √2 + √3, then the value of x2 - 2x - 4 is
a. √6
b. 2√3
c. 3√2
d. 2√6
Ans. d.
Explanation:
Question 10. If ΔABC is an equilateral triangle of side 16 cm, then the length of altitude is
a. 2√3 cm
b. 4√3 cm
c. 8√3 cm
d. 5√3 cm
Ans. c.
Explanation: The length of altitude in equilateral triangle= 3/2*side= 83 cm.
SSC CGL 2018 Tier-I Exam: Preparation Tips and Strategy
Question 11. O is the circumcentre of ΔABC. If AO = 8 cm, then the length of BO is
a. 12 cm
b. 3 cm
c. 6 cm
d. 8 cm
Ans. d.
Explanation: In case of circumcentre, AO=BO=CO= 8 cm.
Question 12. Given that tan(θ + 15°) = √3. Then the value of θ is?
a. 15°
b. 75°
c. 45°
d. 65°
Ans. c.
Explanation: tan(θ + 15) = √3 = tan60;
θ + 15 = 60; θ= 45;
Question 13. The least number of square tiles of side 41 cms required to pave the ceiling of a room of size 15m 17cm long and 9m 2 cm broad is:
a. 902
b. 656
c. 738
d. 814
Ans. d.
Explanation: The length of room = 1517 cm.
The breadth of room = 902 cm.
The no. of tiles to cover the room = (1517*902)/41*41= 814.
Question 14. The average of fruits offered in a temple in a week was 75. The average of fruits offered in six days excluding Tuesday was 72. How many fruits were offered on Tuesday?
a. 90
b. 93
c. 72
d. 92
Ans. b.
Explanation: The average fruits offered in a week = 75;
Hence, the total number of fruits offered in a week = 75*7= 525.
The average fruits offered excluding Tuesday = 72;
Hence, the total number of fruits offered excluding Tuesday= 72*6= 432.
The number of fruits offered in temple on Tuesday= 525 – 432= 93.
Question 15. If a= x1/3y1/3+x-1/3y-1/3, then a3 – 3ay – x is equal to
a. y2 / x
b. x2 / y
c. a2 / x
d. a2/y
Ans.
Explanation: Substitute x=8 and y=1 in the above given equation;
a= 2*1+2-11-1= 5/2;
Put these value in the given equation-
a3 – 3ay – x = (5/2)3-3*(5/2)*1 - 8 = 125/8 -15/2 -8 = 1/8.
Putting all these values in all options, we get option (a.) to be the true answer.
Question 16. If x + 1/5=5, then the value of 6x/(x2+x+1).
a. 1
b. 2
c. 3
d. 4
Ans. a.
Explanation: x= 5-0.20=4.8;
x2+x+1= (4.8)2+4.8+1= 28.8;
Hence, the result will be=(6*4.8)/28.8=28.8/28.8 = 1;
Question 17. If a > b > 0, AB = a - b cm, BC = 2√ab cm, CA=a + b; then ABC is
a. 45°
b. 60°
c. 90°
d. 120°
Ans. c.
Explanation: Looking at the sides of triangle, we can observe that CA2 = AB2 + BC2;
Hence, the triangle is right-angled at point B.
Question 18. PR is a tangent to a circle with centre O and radius 4 cm at point Q. If POR = 90°, OR = 5 cm and OP = 20/3 cm, then the length of PR is:
a. 3 cm
b. 16/3 cm
c. 23/3 cm
d. 25/3 cm
Ans. d.
Explanation: Since, POR triangle is a right-angled triangle;therefore, PR =(5)2+(20/3)2=25/3 cm.
SSC CGL Syllabus 2018: Tier I, II, III and IV with Exam Pattern
Question 19. From a point, 40 m apart from the foot of a tower, the angle of elevation of its top is 60°. The height of the tower is
a. 40√3
b. 40√3 cm
c. 40√3 m
d. 40√2 m
Ans. c.
Explanation: tan60 = Height of tower/base distance;
Tower height = 40√3 m;
Question 20. The radius of a sphere and hemisphere are same. The ratio of their total surface area is
a. 3:1
b. 2:1
c. 3:2
d. 4:3
Ans. d.
Explanation: if the radius are same for sphere and hemisphere.
Then, the surface area of sphere= 4πr2
The total surface area of hemisphere= 3 πr2;
Hence, the ratio = 4:3
Question 21. Two equal arcs of two circles subtend angle of 60° and 75° at the centre. The ratio of the radii of the two circles is
a. 5/4
b. 3/2
c. 4/5
d. 2/3
Ans. a.
Explanation:
Question 22. Study the following graph carefully and answer the questions
Which state has the maximum percentage of electrified villages?
a. B
b. C
c. D
d. F
Ans. c.
Explanation: It is obvious to the figure that villages of state D has the highest electricity.
Question 23. Study the following graph carefully and answer the questions
If the Central Government desires to give aid for speedy electrification starting from states with least electrification, which state will get fourth rank in order of priority?
a. F
b. C
c. E
d. B
Ans. d.
Explanation: The order of electricity in villages of the state is given below-
C < A < F < B < E < D
Question 24. Study the following graph carefully and answer the questions
How many States have at least 60% or more electrified villages?
a. Five
b. Three
c. Four
d. Two
Ans. a.
Explanation: The average % number of electrified villages in all states= 25 +45 + 20 + 60 + 55 + 30 = 235/6.
The 60% of this figure = 23.5. hence, there are five states, which have more electricity than others.
Question 25. Study the following graph carefully and answer the questions
Which state has twice the percentage of villages electrified in comparison to state F?
a. A
b. E
c. D
d. C
Ans. c.
Explanation: it is obvious that state D has the twice electrified villages than F.
Sub-Topics No. of questions Number system 5 Percentages 2 Averages 1 Simple and Compound Interest 1 Profit, Loss, and Discount 2 Mixture and Allegation Time and Distance 1 Geometry 4 Mensuration 3 Trigonometry 2 Data Interpretation 4
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# Circle Theorems: What Are They? Describe The Various Circle Theorems
Assignment Help 9th May 2023
Circle theorems are a set of rules and formulas that apply to circles and their components. They are an essential topic in mathematics and geometry, and form the basis of many mathematical concepts. In this blog post, we will explore the different types of circle theorems and their applications.
Before we dive into the circle theorems Assignment Help themselves, let's first review some basic terminology related to circles. A circle is a two-dimensional shape that consists of all points that are equidistant from a given point called the center. The distance from the center to any point on the circle is called the radius. The diameter is the distance across the circle through the center, and it is twice the length of the radius.
Now let's move on to the circle theorems. There are several types of circle theorems, including angle theorems, tangent theorems, and chord theorems. Each type of theorem is useful in solving different types of problems.
## Angle Theorems:
Angle at the center theorem: The angle at the center of a circle is twice the angle at the circumference that it subtends.
Angles in the same segment theorem: The angles that are subtended by the same arc in a circle are equal.
Angles in a semicircle theorem: The angle subtended by a diameter of a circle is a right angle.
### Tangent Theorems:
Tangents from an external point theorem: If a line is drawn from a point outside a circle to the circle, then the two tangents drawn to the circle from that point are equal in length.
Tangents to a circle from a point theorem: If two tangents are drawn to a circle from an external point, the tangents are equal in length.
### Chord Theorems:
Intersecting chords theorem: If two chords intersect within a circle, the products of the lengths of the segments of one chord are equal to the products of the lengths of the segments of the other chord.
Equal chords theorem: Chords that are equal in length are equidistant from the center of the circle.
Perpendicular bisector theorem: The perpendicular bisector of a chord passes through the center of the circle.
These are just a few examples of the many circle theorems that exist. Each theorem has its own unique application and can be used to solve different types of problems. You can get best cheap assignment help australia from BookMyEssay.
In addition to the circle theorems themselves, there are also several formulas and equations that are useful when working with circles. These include the formula for the circumference of a circle, which is C = 2πr, where r is the radius of the circle, and π is a constant value approximately equal to 3.14. The formula for the area of a circle is A = πr^2, where r is the radius of the circle.
So, why are circle theorems important? They are useful in many different fields, including engineering, physics, and even art. For example, architects use circle theorems to design buildings and structures that are both aesthetically pleasing and structurally sound. Engineers use circle theorems to design bridges and other structures that can withstand high levels of stress and strain. Even artists use circle theorems to create works of art that are based on geometric principles.
In conclusion, circle theorems are an essential topic in mathematics and geometry. They provide a set of rules and formulas that can be used to solve a wide range of problems related to circles and their components. Whether you are an engineer, architect, artist, or mathematician, understanding circle theorems is essential for success in your field. If you need help with circle theorems assignment writing help, there are many resources available online.
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# To differentiate between rational and irrational numbers (Page 2/3)
Page 2 / 3
Step 2: carry on dividing until a pattern becomes visible - the pattern will be indicated by the recurring numbers.
Now try the following:
5.1 $\frac{7}{9}$
5.2 $-5\frac{5}{6}$
5.3 $3\frac{\text{13}}{\text{99}}$
6. What is noticeable about fractions that are recurring decimal numbers (with regard to the denominator)?
7. Now, before we provide the steps for reducing a recurring decimal number to a common fraction, see if you are able to write the following as fractions by making use of the information from no. 6.
8. The following provides complete steps for reducing a recurring decimal number to a common fraction:
Suggestion : Multiply by 10 (if you have one recurring figure). Multiply by 100 (if there are 2 recurring figures), etc.
9. Now try to do no. 7.2 in the way that is discussed in no. 8.
## [lo 1.2.2, 1.2.6, 1.6.1, 1.9.1]
1. What is the meaning of % (percentage)? .....................................................................
2. If you have to reduce any fraction to a percentage, you have to reduce the denominator to 100.
• If this is not possible, you have to x (This principle can be applied in any situation, e.g. when you want to reduce a test that is marked out of 15 to a mark out of 50, you need to multiply by $\frac{\text{50}}{1}$ )
Reduce the following mathematics test marks from a grade 8 class to percentages (to one decimal figure, where necessary):
2.1 $\frac{\text{17}}{\text{20}}$ .......................................
2.2 $\frac{\text{19}}{\text{40}}$ .......................................
2.3 $\frac{\text{38}}{\text{50}}$ .......................................
2.4 $\frac{\text{45}}{\text{60}}$ .......................................
3. Reduce each of the following percentages to a common fraction (or a mixed number):
3.1 55 % .......................................
3.2 15,5% .......................................
3.3 16 $\frac{1}{2}$ % .......................................
3.4 $6\frac{2}{3}$ % .......................................
4. Each South African citizen should have access to some means of transport.
Bolokanang has a community of 25 500 people. Study the accompanying table indicating the number of people that use the given means of transport and answer the questions that follow.
Vehicle Number of users Bicycle $4\frac{1}{8}$ % Car $\frac{3}{5}$ Motorbike 0,085
4.1 Indicate how many inhabitants make use of:
a) a bicycle
b) a car
c) a motorbike
4.2 Express the number of inhabitants that use a car as a fraction of those who travel by bicycle.
4.3 Which percentage of the inhabitants has no vehicle?
4.4 Which other means of transport do farm labourers use to get to the nearest town?
4.5 If the number of job opportunities in rural areas should increase, the fraction of citizens who use cars for transport will double. What fraction of the community will be using cars for transport under such conditions?
## [lo 1.2.2, 1.2.5, 1.2.6, 1.6.2, 1.7.1, 1.7.2, 1.9.1]
1. Reduce each of the following compound numbers to improper fractions.This is very important in addition, subtraction, multiplication and division of fractions.
1.1 5 $\frac{4}{7}$ ................................ 1.2 7 $\frac{7}{9}$ ................................
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
Got questions? Join the online conversation and get instant answers!
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# Comparing and Ordering Integers
Home > By Subject > Integers > Comparing and Ordering Integers
Why Are Big Negative Numbers Less Than Small Ones?
We are used to big positive numbers meaning a big value - the bigger the number, the more, or higher, or longer, or expensive, or whatever the number represents.
With negative numbers, we have to remember that as the digit gets bigger, the number gets smaller. Think what the number represents. Think what the value is.
• The amount of money you have gets lower the more you owe.
• The temperature gets lower, the colder it gets
• The more you take away, the less you have
Looking at numbers on a vertical number line can be useful.
Zero is neither positive or negative
### Examples of Comparing Integers
Which is greater, -5 or -3?
Locate both numbers on the number line. Numbers to the right are greater. -3 is to the right of -5 so -3 is greater.
Put the following numbers in order starting with the least. -23, 17, -32, 2, -4, 0
Start looking at the negative numbers as these are always less than positive numbers. Start with the "biggest" negative number ( -32) as that has the lowest value.
Find the next "biggest" negative number (-23) and continue until the negative numbers are in order.
Zero and positive numbers can then be put into order to give the following complete list:
-32 , -23 , -4 , 0 , 2 , 17
Which of these numbers is greater than -6 but less than 4?
-8, -10, 5, -4, 0
Start be finding all the numbers that are less than -6. This gets rid of -8 and -10 (to the left of -6 on the number line).
Next we'll get rid of any number that is greater (not less) than 4. This means 5 is out leaving:
-4 and 0
Try these worksheets to practice with using integers.
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# Rectangle The area of a rectangle is by multiplying length and height. The perimeter of a rectangle is the distance around the outside of the rectangle.
## Presentation on theme: "Rectangle The area of a rectangle is by multiplying length and height. The perimeter of a rectangle is the distance around the outside of the rectangle."— Presentation transcript:
Rectangle The area of a rectangle is by multiplying length and height. The perimeter of a rectangle is the distance around the outside of the rectangle. A rectangle has a length of 8 centimeters and a width of 3 centimeters. Find the perimeter. Solution 1 - P = 8 cm + 8cm + 3 cm + 3 cm = 22 cm Solution 2 - P = 2(8 cm) + 2(3 cm) = 16 cm + 6 cm = 22 cm Example
Triangle multiply the base by the height, and then divide by 2. Find perimeter add a+b+c Find the perimeter, P, of the triangle shown below. Solution: So, the perimeter is 28 cm.
Parallelogram Example 1: Find the area of a parallelogram with a base of 12 centimeters and a height of 5 centimeters. Solution: A= b*h = (12 cm) · (5 cm) = 60 cm 2 The total distance around the outside of a parallelogram is the perimeter The area of a triangle is A= b*h
Circle Example The radius of a circle is 3 inches. What is the area? Solution: Area is A= 3.14 * r* r Perimeter is 2*3.14* r
Irregular Shape To find the area and perimeter of irregular shapes, the first thing to do is to divide the irregular shape into regular shapes that you can recognize such as triangles, rectangles, circles, squares and so forth... 1. Break down the irregular shapes into smaller shapes. 2. Find the edges of the smaller shapes. 3. Calculate the area of each small shape. 4. Add all of the areas of the small shapes (the sum will be the area of the irregular shape). Example
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# Show that for any integer $n\ge5$, the inequality ${2n \choose n}>3^n$ holds.
I used induction, but got stuck at the last step. By cancelling common factors in the expanded $2n$ choose $n$, I got a simpler equation that didn't make sense.How do i prove this inequality?
• What is this "simpler equation"? Induction is the way to go here, and the algebra is not at all complicated $-$ you just have to show that ${{2n+2}\choose {n+1}} \ge 3{{2n}\choose n}$ for $n \ge 5$. Apr 10, 2017 at 21:21
The main step is to prove for $n+1$ then
$${2(n+1)\choose n+1}=\frac{(2n+2)!}{(n+1)!(n+1)!}=\frac{(2n+2)(2n+1)}{(n+1)^2}{2n\choose n}=\frac{2(2n+1)}{(n+1)}{2n\choose n}$$
Using the hypothesis
$${2(n+1)\choose n+1}>\frac{2(2n+1)}{(n+1)}\cdot 3^n$$
but
$$\frac{2(2n+1)}{n+1}=\frac{4n+2}{n+1}=\frac{3(n+1)+(n-1)}{n+1}=3+\frac{n-1}{n+1}>3$$
so,
$${2(n+1)\choose n+1}>\frac{2(2n+1)}{(n+1)}\cdot 3^n>3\cdot 3^n=3^{n+1}$$
Here's a slightly goofier approach:
By the binomial theorem, we have $4^n=\sum_{k=0}^{2n} \binom{2n}{k}$. Since $\binom{2n}{n}$ is the largest of these $2n+1$ summands, it is at least as large as the average of the summands. So $\binom{2n}{n} \geq \frac{4^n}{2n+1}$.
I claim that $\frac{4^n}{2n+1}>3^n$ for all $n \geq 11$. This can be proved by induction: it is true for $n=11$, and if it is true for some fixed $n\geq 11$, then $$\frac{4^{n+1}}{3^{n+1}}=\frac{4}{3}\frac{4^n}{3^n}>\frac{4}{3}(2n+1)>2n+3$$ (this last inequality holds for all $n \geq 3$).
So the result is true for $n \geq 11$. It only remains to check that it is true for $n=5,6,7,8,9,10$, which is a straightforward computation.
Since $\binom{2m+2n}{m+n}\ge\binom{2m}m\binom{2n}n,$ the set $\{n:\binom{2n}n\ge3^n\}$ is closed under addition. Therefore, in order to show that the inequality holds for all $n\ge5,$ it's enough to verify it for $n=5,6,7,8,9.$
Two useful identities: $$\binom{n}{k} = \frac{n}{k}\binom{n-1}{k-1}$$ $$\binom{n}{k} = \binom{n}{n-k}$$ Applied here. $$\binom{2n+2}{n+1}=\frac{2n+2}{n+1}\binom{2n+1}{n}=\frac{2n+2}{n+1}\binom{2n+1}{n+1}=\frac{(2n+2)(2n+1)}{(n+1)^2}\binom{2n}{n}$$
Then just get lower bounds $$\frac{(2n+2)(2n+1)}{(n+1)^2} \geq 3$$ which is straight forwards.
Let $a=\frac{(2n)!}{n!n!}$. Given that $a>3^n$ show $\frac{(2n+1)(2n+2)}{(n+1)^2}a > 3^{n+1}$.
$\frac{4n^2+7n+2}{n^2+2n+1}a > 3*3^n$. Since $a>3^n$ it is sufficient to show that $\frac{4n^2+7n+2}{n^2+2n+1} > 3$. $$4n^2+7n+2 > 3n^2+6n+3$$$$n^2+n-1 > 0$$ which is clearly true for $n\geq 5$
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# How do you solve the system of equations x+y=14 and -x+y=16?
Apr 4, 2018
$\left(x , y\right) = \left(- 1 , 15\right)$
#### Explanation:
The end goal is to get an equation in terms of one variable, which can be solved.
We can solve using substitution. We'll need to pick one of our equations and solve for $x$ or $y .$ It does not matter which equation is used for this purpose, nor does it matter which variable is solved for.
This is a nice system, solving for either variable in either equation will be equally easy. Let's take $x + y = 14$:
$y = 14 - x$
Now, we can replace all $y$ in the other equation, $- x + y = 16$, with $14 - x :$
$- x + y = 16$
$- x + 14 - x = 16$
$- 2 x = 16 - 14$
$- 2 x = 2$
$x = - 1$
To solve for $y ,$ we can just back-substitute the above result into $y = 14 - x :$
$y = 14 - \left(- 1\right)$
$y = 15$
Thus,
$\left(x , y\right) = \left(- 1 , 15\right)$
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# The Rate of Change
Topics: Analytic geometry, Derivative, Slope Pages: 3 (648 words) Published: February 4, 2013
Integrating Project :The rate of change|
INTEGRATING PROJECT :THE RATE OF CHANGE|
Part:2|
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Pamela Villarreal A01192011Paulina Arizpe A01191939Adrian Covarrubias| 16/11/2012|
During this investigation we will achieve to discover how math can be applied to our everyday lives by using what we have learned in our math class to find out data. |
y=.0125x2
y=.0125x2
Finding the original equation:
y=ax2vertex= (0,0)point= (20,5)
35=a(55)2Domain= (-∞,∞)Range= [0, ∞)
35/25= a
a=.0125
1) Considering the map assigned to your team determine the position of the car when the headlights illuminate the sculpture.
Position= (-65,52.8)
By using the program graphmatica to be more accurate, we found out that in the parabola representing the road (y=.0125x2) the point where the car’s headlights illuminate the sculpture, which the sculpture is in the coordinate (-20,-20), is in the position of (-65, 52.8) because the car is moving to the right and this means that is going to the positive side of the graph.
2) Determine the equation of the line followed by the light of the car when the headlights hit the sculpture
Tangent line=
y=.0125x2
y’=.025x
y’=.025(-65)
y’= -1.625x
y=-1.625x-52.81
To find out the equation of the tangent line we first found the point where the car illuminated the sculpture which is (-65, 52.81), then we found the derivative of the original equation y=.0125x2 and the derivative is y'=.025x . After finding the derivative we needed to find the slope of the tangent line so we used the equation of the derivative and replaced the x with the x of the position and after solving it we got the slope of y’= -1.625x, and to finish the equation we added the y of the position to finally get y=-1.625x-52.81
3) Now determine the position of the car when the tail lights hit the sculpture.
Position (30,11.25)
By using the program graphmatica to get more accurate results, we found out that in...
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# What is the trigonometric form of (-7+2i) ?
Mar 4, 2017
$\sqrt{53} \left(\cos \left(2.86\right) + i \sin \left(2.86\right)\right)$
#### Explanation:
To convert from $\textcolor{b l u e}{\text{standard to trigonometric form}}$
$\text{that is } \left(x , y\right) \to r \left(\cos \theta + i \sin \theta\right)$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{and } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\textcolor{w h i t e}{\times \times} \text{where} - \pi < \theta \le \pi$
$\text{here " x=-7" and } y = 2$
$\Rightarrow r = \sqrt{{\left(- 7\right)}^{2} + {2}^{2}} = \sqrt{53}$
$\text{Since } - 7 + 2 i$ is in the second quadrant, we must ensure that $\theta$ is in the second quadrant.
$\theta = \pi - {\tan}^{-} 1 \left(\frac{2}{7}\right)$
$\Rightarrow \theta = \left(\pi - 0.278\right) = 2.86 \leftarrow \text{ in second quadrant}$
$\Rightarrow - 7 + 2 i \to \sqrt{53} \left(\cos \left(2.86\right) + i \sin \left(2.86\right)\right)$
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# Show that
Question:
$x(\log x)^{2}$
Solution:
$I=\int x(\log x)^{2} d x$
Taking $(\log x)^{2}$ as first function and $x$ as second function and integrating by parts, we obtain
$I=(\log x)^{2} \int x d x-\int\left[\left\{\frac{d}{d x}(\log x)^{2}\right\} \int x d x\right] d x$
$=\frac{x^{2}}{2}(\log x)^{2}-\left[\int 2 \log x \cdot \frac{1}{x} \cdot \frac{x^{2}}{2} d x\right]$
$=\frac{x^{2}}{2}(\log x)^{2}-\int x \log x d x$
Again integrating by parts, we obtain
$I=\frac{x^{2}}{2}(\log x)^{2}-\left[\log x \int x d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x d x\right\} d x\right]$
$=\frac{x^{2}}{2}(\log x)^{2}-\left[\frac{x^{2}}{2} \log x-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x\right]$
$=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{1}{2} \int x d x$
$=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+C$
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# Math comic project
More Options: Make a Folding Card
#### Storyboard Description
This storyboard does not have a description.
#### Storyboard Text
• I am your host, Bella. Let's take a tour!
• Hey guys, welcome to ABC Heights!!!
• ABC Heights
• OMG!!! Mr.H is having a math contest who ever wins will get \$300
• Can you give us an example of one and how to solve it
• Say what now!!!
• ABC Heights
• A multi-step math equation
• Of course!!!
• One example is-4(x-2)=3(x-3) Step 1- Whenever you see parentheses in an equation get ruin of it by distributing. Take the number outside the parentheses and multiply what is inside. Watch out for negatives. Step 2- Identify the like terms of both sides and combine it. You can box the variables and circle the constants with the sign. Step 3- Move the variables to one side. Do the inverse operation of the variable and bring it to the other side. If the variable is positive/add the inverse of that is negative/subtract. Step 4- Move the constant. In order to move the constant to the other side, you have to do the inverse operation of it. The inverse of adding is subtracting. Step 5- Move the coefficient. In order to found what x is you have to divide or multiply.
• -4(x-2)=3(x-3) Distribute : -4x+8=3x-9 Like terms: -4x+8=3x-9 Move the variables : -4x+8=3x-9 +4x =+4 8 = 7x-9 +9= +9 7x = 17 x=17/7-17 over 7
• You have 5 minutes to solve this if you win you will get \$300.00 Good luck!!!!
• Sovle -3(x-6)+4(x+1)=7x-10
• Rose-2nd grade -3(x-6)+4(x+1)=7x-10 -3x+18+4x+1=7x-10 -1x+19=7x-10 +1 = +1 19=8x-10 +10 = +10 8x=29 x=29/8 or 3.625
• 3 weeks later
• Good job!!!
• Cool!!!
• YaYYY!!!
• The winner of this year's math contest is Rose Bestows from 2nd grade!!!!
• The winner is Rose!!!
• CONGRATULATION, Rose
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Combining like terms Whole number coefficients Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Combining like terms Whole number coefficients. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Simplify the following expression by combining like terms:
5a + 4b − 5c + 6a + 9b
Explanation
Step 1:
Combining like terms
5a + 4b − 5c + 6a + 9b = (5a + 6a) + (4b + 9b) – 5c
Step 2:
(5a + 6a) + (4b + 9b) – 5c
= 11a + 13b – 5c
Step 3:
So, 5a + 4b − 5c + 6a + 9b
= 11a + 13b – 5c
Q 2 - Simplify the following expression by combining like terms:
7f + 8g + 4f + 3f + 6f + 2g
Explanation
Step 1:
Combining like terms
7f + 8g + 4f + 3f + 6f + 2g
= (3f + 4f + 6f + 7f) + (8g + 2g)
Step 2:
(3f + 4f + 6f + 7f) + (8g + 2g) = 20f + 10g
Step 3:
So, 7f + 8g + 4f + 3f + 6f + 2g
= 20f + 10g
Q 3 - Simplify the following expression by combining like terms:
12x − 15y − 18x + 17y + 20x
Explanation
Step 1:
Combining like terms
12x − 15y − 18x + 17y + 20x
= (12x −18x + 20x) + (−15y + 17y)
Step 2:
(12x −18x + 20x) + (−15y + 17y) = 14x + 2y
Step 3:
So, 12x − 15y − 18x + 17y + 20x
= 14x + 2y
Q 4 - Simplify the following expression by combining like terms:
10x + 3y + 3x + 4y + 5x + 3z
Explanation
Step 1:
Combining like terms
10x + 3y + 3x + 4y + 5x + 3z
= (10x + 3x + 5x) + (3y + 4y) + 3z
Step 2:
(10x + 3x + 5x) + (3y + 4y) + 3z
= 18x + 7y + 3z
Step 3:
So, 10x + 3y + 3x + 4y + 5x + 3z
= 18x + 7y + 3z
Q 5 - Simplify the following expression by combining like terms:
9x + 7y + 4x − 11y + 10x – 8x
Explanation
Step 1:
Combining like terms
9x + 7y + 4x − 11y + 10x – 8x
= (9x + 4x +10x −8x) + (7y – 11y)
Step 2:
(9x + 4x +10x −8x) + (7y – 11y)
= 15x – 4y
Step 3:
So, 9x + 7y + 4x − 11y + 10x – 8x
= 15x – 4y
Q 6 - Simplify the following expression by combining like terms:
3x + 15y − 7x + 10y + 9z
Explanation
Step 1:
Combining like terms
3x + 15y − 7x + 10y + 9z
= (3x −7x) + (10y + 15y) + 9z
Step 2:
(3x −7x) + (10y + 15y) + 9z
= −4x + 25y + 9z
Step 3:
So, 3x + 15y − 7x + 10y + 9z
= −4x + 25y + 9z
Q 7 - Simplify the following expression by combining like terms:
8a + 7b + 2b + 3a −5c
Explanation
Step 1:
Combining like terms
8a + 7b + 2b + 3a − 5c
= (8a + 3a) + (2b + 7b) − 5c
Step 2:
(8a + 3a) + (2b + 7b) − 5c
= 11a + 9b − 5c
Step 3:
So, 8a + 7b + 2b + 3a − 5c
= 11a + 9b − 5c
Q 8 - Simplify the following expression by combining like terms:
10a + 7b + 6c + 8a + 6b + 9c
Explanation
Step 1:
Combining like terms
10a + 7b + 6c + 8a + 6b + 9c
= (10a + 8a) + (6b + 7b) + (6c + 9c)
Step 2:
(10a + 8a) + (6b + 7b) + (6c + 9c)
= 18a + 13b + 15c
Step 3:
So, 10a + 7b + 6c + 8a + 6b + 9c
= 18a + 13b + 15c
Q 9 - Simplify the following expression by combining like terms:
20x − 2xy2 + 3x − 13xy2
Explanation
Step 1:
Combining like terms
20x − 2xy2 + 3x − 13xy2
= (20x + 3x) − (2xy2 + 13xy2)
Step 2:
(20x + 3x) − (2xy2 + 13xy2)
= 23x − 15xy2
Step 3:
So, 20x − 2xy2 + 3x − 13xy2
= 23x − 15xy2
Q 10 - Simplify the following expression by combining like terms:
2x + 8y + 8x + 9y − 5y − 17z + 5z
Explanation
Step 1:
Combining like terms
2x + 8y + 8x + 9y − 5y − 17z + 5z
= (2x + 8x) + (8y + 9y −5y) + (5z − 17z)
Step 2:
(2x + 8x) + (8y + 9y − 5y) + (5z − 17z)
= 10x + 12y − 12z
Step 3:
So, 2x + 8y + 8x + 9y − 5y − 17z + 5z
= 10x + 12y − 12z
combining_like_terms_whole_number_coefficients.htm
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## Multiple Solutions Part 2: Celebrating Different Methods (Lesson 1)
Last year I taught two lessons on constructing equations to a top set year 8 class. In the first lesson I set 3 questions for the students to attempt in any way they saw fit. It’s important to note that I did not tell them that they could construct equations to solve the problems. I wanted them to do whatever felt natural rather than leading them into a certain strategy. The questions I set are shown below:
1) I think of a number, add 7, multiply by 3, subtract 3, divide by 6, then multiply by 12. The answer is 72. Find the number I was thinking of.
2) The number in each square is the sum of the numbers on either side. Find the missing numbers in the circles.
3) Adrienne has three times as many dollars as Karim. Adrienne gives Karim \$20. She then has \$5 more dollars than Karim. How much does Adrienne have now?
While they were doing the problems I circulated around the room to get a sense of the strategies being used. As I suspected all but one of the pairs started with an algebraic approach. I didn’t anticipate however that some of them would struggle to write down the correct equation. These students reverted to what they had learnt in year 7 and worked back from 72 using inverse operations.
I helped the one pair who persisted with the algebraic approach by leading them into the structure shown below. In this way, the pair would later be able to explain why the inverse operation technique worked to the rest of the class.
([3(x + 7) – 3]/6) × 12 = 72
Once every pair had moved onto question 2 we stopped to discuss the solution to question 1. I praised the entire group for finding some interesting strategies and asked a pair who had used the inverse operation technique to explain their reasoning. Once they had given an adequate explanation I asked the pair who’d struggled on the algebraic technique to explain their strategy on the other side of the board through slowly constructing the equation.
It became immediately obvious to the class that the algebraic technique was in fact identical to the inverse operation technique. The main advantage of the algebraic technique being that it is elegantly communicated and provides an explanation as to why the inverse technique works.
On question 2 most of the pairs used a numerical trial and error technique and every single pair arrived at the correct solution quite quickly. Again, once I saw that everyone had finished I asked one pair who had used a structured table to order their trials to come to the board and then another pair who had successfully used an algebraic strategy shown below.
(30 – x) + (23 – x) = 25
53 – 2x = 25
x = 18 etc.
(Note: I neglected to show the approach of using three different variables and then constructing 3 seperate equations at this point since I wanted the idea of using different algebraic approaches to hit them in the next lesson.)
The pair made a decent attempt to explain why the bottom left and right circles were 30-x and 23-x but struggled to get it across to the rest of the class. I therefore intervened and used a diagram to help explain it.
Once this was verified, which is indeed the most important part of this strategy, the students were easily able to solve the equation and find the three missing values. It’s important to note that we did discuss that the positioning of x could have been in any one of the three circles.
At this point in the lesson it was blindingly obvious that the students were impressed and intrigued by the algebraic strategies and were enthused to try to use an algebraic strategy for the last question.
All pairs with the exception of one started by trying to construct an equation to solve the problem. Many of them became quickly confused with the algebraix approach and reverted back to trial and error until they eventually found a solution. One pair however persisted with an algebraic approach until they derived two simultaneous equations to solve the problem. Incidentally this pair was made up of two girls who’ve recently moved to Brazil from Japan and South Korea (This is something I will come back in an upcoming post). You may think at this point that it’s possible that one of them could have seen a similar question before. I do feel that they struggled on the problem and am confident that this was not the case.
I asked them to explain their strategy on the board which was written as follows:
A = 3K
A – 20 = K + 5
3K – 20 = K + 5
K = 12.5 and therefore A = 37.5
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# How to Find the Median
Last updated Sep 02, 2021
Posted in Yard & Patio
Median is a statistical measure of central tendency, it simply represents the middle value in a set of data. The median is commonly described as the 50th percentile, or 50 percent. In other words, if we have 10 scores in a set and for every 10 scores in the set there are 2 others that are higher than it and for every 10 scores there are 4 others that are lower than it then the score at 50th percentile would be 2 and so on. So when we find a score in a set that is just below 50% then this score is said to be "median".
In statistics, usually what we want to know I’m sure you’ve heard by now is how many people or items should appear at least once in each row or column of data
## How do I calculate the median?
The median is a statistical tool which gives a rough estimate of the middle value between two numbers. It is commonly used in statistics courses to determine which data value has been found to be closest to the middle value. In this topic, we will learn how to calculate the median and compare it with other numbers.
## How do you find the median quickly?
While it is sometimes difficult to find the median of the many data sets that exist in social media, analytics firm Sentient has developed a method to determine the best answer quickly.
## How do you find the median of an example?
In a research, finding the median of an example is important for finding the average level of the data.
The median is a key figure in statistics and it can be found in most examples. For instance, if you want to find the median salary of your employees or if you want to find out if it’s possible to make two groups of people happy at your company (e.g., same income/salary).
Try to find it yourself!
## What is the easiest way to find the median in math?
A common problem for math students is finding the median in a group of numbers. This problem can be solved by using a tool like:
## How do you find the median in math step by step?
A mathematical concept that has been simplified and made popular in the last years. The mean is often used to represent the middle point (or average) of a set of numbers. Often, this term brings to mind the notion of “average” or “middle” in terms of sales, marketing or product development.
## What is the fastest way to find the median?
One of the most popular statistics is the median. It looks like this:
The median is the number that comes between two numbers. For example, if you asked me what is my favorite movie, I would say "Two hours ago" but if you asked "What movie do you like", I may come up with something different - "Two hours ago" or "I don't know".
The median is also known as the middle answer. But now, let’s see what would happen if we ask it with a computer instead of a human.
We could start by asking us to make such a calculation and get an answer: (My favorite movie) at (Timezone). Let's go through some examples: (2/3) at 7pm - (3/5),
## How do I find the median shortcut?
Finding the median shortcut is something most of us wish to do over and over again. But, with so many different ways of doing it, how can we do it efficiently?
The typical approach is getting the median of two numbers or a value. Let’s see how automated approaches can be used to find the median shortcut.
## How do you find the median in math for kids?
Suggestion to use the median in math for kids.
Section topic: What are your recommendations for young people?
Section keywords:
Introduction:
Recommendations for young people.
## How do you find the median for dummies?
In this blog post, I will share some statistical information on how to find the median for a dummy variable. This will be useful when you are trying to find a specific value in a table.
This blog post provides an introduction on the dummies and statistics/facet of dummies for more details can be found at:
## Conclusion
The big data scientists are increasingly focusing on the "how to" rather than the "what to" when it comes to optimization. They are constantly looking for ways to improve the efficiency of their workflows and processes, leading them to focus on finding solutions that can help them accomplish these goals.
A company can use big data analysis in several different ways - for example, by identifying different segments of customers that need more personalized attention or by identifying possible new products or features they could offer.
Most companies do not just want to optimize their processes they also want better outcomes from these processes using data science methods. However, there is a big difference between analytics and machine learning - some people think that some forms of analytics are more appropriate for some businesses while others think that machine learning is not applicable at all
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# Access Point #: MAFS.4.NF.2.AP.4a (Archived Access Point)
This document was generated on CPALMS - www.cpalms.org
Multiply a fraction by a whole number using a visual fraction model.
#### Essential Understandings
Concrete:
• Place fraction manipulatives in groups as indicated by the whole number in a given multiplication expression (e.g., 2 × 1/3 = 2 groups of 1/3 or 3 × 1/4 = 3 groups of 1/4).
• Use repeated addition/skip counting to find the product (e.g., 1/3 + 1/3 = 2/3 or 1/4 + 1/4 + 1/4 = 3/4).
Representation:
• Use a visual representation of a whole divided into equal pieces (each piece may be labeled with the corresponding unit fraction). Shade the number of groups of the fraction (e.g., 3 groups of 1/5 ) as indicated by the whole number.
• Use repeated addition/skip counting to find the product (e.g., 1/5 + 1/5 +1/5 = 3/5).
• Understand the following vocabulary: numerator, denominator.
Number: MAFS.4.NF.2.AP.4a Category: Access Points Cluster: Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers. (Major Cluster) : Clusters should not be sorted from Major to Supporting and then taught in that order. To do so would strip the coherence of the mathematical ideas and miss the opportunity to enhance the major work of the grade with the supporting clusters.
#### Related Standards
Name Description
MAFS.4.NF.2.4: Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.
1. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4).
2. Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 × (2/5) as 6 × (1/5), recognizing this product as 6/5. (In general, n × (a/b) = (n × a)/b.)
3. Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat 3/8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie?
Clarifications:Examples of Opportunities for In-Depth Focus This standard represents an important step in the multi-grade progression for multiplication and division of fractions. Students extend their developing understanding of multiplication to multiply a fraction by a whole number.
#### Related Courses
Name Description 5012060: Mathematics - Grade Four 7712050: Access Mathematics Grade 4 5012065: Grade 4 Accelerated Mathematics 5012015: Foundational Skills in Mathematics 3-5
#### Element Cards
Name Description 4th Grade Math Element Cards: Element Cards are available to assist in planning for instruction. They are designed to promote understanding of how students move toward the academic standards. Element Cards contain one or more access points, essential understandings, suggested instructional strategies and suggested supports.
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## Pages
### Sum of reciprocal squares
Today we will look at a very fascinating proof due to Euler for the following identity: $$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \dots = \frac{\pi^2}{6}$$
The mathematician Euler formulated this proof in 1734 when he was 28 year old.
We will present Euler's proof step by step.
Step 1. Using Taylor series expansion for $f(x) = \sin(x)$
We have
$$f(x) = \sin(x) ~\Rightarrow ~ f(0) = 0$$ $$f'(x) = \cos(x) ~\Rightarrow ~ f'(0) = 1$$ $$f''(x) = -\sin(x) ~\Rightarrow ~ f''(0) = 0$$ $$f^{(3)}(x) = -\cos(x) ~\Rightarrow ~ f^{(3)}(0) = -1$$ $$f^{(4)}(x) = \sin(x) ~\Rightarrow ~ f^{(4)}(0) = 0$$ $$f^{(5)}(x) = \cos(x) ~\Rightarrow ~ f^{(5)}(0) = 1$$ $$f^{(6)}(x) = -\sin(x) ~\Rightarrow ~ f^{(6)}(0) = 0$$ $$f^{(7)}(x) = -\cos(x) ~\Rightarrow ~ f^{(7)}(0) = -1$$
So the Taylor series for the function $f(x) = \sin(x)$ is as follows:
$$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \dots + \frac{f^{(n)}(0)}{n!} x^n + \dots$$
$$\sin(x) = x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 - \frac{1}{7!} x^7 + \dots$$
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$
Step 2. Express $\sin(x)$ as product of linear factors
Since the equation $f(x) = \sin(x) = 0$ has the following roots $$x=0, ~x = \pm \pi, ~x = \pm 2 \pi, ~x = \pm 3 \pi, \dots$$
we have
$$\sin(x) = C x (x - \pi)(x + \pi)(x - 2\pi)(x + 2 \pi)(x - 3\pi)(x + 3\pi) \dots$$
$$\sin(x) = C x (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots$$
Step 3. Combining step 1 and step 2
Comparing the two representations of $\sin(x)$ in step 1 and step 2, we have
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$ $$= C x (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots$$
So
$$1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots$$ $$= C (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots$$
Substitute $x^2$ by $x$, we have
$$1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots$$ $$= C (x - \pi^2)(x - 2^2 \pi^2)(x - 3^2 \pi^2)(x - 4^2 \pi^2) \dots$$
Step 4. Normalization to get rid of the constant factor $C$
If we have an equation of the form
$$f(x) = C g_1(x) g_2(x) g_3(x) \dots$$
we can get rid of the constant factor $C$ by normalization as follows
$$\frac{f(x)}{f(0)} = \frac{g_1(x)}{g_1(0)} \frac{g_2(x)}{g_2(0)} \frac{g_3(x)}{g_3(0)} \dots$$
With the equation that we have in step 3
$$1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots$$ $$= C (x - \pi^2)(x - 2^2 \pi^2)(x - 3^2 \pi^2)(x - 4^2 \pi^2) \dots$$
by normalization, we obtain
$$1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots$$ $$= \frac{(x - \pi^2)}{-\pi^2} \frac{(x - 2^2 \pi^2)}{-2^2 \pi^2} \frac{(x - 3^2 \pi^2)}{-3^2 \pi^2} \frac{(x - 4^2 \pi^2)}{-4^2 \pi^2} \dots$$ $$= (1 - \frac{x}{\pi^2}) (1 - \frac{x}{2^2 \pi^2}) (1 - \frac{x}{3^2 \pi^2}) (1 - \frac{x}{4^2 \pi^2})\dots$$
Step 5. Use Vieta formula
Comparing the coefficients of $x$ in both sides of the equation
$$1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots$$ $$= (1 - \frac{x}{\pi^2}) (1 - \frac{x}{2^2 \pi^2}) (1 - \frac{x}{3^2 \pi^2}) (1 - \frac{x}{4^2 \pi^2})\dots$$
we have $$- \frac{x}{3!} = - \frac{x}{\pi^2} - \frac{x}{2^2 \pi^2} - \frac{x}{3^2 \pi^2} - \frac{x}{4^2 \pi^2} - \dots$$
Therefore, $$\frac{1}{3!} = \frac{1}{\pi^2} + \frac{1}{2^2 \pi^2} + \frac{1}{3^2 \pi^2} + \frac{1}{4^2 \pi^2} + \dots$$
Thus, we obtain
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots =\frac{\pi^2}{3!}= \frac{\pi^2}{6}$$
Let us stop here for now. In the homework section, you can try Euler's method with the function $f(x) = \cos(x)$ to derive another proof of the above identity. Hope to see you again next time.
Homework.
1. Show that the Taylor series for $f(x) = \cos(x)$ is as follows $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$$
2. Using the fact that the equation $\cos(x) = 0$ has the following roots $$x = \pm \frac{\pi}{2}, ~x = \pm \frac{3 \pi}{2}, ~x = \pm \frac{5 \pi}{2}, \dots$$
show that
$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots = (1 - \frac{4 x^2}{\pi^2})(1 - \frac{4 x^2}{3^2 \pi^2})(1 - \frac{4 x^2}{5^2 \pi^2}) \dots$$
3. Prove that
$$\frac{1 }{1^2} + \frac{1 }{3^2} + \frac{1 }{5^2} + \dots = \frac{\pi^2}{8}$$
4. Prove that
$$\frac{1 }{1^2} + \frac{1 }{3^2} + \frac{1 }{5^2} + \dots = \frac{3}{4} \left( \frac{1 }{1^2} + \frac{1 }{2^2} + \frac{1 }{3^2} + \frac{1 }{4^2} + \dots \right)$$
thus deriving
$$\frac{1 }{1^2} + \frac{1 }{2^2} + \frac{1 }{3^2} + \frac{1 }{4^2} + \dots = \frac{\pi^2}{6}$$
5. Go to google.com and search about the Basel problem and the Riemann zeta function.
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# 1.1 Functions and function notation (Page 10/21)
Page 10 / 21
Why does the horizontal line test tell us whether the graph of a function is one-to-one?
When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.
## Algebraic
For the following exercises, determine whether the relation represents a function.
$\left\{\left(a,b\right),\left(b,c\right),\left(c,c\right)\right\}$
function
For the following exercises, determine whether the relation represents $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$
$5x+2y=10$
$y={x}^{2}$
function
$x={y}^{2}$
$3{x}^{2}+y=14$
function
$2x+{y}^{2}=6$
$y=-2{x}^{2}+40x$
function
$y=\frac{1}{x}$
$x=\frac{3y+5}{7y-1}$
function
$x=\sqrt{1-{y}^{2}}$
$y=\frac{3x+5}{7x-1}$
function
${x}^{2}+{y}^{2}=9$
$2xy=1$
function
$x={y}^{3}$
$y={x}^{3}$
function
$y=\sqrt{1-{x}^{2}}$
$x=±\sqrt{1-y}$
function
$y=±\sqrt{1-x}$
${y}^{2}={x}^{2}$
not a function
${y}^{3}={x}^{2}$
For the following exercises, evaluate the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at the indicated values
$f\left(x\right)=2x-5$
$\begin{array}{cccc}f\left(-3\right)=-11;& f\left(2\right)=-1;& f\left(-a\right)=-2a-5;& -f\left(a\right)=-2a+5;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(a+h\right)=2a+2h-5\end{array}$
$f\left(x\right)=-5{x}^{2}+2x-1$
$f\left(x\right)=\sqrt{2-x}+5$
$\begin{array}{cccc}f\left(-3\right)=\sqrt{5}+5;& f\left(2\right)=5;& f\left(-a\right)=\sqrt{2+a}+5;& -f\left(a\right)=-\sqrt{2-a}-5;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(a+h\right)=\end{array}$ $\sqrt{2-a-h}+5$
$f\left(x\right)=\frac{6x-1}{5x+2}$
$f\left(x\right)=|x-1|-|x+1|$
Given the function $\text{\hspace{0.17em}}g\left(x\right)=5-{x}^{2},\text{\hspace{0.17em}}$ evaluate $\text{\hspace{0.17em}}\frac{g\left(x+h\right)-g\left(x\right)}{h},\text{\hspace{0.17em}}h\ne 0.$
Given the function $\text{\hspace{0.17em}}g\left(x\right)={x}^{2}+2x,\text{\hspace{0.17em}}$ evaluate $\text{\hspace{0.17em}}\frac{g\left(x\right)-g\left(a\right)}{x-a},\text{\hspace{0.17em}}x\ne a.$
$\frac{g\left(x\right)-g\left(a\right)}{x-a}=x+a+2,\text{\hspace{0.17em}}x\ne a$
Given the function $\text{\hspace{0.17em}}k\left(t\right)=2t-1:$
1. Evaluate $\text{\hspace{0.17em}}k\left(2\right).$
2. Solve $\text{\hspace{0.17em}}k\left(t\right)=7.$
Given the function $\text{\hspace{0.17em}}f\left(x\right)=8-3x:$
1. Evaluate $\text{\hspace{0.17em}}f\left(-2\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=-1.$
a. $\text{\hspace{0.17em}}f\left(-2\right)=14;\text{\hspace{0.17em}}$ b. $\text{\hspace{0.17em}}x=3$
Given the function $\text{\hspace{0.17em}}p\left(c\right)={c}^{2}+c:$
1. Evaluate $\text{\hspace{0.17em}}p\left(-3\right).$
2. Solve $\text{\hspace{0.17em}}p\left(c\right)=2.$
Given the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}-3x:$
1. Evaluate $\text{\hspace{0.17em}}f\left(5\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=4.$
a. $\text{\hspace{0.17em}}f\left(5\right)=10;\text{\hspace{0.17em}}$ b. or
Given the function $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x+2}:$
1. Evaluate $\text{\hspace{0.17em}}f\left(7\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=4.$
Consider the relationship $\text{\hspace{0.17em}}3r+2t=18.$
1. Write the relationship as a function $\text{\hspace{0.17em}}r=f\left(t\right).$
2. Evaluate $\text{\hspace{0.17em}}f\left(-3\right).$
3. Solve $\text{\hspace{0.17em}}f\left(t\right)=2.$
a. $\text{\hspace{0.17em}}f\left(t\right)=6-\frac{2}{3}t;\text{\hspace{0.17em}}$ b. $\text{\hspace{0.17em}}f\left(-3\right)=8;\text{\hspace{0.17em}}$ c. $\text{\hspace{0.17em}}t=6\text{\hspace{0.17em}}$
## Graphical
For the following exercises, use the vertical line test to determine which graphs show relations that are functions.
not a function
function
function
function
function
function
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(-1\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=3.$
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(0\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=-3.$
a. $\text{\hspace{0.17em}}f\left(0\right)=1;\text{\hspace{0.17em}}$ b. or
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(4\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=1.$
For the following exercises, determine if the given graph is a one-to-one function.
not a function so it is also not a one-to-one function
one-to- one function
function, but not one-to-one
## Numeric
For the following exercises, determine whether the relation represents a function.
$\left\{\left(-1,-1\right),\left(-2,-2\right),\left(-3,-3\right)\right\}$
$\left\{\left(3,4\right),\left(4,5\right),\left(5,6\right)\right\}$
function
$\left\{\left(2,5\right),\left(7,11\right),\left(15,8\right),\left(7,9\right)\right\}$
For the following exercises, determine if the relation represented in table form represents $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x.$
$x$ 5 10 15 $y$ 3 8 14
function
$x$ 5 10 15 $y$ 3 8 8
$x$ 5 10 10 $y$ 3 8 14
not a function
For the following exercises, use the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ represented in [link] .
$x$ $f\left(x\right)$ 0 74 1 28 2 1 3 53 4 56 5 3 6 36 7 45 8 14 9 47
Evaluate $\text{\hspace{0.17em}}f\left(3\right).$
Solve $\text{\hspace{0.17em}}f\left(x\right)=1.$
$f\left(x\right)=1,\text{\hspace{0.17em}}x=2$
For the following exercises, evaluate the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at the values $f\left(-2\right),\text{\hspace{0.17em}}f\left(-1\right),\text{\hspace{0.17em}}f\left(0\right),\text{\hspace{0.17em}}f\left(1\right),$ and $\text{\hspace{0.17em}}f\left(2\right).$
$f\left(x\right)=4-2x$
$f\left(x\right)=8-3x$
$\begin{array}{ccccc}f\left(-2\right)=14;& f\left(-1\right)=11;& f\left(0\right)=8;& f\left(1\right)=5;& f\left(2\right)=2\end{array}$
$f\left(x\right)=8{x}^{2}-7x+3$
$f\left(x\right)=3+\sqrt{x+3}$
$\begin{array}{ccccc}f\left(-2\right)=4;\text{ }& f\left(-1\right)=4.414;& f\left(0\right)=4.732;& f\left(1\right)=4.5;& f\left(2\right)=5.236\end{array}$
$f\left(x\right)=\frac{x-2}{x+3}$
$f\left(x\right)={3}^{x}$
$\begin{array}{ccccc}f\left(-2\right)=\frac{1}{9};& f\left(-1\right)=\frac{1}{3};& f\left(0\right)=1;& f\left(1\right)=3;& f\left(2\right)=9\end{array}$
For the following exercises, evaluate the expressions, given functions $f,\text{\hspace{0.17em}}\text{\hspace{0.17em}}g,$ and $\text{\hspace{0.17em}}h:$
• $f\left(x\right)=3x-2$
• $g\left(x\right)=5-{x}^{2}$
• $h\left(x\right)=-2{x}^{2}+3x-1$
$3f\left(1\right)-4g\left(-2\right)$
$f\left(\frac{7}{3}\right)-h\left(-2\right)$
20
## Technology
For the following exercises, graph $\text{\hspace{0.17em}}y={x}^{2}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
$\left[-100,100\right]$
For the following exercises, graph $\text{\hspace{0.17em}}y={x}^{3}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
For the following exercises, graph $\text{\hspace{0.17em}}y=\sqrt{x}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
For the following exercises, graph $y=\sqrt[3]{x}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
$\left[-0.001,\text{0.001}\right]$
$\left[-0.1,\text{0.1}\right]$
$\left[-1000,\text{1000}\right]$
$\left[-1,000,000,\text{1,000,000}\right]$
## Real-world applications
The amount of garbage, $\text{\hspace{0.17em}}G,\text{\hspace{0.17em}}$ produced by a city with population $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is given by $\text{\hspace{0.17em}}G=f\left(p\right).\text{\hspace{0.17em}}$ $G\text{\hspace{0.17em}}$ is measured in tons per week, and $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is measured in thousands of people.
1. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$
2. Explain the meaning of the statement $\text{\hspace{0.17em}}f\left(5\right)=2.$
The number of cubic yards of dirt, $\text{\hspace{0.17em}}D,\text{\hspace{0.17em}}$ needed to cover a garden with area $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ square feet is given by $\text{\hspace{0.17em}}D=g\left(a\right).$
1. A garden with area 5000 ft 2 requires 50 yd 3 of dirt. Express this information in terms of the function $\text{\hspace{0.17em}}g.$
2. Explain the meaning of the statement $\text{\hspace{0.17em}}g\left(100\right)=1.$
a. $\text{\hspace{0.17em}}g\left(5000\right)=50;$ b. The number of cubic yards of dirt required for a garden of 100 square feet is 1.
Let $\text{\hspace{0.17em}}f\left(t\right)\text{\hspace{0.17em}}$ be the number of ducks in a lake $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ years after 1990. Explain the meaning of each statement:
1. $f\left(5\right)=30$
2. $f\left(10\right)=40$
Let $\text{\hspace{0.17em}}h\left(t\right)\text{\hspace{0.17em}}$ be the height above ground, in feet, of a rocket $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds after launching. Explain the meaning of each statement:
1. $h\left(1\right)=200$
2. $h\left(2\right)=350$
a. The height of a rocket above ground after 1 second is 200 ft. b. the height of a rocket above ground after 2 seconds is 350 ft.
Show that the function $\text{\hspace{0.17em}}f\left(x\right)=3{\left(x-5\right)}^{2}+7\text{\hspace{0.17em}}$ is not one-to-one.
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
|
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Get the free Identify Functions worksheet and other resources for teaching & understanding solving Identify Functions
## How to Determine Functions the Easy Way
In order to Identify Functions from a Graph and Table you must first understand what a function is. A function is a mathematical relationship where every input has one, and only one, output. In other words, all inputs must have exactly one output. If there is more then one output for one input, then the relationship is not a function. When Identifying Functions from a Graph you must look at the graph to determine if each x-value only has one y-value associated with it. An easy way to tell if a graph is a function is to see if it passes the Vertical Line Test. If you can draw a vertical line anywhere on the grid and it crosses the equation in more then one place then it does not pass the test and is not a function. When Identifying Functions from a Table you need to determine if each x-value has only one y-value associated with it. if there is more then one y-value associated with any x-value, then it is not a function.
Common Core Standard: 8.F.A.1
Basic Topics:
## The Concise Function Definition in Math
To answer how to determine Functions from a Graph and Table you need to know that a function is a relationship where each input has one, and just one, output. In the event that there are two outputs for one input, then that relationship is not a function. When Identifying Functions from a Graph you should look at the graph to decide whether every x-value has only one y-value related with it. A simple method to tell if a diagram is a capacity is to check whether it finishes the Vertical Line Test. You can draw a vertical line anyplace on the graph and if it crosses the equation more then once, then it is not a function. When finding how to determine Functions from a Table you have to decide whether every x-value has just a single y-value related with it. Check to make sure each x has only one y. This is the simplest way for finding how to identify a function.
## 3 Steps for Identifying all Relations and Functions Worksheet Problems
1. Functions have one output for each input.
2. If you are given a graph, you can use the vertical line test to see if the equation is a function.
3. If you are given a table, you must check each x value to make sure there is only one y values associated with it.
## Identifying Functions Practice Problems
/5
Triangle Sum Theorem Quiz
Click Start to begin the practice quiz!
1 / 5
Find x for the missing angle measure of the following triangle:
Angle 1: 23 degrees
Angle 2: 132 degrees
Angle 3: x degrees
2 / 5
Find x for the missing angle measure of the following triangle:
Angle 1: 67 degrees
Angle 2: x degrees
Angle 3: 98 degrees
3 / 5
Find x for the missing angle measure of the following triangle:
Angle 1: 45 degrees
Angle 2: 60 degrees
Angle 3: x degrees
4 / 5
Find x for the missing angle measure of the following triangle:
Angle 1: x degrees
Angle 2: 57 degrees
Angle 3: 117 degrees
5 / 5
Find x for the missing angle measure of the following triangle:
Angle 1: 85 degrees
Angle 2: 20 degrees
Angle 3: x degrees
0%
## What our Video Teaching how to Identify Functions
Watch our free video on how to Determine Functions. This video shows how to solve problems that are on our free Relations and Functions worksheet that you can get by submitting your email above. This function or not a functions worksheet will help you figure out how to determine a function from a table or graph.
Watch the free Identify Functions video on YouTube here: How to Identify Functions
Video Transcript:
This video is about how to identify functions. You can get the relations and functions worksheet on how to identify functions for free by clicking on the link in the description below. Questions about functions will ask you to identify the function of a problem and if it is a function or not a function.
In order to identify functions you have to first know what a function is. A function is a special relationship in math where each input into an equation or set of data has one and only one output. In order to check to see if a graph represents a function you can do something called the vertical line test. Now the vertical line test is when you draw an imaginary vertical line on the graph anywhere on the graph that you want and if it only hits the graph of your line one time that means it is a function. The vertical line test will help you figure out how to identify a function from a graph. All the problems we are going to try in this video are on our is it a function worksheet that you can download for free by clicking on the right.
If you happen to draw a vertical line on your graph and it hits the graph of your equation in more than one spot, like in this example, we drew vertical line it hit right there and it hit right there. That’d be two and then it hits here and it hits here. That would also be two times. That means that there are two outputs for the same input.
For example when x equals three. Here’s when X is three Y could be either here or here. So our output could be two separate spots. Because of that that would mean that number two does not represent a function.
The next part of understanding how to identify a function from a table. We already told you that every input into a set of data can only have one output. If you look at our first table here the X’s represent the inputs and the Y’s represent the outputs. If we look all of our inputs are negative 2 negative 1 0 and 1 and each of them have their separate and individual outputs. That each input has one and only one output that means that this does represent a function.
In the case of number two we have our inputs here and we also have our outputs which is our Y column. Our input is three for each row. All of our X’s are three. Now if you look out output for three, three could be ten, three could be 15, three could be twenty, and three could be twenty five. The input of three gives us four separate outputs. That means that this is not a function because each input can have one and only one output. You can try the practice problems in this video by downloading our free identifying functions worksheet with answers.
## worksheet
###### Practice makes Perfect.
We have hundreds of math worksheets for you to master.
|
# 1981 IMO Problems/Problem 5
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Three congruent circles have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle and the point $O$ are collinear.
## Solution 1
Let the triangle have vertices $A,B,C$, and sides $a,b,c$, respectively, and let the centers of the circles inscribed in the angles $A,B,C$ be denoted $O_A, O_B, O_C$, respectively.
The triangles $O_A O_B O_C$ and $ABC$ are homothetic, as their corresponding sides are parallel. Furthermore, since $O_A$ lies on the bisector of angle $A$ and similar relations hold for the triangles' other corresponding points, the center of homothety is the incenter of both the triangles. Since $O$ is clearly the circumcenter of $O_A O_B O_C$, $O$ is collinear with the incenter and circumcenter of $ABC$, as desired.
## Solution 2
Suppose 3 congruent circles with centres P,Q,R lie inside ABC and are such that the circle with centre P touches AB & AC and the circle with centre Q touches CA & BC.an R with remaining 2.
Since O lies in all 3 circles, PO=QO=RO. Therefore, O is circumcentre of PQR. let O' be circumcentre of ABC.
Since BC is tangent to the circles with centers Q & R, the lengths of perpendiculars from Q & R, the lengths are equal. therefore, QR//BC,RP//CA,PQ//AB.
Again, since AB and AC both touch circle with centre P. Therefore P is equidistant from AB & AC. Therefore P lies on the internal bisector of angle A. Similarly Q & R lie internal bisectors of angle B and angle C respectively. Therefore, AP,BQ,CR produced meet at incenter I. Since, QR//BC,RP//CA,PQ//AB, it follows that I is also incentre of PQR, I being the centre of homothety. By the property of enlargements, O and O' must be co-linear with I , the centre of enlargement.
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## The way to measure my lot dimension?
Enter your deal with, define your garden on a Google map, and voilà! See your lot dimension on a display screen! You may draw a number of areas for one deal with; yard, entrance, or facet yard; it’ll summarize the sq. footage and present outcomes for every garden.
## The way to Manually Discover Your Property’s Sq. Footage
When you’d prefer to measure your yard the quaint manner, observe these steps:
1. Stroll the size of your garden, figuring that one tempo equals about 3 ft. Do the identical with the width of the garden.
2. Then multiply the size by the width to reach on the whole. Ensure you subtract the sq. footage of your own home and driveway when calculating the full sq. footage of your garden.
3. In case your garden has an irregular form, attempt dividing it up into quite a lot of sq. areas. Calculate the sq. footage for every space, then add the figures collectively to succeed in your whole garden dimension
## Figuring out the Dimension of Space
Step 1: Examine the Space
The primary piece of knowledge you have to know is the form and dimensions of your garden, notably the width and size. It would be best to measure and multiply the realm size occasions the width in ft till the sq. footage is 1,000 sq. ft. and mark off this space with the assistance of a marking instrument like washable paint or objects to tell apart the remedy areas border line.
Step 2: Convert Measurements
In case your yard is rectangular or sq., measure the size and width then multiply collectively (size X width = sq. footage). For instance, in case your yard has a size of 10 ft and width of 8 ft, you’ll multiply 10 by 8 to get 80 sq. ft.
Bear in mind for triangle formed lawns, you’ll measure the size and width, multiply collectively, then divide by two ( size X width / 2 = sq. footage).
In case your garden is a not an ideal rectangle, break the garden into completely different sections to measure extra simply, then add measurements collectively to get the realm’s whole sq. footage.
For remedy areas with flowerbeds and different obstructions within the yard you’ll measure the sq. footage of the article and subtract out of your yards whole sq. footage.
Instance, you’ve a property that’s 12,000 sq. ft. and within the center you’ve a panorama mattress with a size of three ft and width of two ft. You’ll multiply 3 by 2 to get 6 ft, then subtract out of your yard whole.
• 3 ft. X 2 ft. = 6 sq. ft.
• 12,000 sq. ft. – 6 sq. ft. = 11,994 sq. ft.
Subsequently, you’ll deal with an space with 11,994 sq. ft.
## Irregular Shapes
Geometric shapes are widespread in dwelling lawns, however extra typically, lawns are irregularly formed. Lots of right now’s panorama designers use ideas of movement and stability to interrupt up straight strains. This offers the garden a extra pure and aesthetically pleasing look. In some methods, irregularly formed areas are simpler to measure than geometric shapes. There are two simple to grasp strategies for figuring out the realm of irregular shapes. The offset technique is most frequently used to measure lawns of irregular shapes. One other is the common radius technique.
## Convert amongst sq. inch, sq. foot, sq. yard and sq. meter
You possibly can, for instance, carry out your entire measurements in inches or centimeters, calculate space in sq. inches or sq. centimeters then convert your ultimate reply to the unit you want corresponding to sq. ft or sq. meters.
To transform amongst sq. ft, yards and meters use the next conversion elements. For different models use our calculator for area conversions.
• Sq. Toes to Sq. Inches
• multiply ft2 by 144 to get in2
• Sq. Toes to Sq. Yards
• multiply ft2 by 0.11111 to get yd2
• Sq. Toes to Sq. Meters
• multiply ft2 by 0.092903 to get m2
• Sq. Yards to Sq. Toes
• multiply yd2 by 9 to get ft2
• Sq. Yards to Sq. Meters
• multiply yd2 by 0.836127 to get m2
• Sq. Meters to Sq. Inches
• multiply m2 by 1,550 to get in2
• Sq. Meters to Sq. Toes
• multiply m2 by 10.7639 to get ft2
• Sq. Meters to Sq. Yards
• multiply m2 by 1.19599 to get yd2
## Rectangle
The realm of a rectangle is discovered by multiplying the size (l) by the width (w).
Space = (l)(w)
### Instance 1:
Calculate the realm of an oblong garden with a size of fifty ft and a width of 25 ft.
Space = (l)(w)
Space = (50 ft)(25 ft)
Space = 1,250 ft²
### Feedback
1. Cooper says
I by no means would have thought to make use of Google Maps to measure my garden!! That is SUCH a useful tip! Now I don’t need to exit and purchase a kind of measuring wheels!! Thanks a lot!!!
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### 2.6 Proving Statements about Angles
```2.6 Proving Statements about
Angles
Properties of Angle Congruence
Reflexive
For any angle, A <A <A.
Symmetric
If <A <B, then <B <A.
Transitive
If <A <B and <B <C,
then <A <C.
Right Angle Congruence Theorem
• All right angles are congruent.
A
.
X
B
.C
Y
.
.Z
Congruent Supplements Theorem
•
If two angles are supplementary to the same angle,
then they are congruent
– If m<1 + m<2 = 180° and m<2 + m<3 = 180°,
then m<1 = m<3 or 1 3
Congruent Complements Theorem
•
If two angles are complementary to the same
angle, then the two angles are congruent.
– If m<4 + m<5 = 90° and m<5 + m<6 = 90°,
then m<4 = m<6 or 4 6
Linear Pair Postulate
• If two angles form a linear pair, then they are
supplementary.
1
2
m<1 + m<2 = 180°
Example:
• < 1 and < 2 are a linear pair.
If m<1 = 78°, then find m<2.
Vertical Angles Theorem
•
Vertical angles are congruent.
1
4
2
3
1 3 , 2 4
Example
<1 and <2 are complementary angles.
<1 and <3 are vertical angles.
If m<3 = 49°, find m<2.
Proving the Right Angle
Congruence Theorem
Given: Angle 1 and angle 2 are right angles
1 2
Prove:
Statements
Reasons
1 . 1 and 2 right ' s
1. Given
2 . 1 90 and 2 90
2. Def. of right ’s
3 . m 1 m 2
3. Trans. POE
4. 1 2
4. Def. of ’s
Proving the Vertical Angles Theorem
5
6
7
Given: 5 and 6 are a linear pair. 6 and 7 are a
linear pair.
Prove: 5 7
Statements
Reasons
1. 5 and 6 are a linear pair. 6
and 7 are a linear pair.
1. Given
2. 5 and 6 are supplementary.
6 and 7 are supplementary.
2. Linear Pair Postulate
3. 5 7
3. Supplements
Theorem
Solve for x.
Give a reason for each step of the proof. Choose from the list of reasons
given.
Given: 6 7
Prove: 5 8
Plan for Proof:
First show that 5 6 and 7 8.
Then use transitivity to show that 5 8.)
Statements
Reasons
1. 6 7
1. Given
2. 7 8
2. Vertical ’s Theorem
3. 6 8
3. Trans. POC
4. 5 6
4. Vertical ’s Theorem
5. Trans. POC
5. 5 8
```
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# Chapter 19 Volume and Surface Area of Solids RS Aggarwal Solutions Exercise 19A Class 10 Maths
Chapter Name RS Aggarwal Chapter 19 Volume and Surface Area of Solids Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 19BExercise 19CMCQ Related Study NCERT Solutions for Class 10 Maths
### Volume and Surface Area of Solids Exercise 19A Solutions
1. Two cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid.
Solution
Let the length of the side of cube be ‘a’ cm
Volume of each cube = 27 cm3
Volume of cube = a3
∴ a3 = 27 cm3
⇒ a = (27 cm3)1/3c
⇒ a = 3 cm
Length of a side of cube = 3 cm
Since, two cubes are joined and a cuboid is formed so,
Length of cuboid = l = 2a = 2 × 3 cm = 6 cm
Breadth of cuboid = b = a = 3 cm
Height of cuboid = h = a = 3 cm
Surface area of cuboid = 2 × (l × b + b × h + l × h)
∴ Surface area of resulting cuboid = 2 × (6 × 3 + 3 × 3 + 6 × 3) cm2
= 2 × (18 + 9 + 18) cm2
= 2 × 45 cm2
= 90 cm2
So, surface area of resulting cuboid is 90 cm2
2. The volume of a hemisphere is 2425.1/2 cm3. Find its curved surface area.
Solution
Let the radius of hemisphere be r cm
Volume of hemisphere is given by 2/3Ï€r3
Given, volume of hemisphere = 24251/2 cm3
∴ 2/3Ï€r3 = 24251/2
⇒ r3 = 4851 × 1/2 × 3/2 × 7/22
⇒ r3 = 1157.625 cm3
⇒ r = (1157.625)1/3 cm
⇒ r = 10.5 cm
Curved surface Area of hemisphere = 2Ï€r2
Curved surface Area of hemisphere = 2 × 22/7 × (10.5)2 cm2
= 693 cm2
∴ Curved surface area of hemisphere = 693 cm2
3. If the total surface area of a solid hemisphere is 462 cm2, find its volume.
Solution
Let the radius of solid sphere be r cm
Total surface area of solid hemisphere = 3Ï€r2
Given, total surface area of solid hemisphere = 462 cm2
∴ 3Ï€r2 = 462 cm2
⇒ 3 × 22/7 × r2 = 462 cm2
⇒ r2 = 462 × 1/3 × 7/22 cm2 = 49 cm2
⇒ r = 7 cm
Volume of solid hemisphere = 4/3.Ï€r3
= 2/3 × 22/7 × 73 cm3
= 718.67 cm3
∴ Volume of solid hemisphere is 718.67 cm3
4. A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of Rs 25 per metre.
Solution
Width of cloth used = 5 m
Diameter of conical tent to be made = 14 m
Let the radius of the conical tent be r m
Radius of conical tent = r = diameter ÷ 2 = 142/2 m = 7 m
Height of conical tent = h = 24 m
Let the slant height of conical tent be l
Area of cloth required to make a conical tent = Curved surface area of conical tent
= πrl
= 22/7 × 7 × 25 m2
= 550 m2
Length of cloth used = Area of cloth required ÷ Width of cloth
= 550/5 m
= 110 m
∴ Length of cloth used = 110 m
Cost of cloth used = Rs 25 per meter
Total Cost of cloth required to make a conical tent = 110 × Rs 25
= Rs 2750
∴ Total cost of cloth required to make a conical tent = 2750
5. If the volume of two cones are in the ratio of 1 : 4 and their diameters are in the ratio 4 : 5, find the ratio of their heights.
Solution
Let V1 be the volume of first cone and V2 be the volume of second cone.
Then, V1 : V2 = 1 : 4
Let d1 be the diameter of first cone and d2 be the diameter of second cone.
Then d1 : d2 = 4 : 5
Let h1 be the height of first cone and h2 be the height of second cone.
We know that volume of cone is given by V = 1/3 × Ï€(d2/4)h
V1/V2 = 1/4
⇒ 16/25 × h1/h2 = 1/4
⇒ h1/h2 = 1/4 × 25/26
⇒ h1/h2 = 25/64
∴ h1/h2 = 25 : 64
∴ Ratio of height of two cones is 25 : 64.
6. The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km2. Find the height of mountain.
Solution
Let the radius of base be ‘r’ km and slant height be ‘l’ km
Slant height of conical mountain = 2.5 km
Area of its base = 1. 54 km2
Area of base is given by πr2
∴ Ï€r2 = 1.54 km2
⇒ 22/7 × r2 = 1.54 km2
⇒ r2 = 1.54 × 7/22 km2 = .49 km2
⇒ r = 0.7 km
Let ‘h’ be the height of the mountain
We know,
l2 = r2 + h2
Substituting the values of l and r in the above equation
2.52 = 0.72 + h2
h2 = 2.52 – 0.72
= 6.25 – 0.49 km2
⇒ h2 = 5.76 km2
⇒ h = 2.4 km
∴ Height of the mountain = 2.4 km
7. The sum of the radius of the base and the height of a solid cylinder is 37 metres. If the total surface area of the cylinder be 1628 sq metres, find its volume.
Solution
Let the radius of the solid cylinder be ‘r’ m and its height be ‘h’ m.
Given,
Sum of radius and height of solid cylinder = 37 m
r + h = 37 m
⇒ r = 37 – h
Total surface area of solid cylinder = 1628 m2
Total surface area of solid cylinder is given by 2Ï€r (h + r)
∴ 2Ï€r (h + r) = 1628 m2
Substituting the value of r + h in the above equation
⇒ 2Ï€r × 37 = 1628 m2
⇒ r = 1628 × 7/22 × 1/2 × 1/37 m
⇒ r = 7 m
Since, r + h = 37 m
h = 37 – r m
⇒ h = 37 – 7 m = 30 m
Volume of solid cylinder = πr2h
= 22/7 × 72 × 30 m2
= 4620 m2
8. The surface area of a sphere is 2464 cm2. If the radius be doubled, what will be the surface area of the new sphere?
Solution
Let the radius of sphere ne ‘r’ cm
Surface area of sphere = 2464 cm2
Surface area of sphere is given by 4Ï€r2
∴ 4Ï€r2 = 2664 cm2
⇒ 4 × 22/7 × r2 = 2464 cm2
⇒ r2 = 2464 × 1/4 × 7/22 cm2 = 196 cm2
⇒ r = 14 cm
Let the radius of new sphere be ‘r’ cm
∴ r’ = 2r
r’ = 2 × 14 cm = 28 cm
Surface area of new sphere = 4Ï€r’2
= 4 × 22/7 × 282 cm2
= 9856 cm2
∴ Surface area of new sphere is 9856 cm2.
9. A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of a same base radius. Find the length of canvas used in making the tent, if the breadth of the canvas is 1.5 m.
Solution
The military tent is made as a combination of right circular cylinder and right circular cone oh top.
Total Height of tent = h = 8.25 m
Base diameter of tent = 30 m
Base radius of tent = r = 30/2 = 15 m
Height of right circular cylinder = 5.5 m
Curved surface area of right circular cylindrical part of tent = 2Ï€rh
Height of conical part = total height of tent – height of cylindrical part
hcone = 8.25 – 5.5 m = 2.75 m
Base radius of cone = 15 m
Let l be the slant height of cone
Then, l2 = hcone2 + r2 = 2.752 + 152 m2
⇒ l2 = 7.5625 + 225 m2 = 232.5625
⇒ l = 15.25
Curved surface area of conical part of the tent = πrl
Total surface area of the tent = Curved surface area of cylindrical part + curved surface area of conical part
Total surface area of tent = 2πrh + πrl
= πr(2h + l)
= 22/7 × 15 × (2 × 5.5 + 15.25) m2
= 22/7 × 15 × (11 + 15.25) m2
= 22/7 × 15 × 26.25 m2
= 1237.5 m2
Breadth of canvas used = 1.5 m
Length of canvas used = Total surface are of tent ÷ breadth of canvas used
Length of canvas used = (1237.5)/(1.5)m = 825 m
∴ Length of canvas used is 825 m
10. A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre. [Take π = 22/7.]
Solution
The tent is made as a combination of right circular cylinder and right circular cone on top.
Height of cylindrical part of the tent = h = 3 m
Radius of its base = r = 14 m
Curved surface area of cylindrical part of tent = 2Ï€rh
= 2 × 22/7 × 14 × 3 m2
= 264 m2
Total height of the tent = 13.5 m
Height of conical part of the tent = 13.5 – 3 m = 10.5 m
Let the slant height of the conical part be l
l2 = hcone2 + r2
⇒ l2 = 10.52 + 142 = 110.25 + 196 m2 = 306.25 m2
⇒ l = 17.5 m
Curved surface area of conical part of tent = πrl
= 22/7 × 14 × 17.5 m2
= 770 m2
Total surface area of tent = Curved surface area of cylindrical part of tent + Curved surface area of conical part of tent
Total surface area of tent = 264 m2 + 770 m2 = 1034 m2
Cloth required = Total Surface area of tent = 1034 m2
Cost of cloth = Rs 80/m2
Total cost of cloth required = Total surface area of tent × Cost of cloth
= 1034 × Rs 80
= Rs 82720
Cost of cloth required to make the tent is Rs 82720.
11. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical potion is 53 m, find the area of canvas needed to make the tent. [Take π = 22/7]
Solution
The Circus tent is made as a combination of cylinder and cone on top.
Height of cylindrical part of tent = h = 3 m
Base radius of tent = r = 52.5 m
Area of canvas required for cylindrical part of tent = 2Ï€rh
= 2 × 22/7 × 52.5 × 3 m2
= 990 m2
Slant height of cone = l = 53 m
Area of canvas required for conical part of the tent = πrl
= 22/7 × 52.5 × 53 m2
= 8745 m2
Area of canvas required to make the tent = Area of canvas required for cylindrical part of tent + Area of canvas required for conical part of tent
Are of canvas required to make the tent = 990 + 8745 m2
= 9735 m2
12. A rocket is in the form of a circular cylinder closed at the lower and a cone of the same radius is attached to the top. The radius of the cylinder is 2.5 m, its height is 21 m and the slant height of the cone is 8 m. Calculate the total surface area of the rocket.
Solution
The rocket is in the form of cylinder closed at the bottom and cone on top.
Height of cylindrical part rocket = h = 21 m
Base radius of rocket = r = 2.5 m
Surface area of cylindrical part of rocket = 2πrh + πr2
= 2 × 22/7 × 2.5 × 21 + 22/7 × 2.5 × 2.5 m2
= 330 + 19.64 m2 = 349.64 m2
Slant height of cone = l = 8 m
Surface area of conical part of the rocket = πrl
= 22/7 × 2.5 × 8 m2
= 62.86 m2
Total surface area of the rocket = Surface Area of cylindrical part of rocket + surface area of conical part of rocket
Total surface area of the rocket = 349.64 + 62.86 m2
= 412.5 m2
13. A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid.
Solution
The solid is in the form of a cone surmounted on a hemisphere.
Total height of solid = h = 9.5 m
Radius of solid = r = 3.5 m
Volume of hemispherical part solid = 2/3 × Ï€r3
= 2/3 × 22/7 × 3.53 m3
= 89.83 m3
Height of conical part of solid = hcone = Total height of solid – Radius of solid
Height of conical part of solid = hcone = 9.5 – 3.5 = 6 m
Volume of conical part of solid = 1/3 × Ï€r2h cone
= 1/3 × 22/7 × 3.52 × 6 m3
= 77 m3
Volume of solid = Volume of hemispherical part solid + Volume of conical part solid
Volume of solid = 89.83 + 77 m3 = 166.83 m3
14. A toy is in the form of a cone mounted on a hemisphere of same radius 7 cm. If the total height of the toy is 31 cm, find its total surface area.
Solution
The toy is in the form of a cone mounted on a hemisphere.
Total height of toy = h = 31 cm
Radius of toy = r = 7 cm
Surface area of hemispherical part toy = 2Ï€r2
= 2 × 22/7 × 72 cm2
= 308 cm2
Height of conical part of toy = hcone = Total height of toy – Radius of toy
Height of conical part of toy = hcone = 31 – 7 = 24 cm
Let the slant height of the conical part be l
l2 = hcone2 + r2
⇒ l2 = 242 + 72 = 576 + 49 cm2 = 625 cm2
⇒ l = 25 cm
Surface area of conical part of toy = πrl
= 22/7 × 7 × 25 cm2
= 550 cm2
Total surface area of toy = Surface area of hemispherical part of toy + Surface area of conical part of toy
Total surface area of toy = 308 cm2 + 550 cm2
= 858 cm2
15. A toy is in the shape of a cone mounted on a hemisphere of same base radius. If the volume of the toy is 231 cm3 and its diameter is 7 cm, find the height of the toy.
Solution
A toy is in the shape of a cone mounted on a hemisphere of same base radius.
Volume of Toy = 231 cm3
Base diameter of toy = 7 cm
Base radius of toy = 7/2 cm = 3.5 cm
Volume of hemisphere = 2/3Ï€r3
= 2/3 × 22/7 × 3.53 cm3
= (2 × 22 × 35 × 35 × 35)/(3 × 7 × 10 × 10 × 10)
= 539/6
Volume of cone = Volume of hemisphere – Volume of toy
= 231 – 539/6
= (1386 – 539)/6
= 847/6
Volume of cone is given by 1/3Ï€r2h
Where h is the height of cone
⇒ 1/3Ï€r2h = 847/6
⇒ 22/7(7/2)2h = 847/2
⇒ 77/2.h = 747/2
⇒ h = 11
Height of cone = 11 cm
Height of toy = Height of cone + Height of hemisphere
= 11 cm + 3.5 cm
= 14.5 cm
[Height of hemisphere = Radius of hemisphere]
∴ Height of toy is 14.5 cm.
16. A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, find the radius of the ice-cream cone.
Solution
Radius of cylindrical container = r = 6 cm
Height of cylindrical container = h = 15 cm
Volume of cylindrical container = πr2h
= 22/7 × 6 × 6 × 15 cm3
= 1697.14 cm3
Whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops.
Let the radius of hemisphere and base of cone be r’
Height of cone = h = 4 times the radius of its base
h’ = 4r’
Volume of Hemisphere = 2/3Ï€(r’)3
Volume of cone = 1/3Ï€(r’)2h’ = 1/3Ï€(r’)2 × 4r’
= 2/3 Ï€(r’)3
Volume of ice-cream = Volume of Hemisphere + Volume of cone
= 2/3Ï€(r’)3 + 4/3Ï€(r’)3 = 6/3Ï€(r’)3
Number of ice-creams = 10
∴ Total volume of ice-cream = 10 × Volume of ice-cream
= 10 × 6/3 Ï€(r’)3
= 60/3 Ï€(r’)3
Also, total volume of ice cream = Volume of cylindrical container
⇒ 60/3 Ï€(r’)3 = 1697.14 cm3
⇒ 60/3 × 22/7 × (r’)3 = 1697.14 cm3
⇒ (r’)3 = 1697.14 × 3/60 × 7/22 = 27 cm3
⇒ r = 3 cm
∴ Radius of ice-cream cone = 3 cm
17. A vessel is in the form of a hemispherical bowl surmounted by a hallow cylinder. The diameter of the hemispherical is 21 cm and the total height of the vessel is 14.5 cm. Find its capacity.
Solution
Vessel is in the form of hemispherical bowl surmounted by a hallow cylinder.
Diameter of hemisphere = 21 cm
Radius of hemisphere = 10.5 cm
Volume of hemisphere = 2/3Ï€r3 = 2/3 × 22/7 × 10.53 cm3
= 2425.5 cm3
Total height of vessel = 14.5 cm
Height of cylinder = h = Total height of vessel – Radius of hemisphere
= 14.5 – 10.5
= 4 cm
Volume of cylinder = Ï€r2h = 22/7 × 10.5 × 10.5 × 4 cm3
= 1386 cm3
Volume of vessel = Volume of hemisphere + Volume of cylinder
= 2425.5 cm3 + 1386 cm3
= 3811.5 cm3
∴ Capacity of vessel = 3811.5 cm3
18. A toy is in the form of a cylinder with hemisphere ends. If the whole length of the toy is 90 cm and its diameter is 42 cm, find the cost of painting the toy at the rate of 70 paise per sq cm.
Solution
Toy is in the form of a cylinder with hemisphere ends
Total length of toy = 90 cm
Diameter of toy = 42 cm
Radius of toy = r = 21 cm
Length of cylinder = l = Total length of toy – 2 × Radius of toy
= 90 – 2 × 21 cm
= 48 cm
For cost of painting we need to find out the curved surface area of toy
Curved surface area of cylinder = 2Ï€rl
= 2 × 22/7 × 21 × 48 cm2
= 6336 cm2
Curved surface area of hemispherical ends = 2 × 2Ï€r2
= 2 × 2 × 22/7 × 21 × 21 cm2
= 5544 cm2
Surface area of toy = Curved surface area of cylinder + Curved surface area of hemispherical ends
Surface area of toy = 6336 cm2 + 5544 cm2 = 11880 cm2
Cost of painting = Rs (0.70) cm2
Total cost of painting = Surface area of toy × Cost of painting
= 11880 cm2 × Rs 0.70 cm2 = Rs 8316.00
Total cost of painting the toy = Rs 8316.00
19. A medicine capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution
A medicine capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends.
Total length of entire capsule = 14 mm
Diameter of capsule = 5 mm
Radius of capsule = r = Diameter ÷ 2 = 5/2 mm = 2.5 mm
Length of cylindrical part of capsule = l = Total length of entire capsule – 2 × Radius of capsule
= 14 – 2 × 2.5 mm
= 14 – 5 mm
= 9 mm
Curved surface area of cylindrical part of capsule = 2Ï€rl
= 2 × 3.14 × 9 × 2.5 mm2
= 141.3 mm2
Curved surface area of hemispherical ends = 2 × 2Ï€r2
= 2 × 2 × 3.14 × 2.5 × 2.5 cm2
= 78.5 cm2
Surface area of capsule = Curved surface area of cylindrical part of capsule + Curved surface area of hemispherical ends
Surface area of capsule = 141.3 mm2 + 78.5 mm2 = 219.8 mm2
20. A wooden article was made by scooting out a hemisphere from each end of cylinder, as shown in the figure. If the height of the cylinder is 20 cm and its base is of diameter 7 cm, find the total surface area of the article when it is ready.
Solution
The wooden article was made by scooting out a hemisphere from each end of a cylinder
Let the radius of cylinder be r cm and height be h cm.
Height of cylinder = h = 20 cm
Base diameter of cylinder = 7 cm
Base radius of cylinder = r = diameter ÷ 2 = 7/2 cm = 3.5 cm
Lateral Surface area of cylinder = 2Ï€rh
= 2 × 22/7 × 3.5 × 20 cm2
= 2 × 22/7 × 3.5 × 20 cm2
= 440 cm2
Since, the wooden article was made by scooting out a hemisphere from each end of a cylinder
∴ Two hemisphere are taken out in total
∴ Radius of hemisphere = 3.5 cm
Lateral Surface area of hemisphere = 2Ï€r2
= 2 × 22/7 × 3.5 × 3.5 cm2
= 77 cm2
Total Surface area of two hemispheres = 2 × 77 cm2 = 154 cm2
Total surface area of the article when it is ready = Lateral Surface area of cylinder + Lateral Surface area of hemisphere
Total surface area of the article when it is ready = 440 cm2 + 154 cm2
= 594 cm2
21. solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 2.1 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water. If the radius of the cylinder is 5 cm and its height is 9.8 cm, find the volume of the water left in the tub.
Solution
A solid is in the form of a right circular cone mounted on a hemisphere
Let r be the radius of hemisphere and cone
Let h be the height of the cone
= 2/3 × 22/7 × 2.1 × 2.1 × 2.1 cm3
= 19.404 cm3
Height of cone = h = 4 cm
Radius of cone = r = 2.1 cm
Volume of cone = 1/3Ï€r2h
= 1/3 × 22/7 × 2.1 × 2.1 × 4 cm3
= 18.48 cm3
Volume of solid = Volume of hemisphere + Volume of cone
= 19.404 cm3 + 18.48 cm3
= 37.884 cm3
The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water, so, to find the volume of water left in the tub we need to subtract volume of solid from cylindrical tub.
Radius of cylinder = r’ = 5 cm
Height of cylinder = h’ = 9.8 cm
Volume of cylindrical tub = Ï€r’2h = 22/7 × 5 × 5 × 9.8 cm3
= 770 cm3
Volume of water left in the tub = Volume of cylindrical tub – Volume of solid
Volume of water left in the tub = 770 cm3 – 37.884 cm3
= 732.116 cm3
∴ Volume of water left in the tub is 732.116 cm3
22. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid. [Take π = 3.14]
Solution
Height of solid cylinder = h = 8 cm
Radius of solid cylinder = r = 6 cm
Volume of solid cylinder = πr2h
= 3.14 × 6 × 6 × 8 cm3
= 904.32 cm3
Curved Surface area of solid cylinder = 2Ï€rh
Height of conical cavity = h = 8 cm
Radius conical cavity = r = 6 cm
Volume of conical cavity = 1/3Ï€r2h
= 1/3 × 3.14 × 6 × 6 × 8 cm3
= 301.44 cm3
Let l be the slant height of conical cavity
l2 = r2 + h2
⇒ l2 = (62 + 82) cm2
⇒ l2 = (36 + 64) cm2
⇒ l2 = 100 cm2
⇒ l = 10 cm
Curved surface area of conical cavity = πrl
Since, conical cavity is hollowed out from solid cylinder, So, volume and total surface area of remaining solid will be found out by subtracting volume and total surface area of conical cavity from volume and total surface area of solid cylinder.
Volume of remaining solid = Volume of solid cylinder – Volume of conical cavity
Volume of remaining solid = 904.32 cm3 – 301.44 cm3
= 602.88 cm3
Total surface area of remaining solid = Curved surface area of solid cylinder + Curved surface area of conical cavity + Area of circular base
Total surface area of remaining solid = 2πrh + πrl + πr2
= Ï€r × (2h + l + r)
= 3.14 × 6 × (2 × 8 + 10 + 6) cm2
= 3.14 × 6 × 32 cm2
= 602.88 cm2
23. From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
Solution
Height of solid cylinder = h = 2.8 cm
Diameter of solid cylinder = 4.2 cm
Radius of solid cylinder = r = Diameter ÷ 2 .2 cm
Curved surface area of solid cylinder = 2Ï€rh
= 2 × 22/7 × 2.1 × 2.8 cm2
= 2 × 22/7 × 2.1 × 2.8 cm2
= 36.96 cm2
Height of conical cavity = h = 2.8 cm
Radius conical cavity = r = 2.1 cm
Let l be the slant height of conical cavity
l2 = r2 + h2
⇒ l2 = (2.82 + 2.12) cm2
⇒ l2 = (7.84 + 4.41) cm2
⇒ l2 = 12.25 cm2
⇒ l = 3.5 cm
Curved Surface area of conical cavity = πrl
= 22/7 × 2.1 × 3.5
= 23.1 cm2
Total surface area of remaining = Curved surface area of solid cylinder + Curved surface area of conical cavity + Area of circular base
Total surface area of remaining solid = (36.96 + 23.1 + 22/7 × 2.12) cm2
= (36.96 + 23.1 + 13.86) cm2
= 73.92 cm2
24. From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid.
Solution
Height of solid cylinder = h = 14 cm
Diameter of solid cylinder = 7 cm
Radius of solid cylinder = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm
Volume of solid cylinder = πr2h
= 22/7 × 3.5 × 3.5 × 14 cm3
= 539 cm3
Height of conical cavity = h’ = 4 cm
Radius conical cavity = r’ = 2.1 cm
Volume of conical cavity = 1/3Ï€r’2h’
= 1/3 × 22/7 × 2.1 × 2.1 × 4 cm3
= 18.48 cm3
Since, there are two conical cavities
∴ Volume of two conical cavities = 2 × 18.48 cm3 = 36.96 cm3
Volume of remaining solid = Volume of solid cylinder – Volume of two conical cavity
Volume of remaining solid = 539 cm3 – 36.96 cm3
= 502.04 cm3
25. A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of 3/2 cm and its depth is 8/9 cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.
Solution
Height of metallic cylinder = h = 5 cm
Radius of metallic cylinder = r = 3 cm
Volume of solid cylinder = πr2h
= Ï€ × 3 × 3 × 5 cm3
= 45Ï€ cm3
Height of conical hole = h’ = 8/9 cm.
Radius conical hole = r’ = 3/2 cm
Volume of conical hole = 1/3Ï€r’2h’
= 1/3 × Ï€ × 3/2 × 3/2 × 8/9 cm3
= 2/3Ï€ cm3
Volume of metal left in cylinder = Volume of metallic cylinder – Volume of conical hole
Volume of metal left in cylinder = 45Ï€ – 2/3Ï€ = 133Ï€/3
Ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape = Volume of metal; left in cylinder/Volume of conical hole
Volume of metal left in cylinder: Volume of conical hole = 133Ï€/3 : 2/3Ï€
Volume of metal left in cylinder: Volume of conical hole = 133 : 2
Ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 339: 4
26. A spherical glass vessel has a cylindrical neck 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water it can hold. [Use π= 22/7]
Solution
Length of cylindrical neck = l = 7 cm
Diameter of cylindrical neck = 4 cm
Radius of cylindrical neck = r = Diameter ÷ 2 = 4/2 cm = 2 cm
Volume of cylindrical neck = πr2l
= 22/7 × 2 × 2 × 7 cm3
= 88 cm3
Diameter of spherical part = 21 cm
Radius of spherical part = r’ = Diameter ÷ 2 = 21/2 cm = 10.5 cm
Volume of spherical part = 4/3Ï€r’3
= 4/3 × 22/7 × 10.5 × 10.5 × 10.5 cm3
= 4851 cm3
Quantity of water spherical glass vessel with cylindrical neck can hold = Volume of spherical part + Volume of cylindrical neck
Quantity of water spherical glass vessel with cylindrical neck can hold = 4851 cm3 + 88 cm3 = 4939 cm3
Quantity of water spherical glass vessel with cylindrical neck can hold is 4939 cm3.
27. The given figure represent a solid consisting of cylinder surmounted by a cone at one end a hemisphere at the other. Find the volume of the solid.
Solution
The solid consisting of a cylinder surmounted by a cone at one end a hemisphere at the other.
Length of cylinder = l = 6.5 cm
Diameter of cylinder = 7 cm
Radius of cylinder = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm
Volume of cylinder = πr2l
= 22/7 × 3.5 × 3.5 × 6.5 cm3
= 250.25 cm3
Length of cone = l’ = 12.8 cm – 6.5 cm
= 6.3 cm
Diameter of cone = 7 cm
Radius of cone = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm
Volume of cone = 1/3Ï€r2’l
= 1/3 × 22/7 × 3.5 × 3.5 × 6.3 cm3
= 80.85 cm3
Diameter of hemisphere = 7 cm
Radius of hemisphere = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm
Volume of hemisphere = 2/3Ï€r3
= 2/3 × 22/7 × 3.5 × 3.5 × 3.5 cm3
= 89.83 cm3
Volume of the solid = Volume of cylinder + Volume of cone + Volume of hemisphere
Volume of solid = 250.25 cm3 + 80.85 cm3 + 89.83 cm3
= 420.93 cm3
28. From a cubical piece of wood of side 21 cm, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece. Find the surface area and volume of the remaining piece.
Solution
Length of cubical piece of woof = a = 21 cm
Volume of cubical piece of wood = a3
= 21 × 21 × 21 cm3
= 9261 cm3
Surface area of cubical piece of wood = 6a2
= 6 × 21 × 21 cm2
= 2646 cm2
Since, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece.
So, diameter of hemisphere = length of side of the cubical piece
Diameter of hemisphere = 21 cm
Radius of hemisphere = r = Diameter ÷ 2 = 21/2 cm = 10.5 cm
Volume of hemisphere = 2/3Ï€r3
= 2/3 × 22/7 × 10.5 × 10.5 × 10.5 cm3
= 2425.5 cm3
Surface area of hemisphere = 2Ï€r2
= 2 × 22/7 × 10.5 × 10.5 cm2
= 693 cm2
A hemisphere is carved out from cubical piece of wood
Volume of remaining solid = Volume of cubical piece of wood – Volume of hemisphere
Volume of remaining solid = 9261 cm3 – 2425.5 cm3 = 6835.5 cm3
Surface area remaining piece of solid = Surface area of cubical piece of wood – Area of circular base of hemisphere + Curved Surface area of hemisphere
Surface area remaining piece of solid = 6a2 – Ï€r2 + 2Ï€r2
= (2646 – 22/7 × 10.5 2 + 693) cm2
= 2992.5 cm2
29. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs 5 per sq cm. [use π = 3.14]
Solution
Length of side of cubical block = a = 10 cm
Since a cubical block is surmounted by a hemisphere, so, the largest diameter of hemisphere = 10 cm
Since, hemisphere will be touching the sides of cubical block.
Radius of hemisphere = r = Diameter ÷ 2 = 10/2 cm = 5 cm
Surface area of solid = Surface area of cube – Area of circular part of hemisphere + Curved surface area of hemisphere
Total surface area of solid = 6a2 – Ï€r2 + 2Ï€r2 = 6a2 + Ï€r2
= 6 × 10 × 10 cm2 + 3.14 × 5.5 cm2
= 678.5 cm2
Rate of painting = Rs 5/100 cm2
Cost of painting the total surface area of the solid so formed = Total Surface area of solid × Rate of painting
Cost of painting the total surface area of the solid = Rs 5/100 × 678.5
= Rs 33.925
30. A toy is in the shape of a right circular cylinder with a hemisphere on one end a cone on the other. The radius and height of the cylindrical part area 5 cm and 13 cm respectively. The radii of the hemispherical and the conical parts are the same as that of the cylindrical part. Find the surface area of the toy, if the total height of the toy is 30 cm.
Solution
The toy is in the shape of a right circular cylinder surmounted by a cone at end a hemisphere at the other.
Total height of toy = 30 cm
Height of cylinder = h = 13 cm
Radius of cylinder = r = 5 cm
Curved surface area of cylinder = 2Ï€rh
= 2 × 22/7 × 5 × 13 cm2
Height of cone = h’ = total height of toy – Height of cylinder – Radius of hemisphere
Height of cone = h’ = 30 cm – 13 cm – 5 cm = 12 cm
Radius of cone = r = 5 cm
Let the slant height of cone be l
l2 = h’2 + r2
⇒ l2 = 122 + 52 cm2 = 144 + 25 cm2 = 169 cm2
⇒ l = 13 cm
Curved surface area of cone = πrl
= 22/7 × 5 × 13 cm2
Radius of hemisphere = r = 5 cm
Curved surface area of hemisphere = 2Ï€r2
= 2 × 22/7 × 5 × 5 cm2
Surface area of the toy = Surface area of cylinder + Surface area of cone + Surface area of hemisphere
Surface area of toy = 2πrrh + πrrl + 2πr2
= πr(2h + l + 2r)
= 22/7 × 5 × (2 × 13 + 13 + 2 × 5) cm2
= 22/7 × 5 × 49 cm2
= 770 cm2
Surface area of toy is 770 cm2
31. The inner diameter of a glass is 7 cm and it has a raised portion in the bottom in the shape of a hemisphere, as shown in the figure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass.
Solution
Inner diameter of a glass = 7 cm
Inner radius of glass = r = 7/2 cm = 3.5 cm
Height of glass = h = 16 cm
Apparent capacity of glass = πr2h
= 22/7 × 3.5 × 3.5 × 16 cm3
= 616 cm3
Volume of the hemisphere in the bottom = 2/3Ï€r3
= 2/3 × 22/7 × 3.53 cm3
= 89.83 cm3
Actual capacity of the class=Apparent capacity of glass – Volume of the hemisphere
Actual capacity of the class=616 cm3 – 89.83 cm3 = 526.17 cm3
32. A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure. The total height of the toy is 26 cm, while the height of the conical part is 6 cm. The diameter of the base of the conical part is 5 cm and that of the cylindrical part in 4 cm. the conical part and the cylindrical part are respectively painted red and white. Find the area to be painted by each of them colours. [Take π = 22/7]
Solution
The wooden toy is in the shape of a cone mounted on a cylinder
Total height of the toy = 26 cm
Height of conical part = H = 6 cm
Height of cylindrical part = Total height of the toy – Height of conical part
h = 26 cm – 6 cm = 20 cm
Diameter of conical part = 5 cm
Radius of conical part = R = Diameter/2 = 5/2 cm = 2.5 cm
Let L be the slant height of the cone
L2 = H2 + R2
⇒ L2 = 62 + 2.52 cm2 = 36 + 6.25 cm2 = 42.25 cm2
⇒ L = 6.5 cm
Diameter of cylindrical part = 4 cm
Radius of cylindrical part = r = Diameter/2 = 4/2 cm = 2 cm
Area to be painted red = Curved surface area of cone + Base are of cone – base area of cylinder
Area to be painted red = Ï€RL + Ï€R2 – Ï€r2 = Ï€(RL + R2 – r2)
= 22/7 × (2.5 × 6.5 + 2.5 × 2.5 – 2 × 2) cm2
= 22/7 × (16.25 + 6.25 – 4) cm2
= 22/7 × 18.5 cm2
= 58.143 cm2
Area to be painted white = Curved Surface area of cylinder + Base area of cylinder
Area to be painted White = 2πrh + πr2 = πr(2h + r)
= 22/7 × 2 × (2 × 20 + 2) cm2
= 22/7 × 2 × (40 + 2) cm2
= 22/7 × 2 × 42 cm2
= 264 cm2
∴ Area to be painted red is 58.143 cm2 and area to be painted white is 264 cm2.
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com
Class XII
## Chapter 3 Current Electricity
Physics
Question 3.1:
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery
is 0.4, what is the maximum current that can be drawn from the battery?
Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4
Maximum current drawn from the battery = I
According to Ohms law,
## The maximum current drawn from the given battery is 30 A.
Question 3.2:
A battery of emf 10 V and internal resistance 3 is connected to a resistor. If the
current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal
voltage of the battery when the circuit is closed?
Emf of the battery, E = 10 V
Internal resistance of the battery, r = 3
Current in the circuit, I = 0.5 A
Resistance of the resistor = R
The relation for current using Ohms law is,
Page 1 of 26
Class XII
Physics
## According to Ohms law,
V = IR
= 0.5 17
= 8.5 V
Therefore, the resistance of the resistor is 17 and the terminal voltage is
8.5 V.
Question 3.3:
(a) Three resistors 1 , 2 , and 3 are combined in series. What is the total
resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal
resistance, obtain the potential drop across each resistor.
(a) Three resistors of resistances 1 , 2 , and 3 are combined in series. Total
resistance of the combination is given by the algebraic sum of individual resistances.
Total resistance = 1 + 2 + 3 = 6
(b) Current flowing through the circuit = I
Emf of the battery, E = 12 V
Total resistance of the circuit, R = 6
The relation for current using Ohms law is,
## Potential drop across 1 resistor = V1
From Ohms law, the value of V1 can be obtained as
V1 = 2 1= 2 V (i)
Potential drop across 2 resistor = V2
Again, from Ohms law, the value of V2 can be obtained as
V2 = 2 2= 4 V (ii)
Potential drop across 3 resistor = V3
Again, from Ohms law, the value of V3 can be obtained as
V3 = 2 3= 6 V (iii)
Page 2 of 26
Class XII
Physics
## Therefore, the potential drop across 1 , 2 , and 3 resistors are 2 V, 4 V, and 6 V
respectively.
Question 3.4:
(a) Three resistors 2 , 4 and 5 are combined in parallel. What is the total
resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal
resistance, determine the current through each resistor, and the total current drawn
from the battery.
(a) There are three resistors of resistances,
R1 = 2 , R2 = 4 , and R3 = 5
They are connected in parallel. Hence, total resistance (R) of the combination is given
by,
## (b) Emf of the battery, V = 20 V
Current (I1) flowing through resistor R1 is given by,
Page 3 of 26
Class XII
Physics
## Current (I3) flowing through resistor R3 is given by,
Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A
Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the
total current is 19 A.
Question 3.5:
At room temperature (27.0 C) the resistance of a heating element is 100 . What is the
temperature of the element if the resistance is found to be 117 , given that the
temperature coefficient of the material of the resistor is
Room temperature, T = 27C
Resistance of the heating element at T, R = 100
Let T1 is the increased temperature of the filament.
Resistance of the heating element at T1, R1 = 117
Temperature co-efficient of the material of the filament,
Page 4 of 26
Class XII
## Chapter 3 Current Electricity
Physics
Question 3.6:
A negligibly small current is passed through a wire of length 15 m and uniform crosssection 6.0 107 m2, and its resistance is measured to be 5.0 . What is the resistivity
of the material at the temperature of the experiment?
Length of the wire, l =15 m
Area of cross-section of the wire, a = 6.0 107 m2
Resistance of the material of the wire, R = 5.0
Resistivity of the material of the wire =
Resistance is related with the resistivity as
## Therefore, the resistivity of the material is 2 107 m.
Question 3.7:
A silver wire has a resistance of 2.1 at 27.5 C, and a resistance of 2.7 at 100 C.
Determine the temperature coefficient of resistivity of silver.
Temperature, T1 = 27.5C
Resistance of the silver wire at T1, R1 = 2.1
Temperature, T2 = 100C
Resistance of the silver wire at T2, R2 = 2.7
Temperature coefficient of silver =
It is related with temperature and resistance as
Page 5 of 26
Class XII
## Chapter 3 Current Electricity
Physics
Question 3.8:
Aheating element using nichrome connected to a 230 V supply draws an initial current of
3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady
temperature of the heating element if the room temperature is 27.0 C? Temperature
coefficient of resistance of nichrome averaged over the temperature range involved is
1.70 104 C
Supply voltage, V = 230 V
Initial current drawn, I1 = 3.2 A
Initial resistance = R1, which is given by the relation,
## Steady state value of the current, I2 = 2.8 A
Resistance at the steady state = R2, which is given as
## Initial temperature of nichrome, T1= 27.0C
Study state temperature reached by nichrome = T2
T2 can be obtained by the relation for ,
Page 6 of 26
Class XII
## Chapter 3 Current Electricity
Physics
Question 3.9:
Determine the current in each branch of the network shown in fig 3.30:
Current flowing through various branches of the circuit is represented in the given figure.
## I1 = Current flowing through the outer circuit
I2 = Current flowing through branch AB
I3 = Current flowing through branch AD
I2 I4 = Current flowing through branch BC
I3 + I4 = Current flowing through branch CD
I4 = Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
10I2 + 5I4 5I3 = 0
2I2 + I4 I3 = 0
Page 7 of 26
Class XII
## Chapter 3 Current Electricity
Physics
I3 = 2I2 + I4 (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I2 I4) 10(I3 + I4) 5I4 = 0
5I2 + 5I4 10I3 10I4 5I4 = 0
5I2 10I3 20I4 = 0
I2 = 2I3 + 4I4 (2)
For the closed circuit ABCFEA, potential is zero i.e.,
10 + 10 (I1) + 10(I2) + 5(I2 I4) = 0
10 = 15I2 + 10I1 5I4
3I2 + 2I1 I4 = 2 (3)
From equations (1) and (2), we obtain
I3 = 2(2I3 + 4I4) + I4
I3 = 4I3 + 8I4 + I4
3I3 = 9I4
3I4 = + I3 (4)
Putting equation (4) in equation (1), we obtain
I3 = 2I2 + I4
4I4 = 2I2
I2 = 2I4 (5)
It is evident from the given figure that,
I1 = I3 + I2 (6)
Putting equation (6) in equation (1), we obtain
3I2 +2(I3 + I2) I4 = 2
5I2 + 2I3 I4 = 2 (7)
Putting equations (4) and (5) in equation (7), we obtain
5(2 I4) + 2( 3 I4) I4 = 2
10I4 6I4 I4 = 2
17I4 = 2
I3 = 3(I4)
Page 8 of 26
Class XII
Physics
## Therefore, current in branch
In branch BC =
In branch CD =
In branch BD =
Total current =
Question 3.10:
(a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the
end A, when the resistor Y is of 12.5 . Determine the resistance of X. Why are the
connections between resistors in a Wheatstone or meter bridge made of thick copper
strips?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
Page 9 of 26
Class XII
## Chapter 3 Current Electricity
Physics
(c) What happens if the galvanometer and cell are interchanged at the balance point of
the bridge? Would the galvanometer show any current?
A metre bridge with resistors X and Y is represented in the given figure.
## (a) Balance point from end A, l1 = 39.5 cm
Resistance of the resistor Y = 12.5
Condition for the balance is given as,
## Therefore, the resistance of resistor X is 8.2 .
The connection between resistors in a Wheatstone or metre bridge is made of thick
copper strips to minimize the resistance, which is not taken into consideration in the
bridge formula.
(b) If X and Y are interchanged, then l1 and 100l1 get interchanged.
The balance point of the bridge will be 100l1 from A.
100l1 = 100 39.5 = 60.5 cm
Therefore, the balance point is 60.5 cm from A.
(c) When the galvanometer and cell are interchanged at the balance point of the bridge,
the galvanometer will show no deflection. Hence, no current would flow through the
galvanometer.
Page 10 of 26
Class XII
## Chapter 3 Current Electricity
Physics
Question 3.11:
A storage battery of emf 8.0 V and internal resistance 0.5 is being charged by a 120 V
dc supply using a series resistor of 15.5 . What is the terminal voltage of the battery
during charging? What is the purpose of having a series resistor in the charging circuit?
Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5
Effective voltage in the circuit = V1
R is connected to the storage battery in series. Hence, it can be written as
V1 = V E
V1 = 120 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
## Voltage across resistor R given by the product, IR = 7 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R
Terminal voltage of battery = 120 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
Question 3.12:
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm
length of the wire. If the cell is replaced by another cell and the balance point shifts to
63.0 cm, what is the emf of the second cell?
Emf of the cell, E1 = 1.25 V
Balance point of the potentiometer, l1= 35 cm
The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm
Page 11 of 26
Class XII
Physics
## Therefore, emf of the second cell is 2.25V.
Question 3.13:
The number density of free electrons in a copper conductor estimated in Example 3.1 is
8.5 1028 m3. How long does an electron take to drift from one end of a wire 3.0 m
long to its other end? The area of cross-section of the wire is 2.0 106 m2 and it is
carrying a current of 3.0 A.
Number density of free electrons in a copper conductor, n = 8.5 1028 m3 Length of
the copper wire, l = 3.0 m
Area of cross-section of the wire, A = 2.0 106 m2
Current carried by the wire, I = 3.0 A, which is given by the relation,
I = nAeVd
Where,
e = Electric charge = 1.6 1019 C
Vd = Drift velocity
Therefore, the time taken by an electron to drift from one end of the wire to the other is
2.7 104 s.
Page 12 of 26
Class XII
## Chapter 3 Current Electricity
Physics
Question 3.14:
The earths surface has a negative surface charge density of 109 C m2. The potential
difference of 400 kV between the top of the atmosphere and the surface results (due to
the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire
globe. If there were no mechanism of sustaining atmospheric electric field, how much
time (roughly) would be required to neutralise the earths surface? (This never happens
in practice because there is a mechanism to replenish electric charges, namely the
continual thunderstorms and lightning in different parts of the globe). (Radius of earth =
6.37 106 m.)
Surface charge density of the earth, = 109 C m2
Current over the entire globe, I = 1800 A
Radius of the earth, r = 6.37 106 m
Surface area of the earth,
A = 4r2
= 4 (6.37 106)2
= 5.09 1014 m2
Charge on the earth surface,
q=A
= 109 5.09 1014
= 5.09 105 C
Time taken to neutralize the earths surface = t
Current,
Page 13 of 26
Class XII
## Chapter 3 Current Electricity
Physics
Question 3.15:
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015
are joined in series to provide a supply to a resistance of 8.5 . What are the current
drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of
380 . What maximum current can be drawn from the cell? Could the cell drive the
starting motor of a car?
(a) Number of secondary cells, n = 6
Emf of each secondary cell, E = 2.0 V
Internal resistance of each cell, r = 0.015
series resistor is connected to the combination of cells.
Resistance of the resistor, R = 8.5
Current drawn from the supply = I, which is given by the relation,
## Terminal voltage, V = IR = 1.39 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is
11.87 A.
(b) After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380
## Hence, maximum current
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is
required to start the motor of a car, the cell cannot be used to start a motor.
Question 3.16:
Two wires of equal length, one of aluminium and the other of copper have the same
resistance. Which of the two wires is lighter? Hence explain why aluminium wires are
Page 14 of 26
Class XII
## Chapter 3 Current Electricity
Physics
preferred for overhead power cables. (Al = 2.63 108 m, Cu = 1.72 108 m,
Relative density of Al = 2.7, of Cu = 8.9.)
Resistivity of aluminium, Al = 2.63 108 m
Relative density of aluminium, d1 = 2.7
Let l1 be the length of aluminium wire and m1 be its mass.
Resistance of the aluminium wire = R1
Area of cross-section of the aluminium wire = A1
Resistivity of copper, Cu = 1.72 108 m
Relative density of copper, d2 = 8.9
Let l2 be the length of copper wire and m2 be its mass.
Resistance of the copper wire = R2
Area of cross-section of the copper wire = A2
The two relations can be written as
It is given that,
And,
## Mass of the aluminium wire,
m1 = Volume Density
Page 15 of 26
Class XII
Physics
## = A1l1 d1 = A1 l1d1 (3)
Mass of the copper wire,
m2 = Volume Density
= A2l2 d2 = A2 l2d2 (4)
Dividing equation (3) by equation (4), we obtain
It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter
than copper.
Since aluminium is lighter, it is preferred for overhead power cables over copper.
Question 3.17:
What conclusion can you draw from the following observations on a resistor made of
alloy manganin?
Current
Voltage
Current
Voltage
0.2
3.94
3.0
59.2
0.4
7.87
4.0
78.8
0.6
11.8
5.0
98.6
0.8
15.7
6.0
118.5
1.0
19.7
7.0
138.2
Page 16 of 26
Class XII
2.0
## Chapter 3 Current Electricity
39.4
8.0
Physics
158.0
It can be inferred from the given table that the ratio of voltage with current is a
constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy
obeys Ohms law. According to Ohms law, the ratio of voltage with current is the
resistance of the conductor. Hence, the resistance of manganin is 19.7 .
Question 3.18:
(a) A steady current flows in a metallic conductor of non-uniform cross- section. Which
of these quantities is constant along the conductor: current, current density, electric
field, drift speed?
(b) Is Ohms law universally applicable for all conducting elements?
If not, give examples of elements which do not obey Ohms law.
(c) A low voltage supply from which one needs high currents must have very low
internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance.
Why?
(a) When a steady current flows in a metallic conductor of non-uniform cross-section,
the current flowing through the conductor is constant. Current density, electric field, and
drift speed are inversely proportional to the area of cross-section. Therefore, they are
not constant.
(b) No, Ohms law is not universally applicable for all conducting elements. Vacuum
diode semi-conductor is a non-ohmic conductor. Ohms law is not valid for it.
(c) According to Ohms law, the relation for the potential is V = IR
Voltage (V) is directly proportional to current (I).
R is the internal resistance of the source.
If V is low, then R must be very low, so that high current can be drawn from the source.
Page 17 of 26
Class XII
## Chapter 3 Current Electricity
Physics
(d) In order to prohibit the current from exceeding the safety limit, a high tension
supply must have a very large internal resistance. If the internal resistance is not large,
then the current drawn can exceed the safety limits in case of a short circuit.
Question 3.19:
Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent
metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than
pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with
increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a
factor of the order of (1022/103).
(a) Alloys of metals usually have greater resistivity than that of their constituent
metals.
(b) Alloys usually have lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy, manganin, is nearly independent of increase of
temperature.
(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the
order of 1022.
Question 3.20:
(a) Given n resistors each of resistance R, how will you combine them to get the (i)
maximum (ii) minimum effective resistance? What is the ratio of the maximum to
minimum resistance?
(b) Given the resistances of 1 , 2 , 3 , how will be combine them to get an
equivalent resistance of (i) (11/3) (ii) (11/5) , (iii) 6 , (iv) (6/11) ?
(c) Determine the equivalent resistance of networks shown in Fig. 3.31.
Page 18 of 26
Class XII
## Chapter 3 Current Electricity
Physics
(a) Total number of resistors = n
Resistance of each resistor = R
(i) When n resistors are connected in series, effective resistance R1is the maximum,
given by the product nR.
Hence, maximum resistance of the combination, R1 = nR
(ii) When n resistors are connected in parallel, the effective resistance (R2) is the
## Hence, minimum resistance of the combination, R2 =
(iii) The ratio of the maximum to the minimum resistance is,
## (b) The resistance of the given resistors is,
R1 = 1 , R2 = 2 , R3 = 3 2
i.
Equivalent resistance,
Page 19 of 26
Class XII
ii.
## Chapter 3 Current Electricity
Physics
Equivalent resistance,
## (iii) Equivalent resistance, R = 6
Consider the series combination of the resistors, as shown in the given circuit.
## Equivalent resistance of the circuit is given by the sum,
R = 1 + 2 + 3 = 6
## (iv) Equivalent resistance,
Consider the series combination of the resistors, as shown in the given circuit.
## Equivalent resistance of the circuit is given by,
(c) (a) It can be observed from the given circuit that in the first small loop, two
resistors of resistance 1 each are connected in series.
Hence, their equivalent resistance = (1+1) = 2
It can also be observed that two resistors of resistance 2 each are connected in series.
Hence, their equivalent resistance = (2 + 2) = 4 .
Therefore, the circuit can be redrawn as
Page 20 of 26
Class XII
## Chapter 3 Current Electricity
Physics
It can be observed that 2 and 4 resistors are connected in parallel in all the four
loops. Hence, equivalent resistance (R) of each loop is given by,
## Hence, equivalent resistance of the given circuit is
(b) It can be observed from the given circuit that five resistors of resistance R each are
connected in series.
Hence, equivalent resistance of the circuit = R + R + R + R + R
=5R
2
Question 3.21:
Determine the current drawn from a 12 V supply with internal resistance 0.5 by the
infinite network shown in Fig. 3.32. Each resistor has 1 resistance.
The resistance of each resistor connected in the given circuit, R = 1
Equivalent resistance of the given circuit = R
The network is infinite. Hence, equivalent resistance is given by the relation,
Page 21 of 26
Class XII
Physics
## Internal resistance of the circuit, r = 0.5
Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23
Supply voltage, V = 12 V
According to Ohms Law, current drawn from the source is given by the ratio,
3.72 A
Question 3.22:
Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40
maintaining a potential drop across the resistor wire AB. A standard cell which maintains
a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance
point at 67.3 cm length of the wire. To ensure very low currents drawn from the
standard cell, a very high resistance of 600 k is put in series with it, which is shorted
close to the balance point. The standard cell is then replaced by a cell of unknown emf
and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
Page 22 of 26
Class XII
Physics
## (a) What is the value ?
(b) What purpose does the high resistance of 600 k have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer
(f ) Would the circuit work well for determining an extremely small emf, say of the order
of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify
the circuit?
(a) Constant emf of the given standard cell, E1 = 1.02 V
Balance point on the wire, l1 = 67.3 cm
A cell of unknown emf, ,replaced the standard cell. Therefore, new balance point on the
wire, l = 82.3 cm
The relation connecting emf and balance point is,
## The value of unknown emfis 1.247 V.
(b) The purpose of using the high resistance of 600 k is to reduce the current through
the galvanometer when the movable contact is far from the balance point.
Page 23 of 26
Class XII
## Chapter 3 Current Electricity
Physics
(c) The balance point is not affected by the presence of high resistance.
(d) The point is not affected by the internal resistance of the driver cell.
(e) The method would not work if the driver cell of the potentiometer had an emf of 1.0
V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is
less than the emf of the other cell, then there would be no balance point on the wire.
(f) The circuit would not work well for determining an extremely small emf. As the
circuit would be unstable, the balance point would be close to end A. Hence, there would
be a large percentage of error.
The given circuit can be modified if a series resistance is connected with the wire AB. The
potential drop across AB is slightly greater than the emf measured. The percentage error
would be small.
Question 3.23:
Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance
point with a standard resistor R = 10.0 is found to be 58.3 cm, while that with the
unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you
failed to find a balance point with the given cell of emf ?
Resistance of the standard resistor, R = 10.0
Balance point for this resistance, l1 = 58.3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E1 = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l2 = 68.5 cm
Page 24 of 26
Class XII
Physics
## Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,
## Therefore, the value of the unknown resistance, X, is 11.75 .
If we fail to find a balance point with the given cell of emf, , then the potential drop
across R and X must be reduced by putting a resistance in series with it. Only if the
potential drop across R or X is smaller than the potential drop across the potentiometer
wire AB, a balance point is obtained.
Question 3.24:
Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal
resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a
resistor of 9.5 is used in the external circuit of the cell, the balance point shifts to 64.8
cm length of the potentiometer wire. Determine the internal resistance of the cell.
Internal resistance of the cell = r
Balance point of the cell in open circuit, l1 = 76.3 cm
Page 25 of 26
Class XII
Physics
## An external resistance (R) is connected to the circuit with R = 9.5
New balance point of the circuit, l2 = 64.8 cm
Current flowing through the circuit = I
The relation connecting resistance and emf is,
Page 26 of 26
|
## The Lesson
We can find the mean of a set of numbers. Imagine a teacher wanted to find the class's average test score in mathematics.
## How to Find the Mean
The mean is found by adding all the numbers together and dividing by how many numbers there are.
## Question
What is the mean of the test scores below?
# 1
Add up the numbers.
7 + 10 + 8 + 6 + 4 = 35
# 2
Divide the answer (35) by how many numbers there are. In our example, there are 5 students, so there are 5 test scores.
35 ÷ 5 = 7
The mean of the test scores is 7.
## Lesson Slides
The slider below gives another example of finding the mean.
## A Formula to Find the Mean
The formula for finding the mean is shown below:
In this formula,
• is the symbol for the mean. It is said "x bar".
• xi is each value, where i = 1, 2... n. n is how many numbers there are. In our example of the test scores:
x1 = 7, x2 = 10, x3 = 8, x4 = 6, x5 = 4
• Σxi means "sum of xi" from i = 1 (below the Σ) to n (above the Σ).
Σxi = x1 + x2 + ... + xn
In our example, n = 5
Σxi = x1 + x2 + x3 + x4 + x5
Σxi = 7 + 10 + 8 + 6 + 4
Σxi = 35
• This is all written above a line, with n under it. This means divide by n. In our example, n = 5:
x̄ = Σxi / n = 35 ÷ 5 = 7
The mean of the test scores is 7.
## Sum Over Count
The mean is the sum of the number divided by a count of the numbers.
## Outliers
Sometimes in a group of numbers, a few of them are much larger...
3, 5, 2, 7, 150
...or much smaller than the rest:
1, 502, 847, 564, 980
These relatively large (150) or small (1) numbers are called outliers. The mean is very affected by outliers, unlike the median. For example, consider the numbers:
1, 2, 3
The mean is 2 and the median is 2. If instead the numbers have a large outlier:
1, 2, 300
The mean is now 101 while the median is still 2. If you read about the average salary in a country, ask if it is the mean or median average? The mean average may dragged up by some very rich people, while the median won't be.
## How to Find the Mean from a Frequency Table
Sometimes data is presented in frequency tables. A frequency table representing the test scores is shown below:
It is possible to find that the mean test score is 7.
Read more about finding the mean from a frequency table
|
WAEC General Mathematics Question And Answer
2021 WAEC Mathematics Question And Answer
1. Which of these numbers is not less than -2?
-1,
-3,
-4,
-5.
-6
Solution Solving: -2 is greater than -2, -3, -4, -5, -6 and so on. But -1 is greater than -2.
The Correct Answer is A): Because it is greater than -2.
2. -6 is greater than -2 but less than -7? True or false
True
Both
False
None of the above
Solving for the solution: -6 is greater than any number from -7 to the negative infinity but less than any number from -5 to the positive infinity.
The Correct Answer is C): Because -6 is not greater than -2 and it is not less than -7.
3. Add 26b + 12a + 16a – 4b
22b + 28a
-22b + 28a
22b – 28a
-22b – 28a
Solution Solving for Questions 3: Firstly, you have to Collect like terms. Therefore, we have;
26b – 4b + 12a + 16a.
Next, we have 26b – 4b = 22b
And 12a + 16a = 28a
Therefore, we have 22b + 28a
The Correct Answer is A): 22b + 28a
4. Multiply this equation: (x – 6)(2x + 7)?
2×2 + 5x – 42
2×2 – 5x – 42
2×2 – 5x + 42
2×2 + 5x + 42
-2×2 + 5x – 42
Solving for question 4 solution: Here, we start by removing the bracket
Thus; 2×2 + 7x – 12x – 42
Hence, 2×2 – 5x – 42
The Correct Answer is B): 2×2 – 5x – 42
5. Factorise the following: 5×2 – 15x – 20?
5(x+4)(x+1).
-2(x-4)(x+5)
-5(x+4)(x-1)
5(x-4)(x+1)
Solution Solving to question 5: Firs find the L.C.M of 5, 15 and 20 = 5(x2 – 3x – 4).
= 5(x2 – 4x + x – 4).
= 5{x(x – 4) +1(x – 4)}.
= 5(x-4)(x+1).
The Correct Answer is D): 5(x-4)(x+1)
6. In a school, 180 students offer mathematics or physics or both. If 125 offer mathematics and 105 offer physics. How many students offer mathematics only?
A.75 B. 80 C. 55 D. 125.
7. Find the value of x for which 3(24x + 3) = 96
A. 2 B. -2 C. ½ D. -1/2
8. The cost of renovating a 5m square room is N500. What is the cost of renovating a 10m square room?
A. N1,000 B. N2,500 C. 2,000 D. N10,000
9. Find the rate of change of the total surface area S of a sphere with respect to its radius r when r = 2
A. 8 B. 16 C. 10 D. 14
10. Differentiate (cosӨ + sinӨ)2 with respect to Ө.
A. 2cos2Ө B. 2sin2Ө C. -2cos2Ө D. -2sin2Ө
11. A binary operation * on the set of rational numbers is defined as x * y = 2x + [x3 – y3/x + y]. find -1*2
A. 11 B. -11 C. 8 D. -8
12. A polynomial in x whose zeroes are 2, 1 and -3 is ______
A. x3 – 7x + 6
B. x3 + 7x – 6
C. x3 – 7x – 6
D. x3 + 7x + 6
13. Find the range of values of x for which 7x – 3 > 3x + 4
A. x < 7/4
B. x > 7/4
C. 7x < 4
D. -4x < 7
14. Some red balls were put in a basket containing 12 white balls and 16 blue balls. If the probability of picking a red ball from the basket is 3/7, how many red balls were introduced?
A. 13 B.20 C. 12 D. 21
15. Convert 12314 to a number in base 6.
A. 1056
B. 3016
C. 1036
D. 5016
16. Find the slope of the curve y = 3×3 + 5×2 – 3 at (-1, 5).
A. 1 B. -1 C. 19 D. -19
17. Find the area of the region bounded by y = x2 + x – 2 and x-axis
A. 9/2 B. -39/6 C. 8/3 D. 16/3
18. The minimum value of y = x2 – 4x – 5 is ______
A. 2 B. -2 C.13 D. -13
Waec General Mathematics Theory and Answer Paper 2, WASSCE
Copy and complete the table of values for the equation y = 2×2 – 7x – 9 for -3 ≤ x ≤ 6.
X -3 – 2 -1 0 1 2 3 4 5 6
Y 13 -9 -14 -12 6
Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 4 units on the y – axis, draw the graph of y = 2×2 – 7x – 9 for -3 ≤ x ≤ 6.
Use the graph to estimate the:
roots of the equation 2×2 – 7x = 26;
(ii) coordinates of the minimum point of y;
(iii) range of values for which 2×2 – 7x < 9.
Solution to Question 1
The Chief Examiner reported that the majority of the Candidates’ who attempted this question got right the complete table of values for which y = 2×2 – 7x – 9. However, many failed to plot the points on the graph correctly on the graph thereby losing marks. Also, they could not read up values from the graph to answer subsequent questions.
In part (a), the Candidates were expected to complete the table as follows:
X -3 – 2 -1 0 1 2 3 4 5 6
Y 30 13 0 -9 -14 -15 -12 -5 6 21
In part (b), Candidates would then use this table and the scale provided to draw the graph of the relation y = 2×2 – 7x – 9 and use this graph to answer the questions that followed.
In part (c), from the graph, the roots of the equation 2×2 – 7x = 26 were obtained by drawing a line through y = 17 to meet the curve y = 2×2 – 7x – 9. The required roots are – 2.2 ± 0.1 and 5.8 ± 0.1. The coordinates of the minimum point of y is 1.8 ± 0.1, -15 ± 0.4. The ranges -10.1<x<4.5
Question 2
Marks 1 2 3 4 5
Number of students m + 2 m – 1 2m – 3 m + 5 3m – 4
The table shows the distribution of marks scored by some students in a test.
If the mean mark is, find the value of m.
Find the:
(i) inter-quartile range;
(ii) probability of selecting a student who scored at least 4 marks in the test.
The solution to Question 2
x 1 2 3 4 5 Total
f m+2 m-1 2m-3 m+5 3m-4 8m-1
fx m+2 2m-2 6m-9 4m+20 15m-20 28m-9
From the table, and, so that and solving yields m = 3.
In part (b) (i), since m has been gotten to be 3, the table then becomes:
x 1 2 3 4 5
f 5 2 3 8 5
Cf 5 7 10 18 23
Therefore, interquartile range = 4 – 2 = 2.
Finally, in part (b) (ii), the probability of selecting a student who scored at least 4 marks is whereas most Candidates’ made use of at most 4 marks to obtain.
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## GOODMAN 3 years ago Two cards are drawn from a standard deck of 52 cards without replacement. What is the probability that both cards are greater than 2 and less than 9?
1. ParthKohli
2. GOODMAN
Thanks little bro, lol. I am studying and forgot this all.
3. ParthKohli
What would be the probability of getting a card between 2 and 9 in the first pick?
4. GOODMAN
2/52 and 9/52
5. ParthKohli
No...it's saying that it's greater than 2 and less than 9. 3,4,5,6,7,8 are the numbers that qualify for me :)
6. GOODMAN
So we have to get the ones in between?
7. ParthKohli
Yes.. exactly
8. ParthKohli
Since there are 3,4,5,6,7,8 in four suits, the probability would be: $$\color{Black}{\Rightarrow \Large {4 \times 6 \over 52} }$$ For the first pick.
9. ParthKohli
Simplifying further: $$\color{Black}{\Rightarrow \Large {24 \over 52} = {12 \over 26} = {6 \over 13} }$$
10. ParthKohli
Now we've picked the first. We have 1 card LESS in the deck because we haven't replaced the cards.
11. ParthKohli
So this time the denominator of the probability would become 51. The numerator will also get one less because we have assumed that we have picked one card which is 3,4,5,6,7 or 8.
12. ParthKohli
There are 23 cards left that we want. 51 total cards left. $$\color{Black}{\Rightarrow \Large {23 \over 51} }$$
13. GOODMAN
Okay...so since we had not replaced, we have to subtract 1?
14. ParthKohli
Now if we want two things to happen at the same time, we shall multiply the probabilities. $$\color{Black}{\Rightarrow \Large {13 \over 26} \times {23 \over 51}}$$
15. ParthKohli
We have to subtract 1 from both numerator and denominator.
16. GOODMAN
Thats it?
17. ParthKohli
Nope
18. GOODMAN
Wait. so after we multiply, we have to simplify further too?
19. ParthKohli
Oops.. I meant this: $$\color{Black}{\Rightarrow \Large {6 \over 13} \times {23 \over 51}}$$
20. ParthKohli
Multiply the fractions. The fractions are in their simplest forms so when we'll multiply it'd be in the simplest form.
21. GOODMAN
Yes, i did that. Okay, so 46/221 is final answer?
22. ParthKohli
How did you get that?
23. ParthKohli
Oh yes
24. ParthKohli
$$\color{Black}{\Rightarrow \Large {2 \over 13} \times {23 \over 17} }$$
25. ParthKohli
Correct! :D
26. GOODMAN
Okay, thats awesome!! Thanks soo much Parth!! Probability is my weak spot :/
27. ParthKohli
Haha probability is easy...getting the hang of what it involves is important
28. GOODMAN
Yea, see, im reviewing this since last semester, lol.
29. ParthKohli
And I've helped someone on this site after a long time. we usually just answer questions ://
30. GOODMAN
Thanks, I actually learned :D
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# Class 11 NCERT Solutions- Chapter 14 Mathematical Reasoning – Exercise 14.5
• Last Updated : 05 May, 2021
### (i) direct method
Solution:
p: ‘If x is a real number such that x3 + 4x = 0, then x is 0’.
Let q: x is a real number such that x3 + 4x = 0
r: x is 0.
(i) To show that statement p is true, we assume that q, is true and then show that r is true.
Therefore, let statement q be true.
x3 + 4x = 0
x(x2 + 4) = 0
x = 0 or x2 + 4 = 0
However, since x is real, it is 0.
Thus, statement r is true.
Therefore, the given statement is true.
Solution:
To show statement p to be true by contradiction, we assume that p is not true.
Let x be a real number such that x3 + 4x = 0 and let x is not 0.
Therefore, x3 + 4x = 0
x (x2 + 4) = 0
x = 0 or x2 + 4 = 0
x = 0 or x2 = –4
However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0.
### (iii) method of contrapositive
Solution:
To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false.
Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement.
~ r: x is not 0
I can be seen that (x2 + 4) will always be positive
x = 0 implies that the product of any positive real number with x is not zero.
Let us consider the product of x with (x2 + 4)
⸫ x(x2 + 4) = 0
x3 + 4x = 0
This shows that statement q is not true.
Thus, it has been proved that
~ r
⇒ ~ q
Therefore, the given statement p is true.
### Question 2: Show that the statement ‘For any real numbers a and b, a2 = b2 implies that a = b’ is not true by giving a counter-example.
Solution:
The given statement can be written in the form of ‘if-then’ as follows.
If a and b are real numbers such that a2 = b2 , then a = b.
Let p: a and b are real numbers such that a2 = b2.
q: a = b
The given statement has to be proved false. For this purpose, it has to be proved that if p, then ~ q.
To show this, two real numbers, a and b, with a2 = b2 are required such that a ≠b.
Let a = 1 and b = –1
a2 = (1)2 and b2 = (–1)2 = 1
a2 = b2
However, a = b
Thus, it can be concluded that the given statement is false.
### Question 3: Show that the following statement is true by the method of contrapositive. p: If x is an integer and x2 is even, then x is also even.
Solution:
p: If x is an integer and x2 is even, then x is also even.
Let q: x is an integer and x2 is even.
r: x is even.
To prove that p is true by contrapositive method, we assume that r is false, and prove that q is also false.
Let x is not even.
To prove that q is false, it has to be proved that x is not an integer or x2 is not even.
x is not even implies that x2 is also not even.
Therefore, statement q is false.
Thus, the given statement p is true.
### (i) p: If all the angles of a triangle are equal, then the triangle is an obtuse-angled triangle.
Solution:
The given statement is of the form ‘if q then r’.
q: All the angles of a triangle are equal.
r: The triangle is an obtuse-angled triangle.
The given statement p has to be proved false. For this purpose, it has to be proved that if q, then ~ r.
To show this, angles of a triangle are required such that none of them is an obtuse angle.
It is known that the sum of all angles of a triangle is 180°. Therefore, if all the three angles
are equal, then each of them is of measure 60°, which is not an obtuse angle.
In an equilateral triangle, the measure of all angles is equal. However, the triangle is not an obtuse-angled triangle.
Thus, it can be concluded that the given statement p is false.
### (ii) q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.
Solution:
The given statement is as follows.
q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.
This statement has to be proved false. To show this, a counter example is required.
Consider x2– 1 = 0
x2 = 1
x = ±1
One root of the equation x2 – 1 = 0, i.e. the root x = 1, lies between 0 and 2.
Thus, the given statement is false.
### (i) p: Each radius of a circle is a chord of the circle.
Solution:
The given statement p is false.
According to the definition of chord, it should intersect the circle at two distinct points.
### (ii) q: The centre of a circle bisects each chord of the circle.
Solution:
The given statement q is false.
If the chord is not the diameter of the circle, then the centre will not bisect that chord.
In other words, the centre of a circle only bisects the diameter, which is the chord of the
circle.
### (iii) r: Circle is a particular case of an ellipse.
Solution:
The equation of an ellipse is,
X2/a2 + Y2/b2 = 1
If we put a = b = 1, then we obtain
x2 + y2 = 1,which is an equation of a circle
Therefore, circle is a particular case of an eclipse.
Thus, statement r is true.
### (iv) s: If x and y are integers such that x > y, then – x < – y.
Solution:
x > y
⇒ – x < – y (By a rule of inequality)
Thus, the given statement s is true.
### (v) t: √11 is a rational number.
Solution:
11 is a prime number, and we know that the square root of any prime number is an irrational number.
Therefore,
√11 is an irrational number.
Thus, the given statement t is false.
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Q: Probability of selecting balls from an urn. ( Answered , 1 Comment )
Question
Subject: Probability of selecting balls from an urn. Category: Science > Math Asked by: wobbleone-ga List Price: \$10.00 Posted: 10 Sep 2006 17:25 PDT Expires: 10 Oct 2006 17:25 PDT Question ID: 763991
```An urn contains one red ball and four with balls. You and a friend take turns selecting a ball from the urn, one at a time. The first person to select the red ball wins. If you draw first, what is the probability that you will win if sampling is done with replacement and the probability without replacement?```
```Hello! Let's start with the case in which the sampling is done without replacement. Since there is a total of 5 balls, you can draw balls a maximum of three times: in the first turn, in the third turn (if your friend doesn't win in the 2nd one) and in the fifth turn (if your friend doesn't win in the 2nd or 4th one). Therefore, we can write the probability of winning as: P(Win) = P(Win in 1st turn) + P(Win in 3rd turn) + P(Win in 5th turn) In the first turn, there's one red ball out of a total of 5 balls. Therefore, the probability of winning in this turn is 1/5 = 0.2: P(Win in 1st turn) = 0.2 Now, in order to win in the 3rd turn, it must happen that: - You draw a white ball in the 1st turn - Your friend draws a white ball in the 2nd turn - You draw the red ball in the third turn The probability that you draw a white ball in the 1st turn is clearly 4/5 = 0.8. After this turn, there are 4 balls left, one of which is red. Therefore, the probability that your friend draws a white ball in the 2nd turn is 3/4 = 0.75. In the third turn, there are 3 balls left, one of which is red. Therefore, the probability of drawing the red ball is 1/3 = 0.333... We thus conclude that: P(Win in 3rd turn) = (4/5)*(3/4)*(1/3) = 0.2 Finally, we have to calculate the probability that you win in the 5th turn. Using the same reasoning as above, it must happen that neither you or your friend draw the red ball in the first four turns. Therefore, the probability of this is: P(Win in 5th turn) = (4/5)*(3/4)*(2/3)*(1/2) = 0.2 We conclude that: P(Win) = 0.2 + 0.2 + 0.2 = 0.6 If you draw first, you have a 0.6 probability of winning. Let's now see the case with replacement. In this case, theoretically, there can be any number of turns before someone draws the red ball. Since there is replacement, the probability of drawing the red ball is always 1/5 (as there are always 5 balls), so the probability of drawing a white one is 4/5. The probability of winning can be written as: P(Win) = P(Win in 1st turn) + P(Win in 3rd turn) + P(Win in 5th turn) + + P(Win in 7th turn) + ... Now, the probability of winning in the 1st turn is clearly 1/5. In order to win in the 3rd turn, both you and your friend must draw white balls in the 1st and 2nd turn, and you must draw the red one in the 3rd turn. Therefore, the probability of this event is (4/5)*(4/5)*(1/5) = [(4/5)^2]*(1/5). In order to win in the 5th turn, both you and your friend must draw white balls in the 1st, 2nd, 3rd and 4th turn, and you must draw the red one in the fifth turn. Therefore, the probability of this event is: (4/5)*(4/5)*(4/5)*(4/5)*(1/5) = [[(4/5)^2]^2]*(1/5) The pattern should now be clear. The probability of winning in the 7th turn would be [[(4/5)^2]^3]*(1/5); in the 9th turn it would be [[(4/5)^2]^4]*(1/5), and so on. Therefore, P(Win) = 1/5 + [(4/5)^2]*(1/5) + [[(4/5)^2]^2]*(1/5) + [[(4/5)^2]^3]*(1/5) + + ... P(Win) = (1/5)*(1 + [(4/5)^2] + [[(4/5)^2]^2] + [[(4/5)^2]^3] + ...) The infinite sum in the above expression has the form 1 + r + r^2 + r^3 + ... This sum (as long as |r|<1) is known to converge to 1/(1-r). Therefore, we get: P(Win) = (1/5)/(1 - (4/5)^2) = 5/9 = 0.555... Therefore, the probability of winning if you draw first and the sampling is with replacement is 5/9, a bit smaller than in the other case. I hope this helps! If you have any doubt regarding my answer, please don't hesitate to request clarification before rating it. Otherwise, I await your rating and final comments. Best wishes! elmarto```
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# Simplifying with Exponents Answers and Explanations
## Power of a Quotient Rule
Practice simplifying exponents using the Power of a Quotient Rule.
### 1. Simplify.
(36/49)1/2= 6/7
Raise the numerator and denominator to the power indicated by the exponent:
• 361/2
• 491/2
Simplify the numerator and denominator:
• 361/2 = 6
• 491/2 = 7
Bring the numerator and denominator together:
6/7
### 2. Simplify.
(ab ÷ 180)3= a3b3/5,832,000
Raise the numerator and denominator to the power indicated by the exponent:
• a3b3
• 1803
Simplify the numerator and denominator:
• a3b3= a3b3
• 1803= 5,832,000
Bring the numerator and denominator together:
a3b3/5,832,000
### 3. Simplify.
(-5/p)3= -125/p3
Raise the numerator and denominator to the power indicated by the exponent:
• -53
• p3
Simplify the numerator and denominator:
• -53= -125
• p3= p3
Bring the numerator and denominator together:
-125/p3
### 4. Simplify.
(-f/g)7= -f 7/g 7
Raise the numerator and denominator to the power indicated by the exponent:
• -f 7
• g 7
Simplify the numerator and denominator:
• -f 7= -f 7
• g 7 = g 7
Bring the numerator and denominator together:
-f 7/g 7
### 5. Simplify.
(e/20)-1= 20/e
Raise the numerator and denominator to the power indicated by the exponent:
• e-1
• 20-1
Simplify the numerator and denominator:
• e-1 = 1/e1
• 20-1= 1/201
Bring the numerator and denominator together:
20/e
### 6. Simplify.
(xy/Π)10= x10y10 /Π10
Raise the numerator and denominator to the power indicated by the exponent:
• x10y10
• Π10
Bring the numerator and denominator together:
x10y10/Π10
##### Exponent Worksheets
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It's All in the Numbers
# 6. Number Properties
Suggested Learning Intentions
• To explore a range of number properties and use these properties to arrange numbers into sets
• To recognise that some pairs of sets can have common numbers, and some pairs of sets do not have common numbers
Sample Success Criteria
• I can identify properties of numbers
• I can organise numbers into sets with common properties
• I can explain my thinking using manipulatives
The purpose of this activity is review some concepts students may have been previously exposed to, either as part of the activities in this sequence or in previous years. Should some concepts be unfamiliar to some students, you can use this activity to teach these specific concepts ‘in the moment’.
Present students with a series of cards, with assorted numbers on them. For example:
Ask students to use these cards to make different sets of numbers, using these cards only.
Depending on the prior knowledge of your students regarding number properties, you could ask them to produce sets of:
• Even numbers, or odd numbers (two different sets)
• Factors of 24, or 60 (two different sets)
• Multiples of 3, or 7 (two different sets)
• Prime Numbers
• Composite Numbers
• Square Numbers
• Triangular Numbers
Use this opportunity to teach any specific concepts that individual or small groups of students may be unfamiliar with.
Other questions can be posed to further gauge student understanding of these concepts, including:
• Which numbers may appear in more than one set? (This question can be used to gauge understanding of common factors and multiples).
• Which sets couldn’t have any numbers in common? (For example, evens and odds, or primes and composites).
• Find a number that would fit into as many, or as few of the sets as possible (for example, the 12 card would fit into the set of even cards, the factors of 24, the factors of 60, the multiples of 3, and the composite numbers set).
• Write your own description of a number property, such as ‘factors of 50’ or ‘composite numbers greater than 20’ and use the cards to create this set.
Show students a grid such as this one below:
Select one number for discussion with the class, such as 25. Invite students to explain which squares they believe 25 could be placed in. For example, 25 could be placed in the bottom right corner, because it is both an odd number and a square number. Facilitate similar discussions using other numbers.
Ask students to identify nine different numbers that could be used to complete the grid.
Enable students requiring further support by providing a 2 x 2 grid to begin with, and/or a set of numbers to select from to fill the grid with (such as the cards used in the initial task).
Question students to draw out their understanding of the different sets, and the relationships that exist between them. Sample questions include:
• Are there any squares in the grid that can only be filled by one number? Why is this? A clear explanation of why 2 must go in the top left corner (because 2 is the only even prime) will help consolidate student understanding of prime numbers.
• If the headings were placed in different parts of the grid, would you still be able to fill it entirely? Students should recognise that even and odd numbers must go on the same ‘side’ of the table, as they have no common values. The same is true for primes and squares (remember 1 is not a prime!)
A sample solution is provided below. The 2 and the 5 are highlighted in red to indicate that they are the only possible numbers that can go in this part of the table.
Extend students by challenging them to construct their own grid. This could be undertaken in different ways:
• Construct a grid with the headings included and the spaces blank and ask a classmate to fill it out.
• Construct a grid where the numbers are included but the headings are blank and ask a classmate to work out what the headings should be.
• Construct a grid that can be filled with consecutive numbers. A sample solution is provided above, using the numbers 1 - 9. Note the use of ‘Factors of 168’ which reflects the challenge present in what appears a simple task.
Areas for further exploration
Present students with the ‘Factors and Multiples’ puzzle on NRICH, published by the University of Cambridge.
Read through the Teachers’ Resources section on NRICH to know how to present this task to students, or use the resources provided, such as the printable worksheets. Note that the website has sample solutions provided by students, so it is best not to direct students to the website themselves.
This puzzle is like the problems presented earlier in this stage but is larger and more complex. The puzzle includes a 5 x 5 grid and students are provided with 10 different headings and 25 numbers with which to complete the grid.
Encourage students to reflect on the strategies they have used in the previous activities to help determine any headings that need to go on the ‘same side’ of the table.
Enable students requiring further support by providing prompts such as:
• Can you think of any numbers that are even and odd? What does this tell you about where the ODD NUMBER and EVEN NUMBER headings should go?
• Can you think of any numbers that are less than 20 AND more than 20?
This is a particularly challenging puzzle and students will require a lot of time to engage with it.
This stage exposes students to a range of mathematical skills and concepts which you may wish to review and consolidate with students. Consider the progress made by students when determining which skills and concepts you select to review.
Two suggested concepts for review are presented below.
1. Square numbers can be found by multiplying a value by itself
You may wish to emphasise the following concepts:
• square numbers can be located on the diagonal of a common multiplication grid, found in many classrooms
• square numbers can be represented visually by an array of square dots
• working ‘backwards’ from a square number is called ‘finding the square root’ of the number.
2. All numbers have a set of factors and a set of multiples
You may wish to emphasise the following concepts:
• there are a range of approaches we can use to find factors of numbers; some are more systematic than others
• 1 is a factor of all whole numbers
• a number is both a factor and a multiple of itself
• multiples of a number can be found by ‘counting by’ that number.
You could support students to consolidate their understanding of factors and multiples by playing the ‘Factors and Multiples Game’ on nRich, published by the University of Cambridge. This game uses a 1-100 number grid and can be played in different ways.
Students can play competitively in pairs; the first player chooses a starting number and crosses it off the grid. The second player then chooses a factor or multiple of that number and crosses it off. Play alternates, with each player selecting a factor or multiple of the last number crossed off, until one player cannot choose a number, at which point the other player is the winner.
Alternatively, players can play collaboratively, and try and form the longest chain possible of connected factors and multiples, using the rules of the game. This chain could be submitted to help assess student understanding of factors and multiples.
The game can be played either online or on printable grids provided by nRich. The Teachers Resources section provides further advice as to how you can best use this with students.
University of Cambridge, n.d. Factors and Multiples Game. [Online]
Available at: nrich.maths.org/factorsandmultiples
[Accessed 15 March 2022].
University of Cambridge, n.d. Factors and Multiples Puzzle. [Online]
Available at: nrich.maths.org/5448
[Accessed 15 March 2022].
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Area of a Trapezoid
The area of a polygon is the number of square units inside that polygon. Area is 2-dimensional like a carpet or an area rug.
A trapezoid is a 4-sided figure with one pair of parallel sides. For example, in the diagram to the right, the bases are parallel. To find the area of a trapezoid, take the sum of its bases, multiply the sum by the height of the trapezoid, and then divide the result by 2, The formula looks like this:
or
Where is is is height and · means multiply.
Each base of a trapezoid must be perpendicular to the height. In the diagram above, both bases are sides of the trapezoid. However, since the lateral sides are not perpendicular to either of the bases, a dotted line is drawn to represent the height.
In Examples 1 and 3 below, the height is a side of the trapezoid since it is perpendicular to the base. In Example 2, the lateral sides are not perpendicular to the base, so a dotted line is drawn to represent the height.
Example 1: Find the area of a trapezoid with bases of 10 inches and 14 inches, and a height of 5 inches.
Solution:
= · (10 in + 14 in) · 5 in
= · (24 in) · (5 in)
= · 120 in2
= 60 in2
Example 2: Find the area of a trapezoid with bases of 9 centimeters and 7 centimeters, and a height of 3 centimeters.
Solution:
= · (9 cm + 7 cm) · 3 cm
= · (16 cm) · (3 cm)
= · 48 cm2
= 24 cm2
Example 3: The area of a trapezoid is 52 square inches and the bases are 11 inches and 15 inches. Find the height.
Solution:
52 in2 = · (11 in + 15 in) ·
52 in2 = · (26 in) ·
52 in2 = (13 in) ·
52 in2 ÷ (13 in) =
= 4 in
Summary: To find the area of a trapezoid, take the sum of its bases, multiply the sum by the height of the trapezoid, and then divide the result by 2, The formula looks like this:
or
Where is is is height.
Exercises
Directions: Read each question below. Click once in an ANSWER BOX and type in your answer; then click ENTER. Your answers should be given as whole numbers greater than zero. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR.
1 Find the area of a trapezoid with bases of 5 feet and 7 feet, and a height of 3 feet. ANSWER BOX: A = ft2 RESULTS BOX:
2 Find the area of a trapezoid with bases of 12 meters and 6 meters, and a height of 5 meters. ANSWER BOX: A = m2 RESULTS BOX:
3 Find the area of a trapezoid with bases of 17 centimeters and 19 centimeters, and a height of 2 centimeters. ANSWER BOX: A = cm2 RESULTS BOX:
4 If the trapezoid's area is 48 square inches and the bases are 4 inches and 8 inches. What is the height? ANSWER BOX: H = in RESULTS BOX:
5 If the trapezoid's area is 77 square centimeters and the bases are 9 centimeters and 13 centimeters. What is the height? ANSWER BOX: H = cm RESULTS BOX:
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# What is the perimeter of 6m in a square?
012
###### 2014-09-12 21:38:16
The perimeter is four times the length of any of the sides.
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## Related Questions
A square has four sides of equal length. If the total of all four sides is 24m then each side must measure 6m. There area is therefore 6m x 6m which equals 36 square metres.
By definition, the sides of a square MUST be the same. It is not possible to have a square with sides of 12m and 6m.
A square with a side length of 6m has an area of 36m2
In order to find the perimeter of a rectangle, you will need to add the measurements of all 4 sides. For a rectangle that is 7m x 6m, you would have two sides that are 7m and two sides that are 6m. That means the perimeter of the rectangle would be 26m.
1m and 11m 2m and 10m 3m and 9m 4m and 8m 5m and 7m. The next one, 6m and 6m, would be a square.
There is not enough information to answer the question.
a square has a perimeter but a square isnt a perimeter. so no a square isnt a perimeter
Only if they are congruent. One of them could be the a rotation of the other - eg 4m x 6m and 6m x 4m.
The area of a square is a function of the perimeter of the square.
Perimeter is in feet and not in square feet
36 square units. You can't express a perimeter in square units; a perimeter is a length expressed in ordinary units. If the perimeter of this square is 24 units then the answer above is correct.
perimeter of a square is 4x square that term (4x)(4x)and it gives you 16x2
The perimeter of a square is the sum of the four equal sides.
If the area is a square the perimeter is 508.87m.
The formula for the perimeter of a square is P equals 4 times a. 'P' represents the perimeter, and 'a' represents a side of the square.
Perimeter of a square = 4*side. So perimeter of 6x6 square = 4*6 = 24 units.
###### Math and ArithmeticScienceGeometryAlgebra
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Chapter 8 Class 10 Introduction to Trignometry
Class 10
Important Questions for Exam - Class 10
### Transcript
Example 2 If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q. Given: sin B = sin Q To prove: ∠ B = ∠ Q Proof: Let’s take two right angle triangles ABC & PQR Since, sin B = sin Q sin B = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝐵)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 sin B = 𝑨𝑪/𝑨𝑩 sin Q = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑄)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 sin Q = 𝑷𝑹/𝑷𝑸 𝐴𝐶/𝑃𝑅=𝐴𝐵/𝑃𝑄 Let 𝑨𝑪/𝑷𝑹=𝑨𝑩/𝑷𝑸= k So, AC = k PR & AB = k PQ Now, Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 In right Δ ABC, (AB)2 = AC2 + BC2 BC2 = AB2 – AC2 BC = √(𝐴𝐵2 −𝐴𝐶2) In right Δ PQR, (PQ)2 = (PR)2 + (QR)2 QR2 = PQ2 – PR2 QR = √(𝑃𝑄2−𝑃𝑅2) Now, 𝑩𝑪/𝑸𝑹 = √(𝑨𝑩^𝟐 − 𝑨𝑪^𝟐 )/√(𝑷𝑸^𝟐 − 𝑷𝑹^𝟐 ) From (1) AB = k(PQ) and AC = k(PR) From (1) and (2) 𝐴𝐶/𝑃𝑅 = 𝐴𝐵/𝑃𝑄 = 𝐵𝐶/𝑄𝑅 = k 𝑨𝑪/𝑷𝑹 = 𝑨𝑩/𝑷𝑸 = 𝑩𝑪/𝑸𝑹 Since corresponding sides of Δ ABC & Δ PQR are in the same ratio Thus, ∆ ABC ~ ∆ PQR Since corresponding angles of similar triangles are equal ∴ ∠ B = ∠ Q Hence proved 𝐵𝐶/𝑄𝑅 = √((𝑘𝑃𝑄)^2 − (𝑘𝑃𝑅)^2 )/√(𝑃𝑄^2 − 𝑃𝑅^2 ) 𝑩𝑪/𝑸𝑹= √(𝒌^𝟐 𝑷𝑸^𝟐− 𝒌^𝟐 𝑷𝑹^𝟐 )/√(𝑷𝑸^𝟐 −𝑷𝑹^𝟐 ) 𝐵𝐶/𝑄𝑅 = (𝑘√(𝑃𝑄^2 −𝑃𝑅^2 ))/√(𝑃𝑄^2 −𝑃𝑅^2 ) 𝑩𝑪/𝑸𝑹 = k
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#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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# Synthetic Long Division
## Synthetic Long Division
### Is the long division the same as the synthetic division?
Synthetic division is another technique for dividing polynomials. This is a brief description of long division, which only works when dividing with a degree 1 polynomial.
### I also asked what is the difference between synthetic division and long division?
Synthetic Polynomial Division: The Method Instead of the typical division support as in long division, synthetic division uses vertical straight lines and creates space for several subseries. If there is a remainder, the zero of the test is not a real zero of the polynomial.
### So the question is, when can I use synthetic division?
Synthetic division is an abbreviation that can be used when the divisor is a binomial of the form x - k. Synthetic division uses only coefficients in the division process.
### How is a long synthetic joint made here?
Synthetic division is another way of dividing a polynomial by the binomial x c, where c is a constant.
1. Step 1: Set up the synthetic department.
2. Step 2: Lower the leader in the bottom row.
3. Step 3: Multiply c by the value that's right on the bottom line.
4. Step 4: Add the column created in Step 3.
What is the denominator in the synthetic division?P (x) is the yield, Q (x) is the quotient and R (x) is the remainder How to solve this problem by synthetic division. To use synthetic division, the divisor must be of the first degree and take the form x - a. In this example the divisor is x - 2 with a = 2.
### What is synthetic division and examples?
Synthetic division is a shortened method of dividing polynomials for the special case of division by a linear factor whose dominant coefficient is 1. To illustrate the process, refer to the example at the beginning of this section. Divide 2x3−3x2 + 4x + 5 2 x 3-3 x 2 + 4 x + 5 by x + 2 using the long division algorithm.
### What do you do with the long division remnants?
How to do a long split with Reester? Solve the division problem with the long division or long division symbol. Divide the first number by the dividend, 4 by the divisor, 32. Put 0 above the division bracket. Draw a line under 0 and subtract 0 from 4.
### What is the purpose of dividing polynomials?
In algebra, long division of a polynomial is an algorithm for dividing a polynomial by another polynomial of equal or less degree, a generalized version of the well-known arithmetic technique called long division. This can easily be done by hand as it breaks down an otherwise complex subproblem into smaller problems.
### How does synthetic substitution work?
In mathematics, synthetic substitution offers us a way to evaluate a polynomial for a given value of the variable. It is based on the remainder of the polynomial theorem, which says that the remainder of P (x) x - a P (x) x - a, where P (x) is a polynomial function, is equal to P (a), or P evaluated in x = a.
### Is it possible to synthetically divide with a fraction?
Since you are dividing by a polynomial of degree 1, the degree of resolution is 1 less than the degree of efficiency. For this problem, the answer starts with a force of 2, then a force of 1, then a force of 0 (the constant). The last value on the bottom line is the remainder and is written as a fraction.
### Why is synthetic division important?
Basically, the explanation is that we are using synthetic division to find polynomial factors, which is actually division. If the remainder of the synthetic division is zero, the divisor is a factor. It is important that synthetic division divides a polynomial only by a linear factor.
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Section 6.4
# Section 6.4
Télécharger la présentation
## Section 6.4
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##### Presentation Transcript
1. Section 6.4 Solving Polynomial Equations
2. y = 3x + 1 y = –2x + 6 –2x + 3y = 0 x + 3y = 3 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (For help, go to Lessons 6-2 and 6-3.) Graph each system. Find any points of intersection. 1. 2. Factor each expression. 3.x2 – 2x – 15 4.x2 – 9x + 14 Check Skills You’ll Need 6-4
3. y = 3x + 1 y = –2x + 6 –2x + 3y = 0 x + 3y = 3 2 3 y = x y = – x + 1 1 3 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solutions 1. 2. intersection: (1, 4) Rewrite equations: intersection: (1, ) 3. Factors of –15 with a sum of –2: –5 and 3 x2 – 2x – 15 = (x – 5)(x + 3) 4. Factors of 14 with a sum of –9: –7 and –2 x2 – 9x + 14 = (x – 7)(x – 2) 2 3 6-4
4. Step 1: Graph y1 = x3 – 19x and y2 = –2x2 + 20 on a graphing calculator. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Graph and solve x3 – 19x = –2x2 + 20. Step 2: Use the Intersect feature to find the x values at the points of intersection. The solutions are –5, –1, and 4. 6-4
5. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (continued) Check: Show that each solution makes the original equation a true statement. x3 – 19x = –2x2 + 20 x3 – 19x = –2x2 + 20 (–5)3 – 19(–5) –2(–5)2 + 20 (–1)3 – 19(–1) –2(–1)2 + 20 –125 + 95 –50 + 20 –1 + 19 –2 + 20 –30 = –30 18 = 18 x3 – 19x = –2x2 + 20 (4)3 – 19(4) –2(4)2 + 20 64 – 76 –32 + 20 –12 = –12 Quick Check 6-4
6. 15.9 ft3 • = 27475.2 in.3Convert the volume to cubic inches. Graph y1 = 27475.2 and y2 = (x + 4)x(x – 3). Use the Intersect option of the calculator. When y = 27475.2, x 30. So x – 3 27 and x + 4 34. 123 in.3 ft3 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Quick Check The dimensions in inches of the cubicle area inside a doghouse can be expressed as width x, length x + 4, and height x – 3. The volume is 15.9 ft3. Find the dimensions of the doghouse. V = l • w • hWrite the formula for volume. 27475.2 = (x + 4)x(x – 3) Substitute. The dimensions of the doghouse are about 30 in. by 27 in. by 34 in. 6-4
7. Sum and Difference of Cubes
8. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Factor x3 – 125. x3 – 125 = (x)3 – (5)3Rewrite the expression as the difference of cubes. = (x – 5)(x2 + 5x + (5)2) Factor. = (x – 5)(x2 + 5x + 25) Simplify. Quick Check 6-4
9. 8x3 + 125 = (2x)3 + (5)3Rewrite the expression as the difference of cubes. = (2x + 5)((2x)2 – 10x + (5)2) Factor. = (2x + 5)(4x2 – 10x + 25) Simplify. 5 2 Since 2x + 5 is a factor, x = – is a root. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solve 8x3 + 125 = 0. Find all complex roots. The quadratic expression 4x2 – 10x + 25 cannot be factored, so use the Quadratic Formula to solve the related quadratic equation 4x2 – 10x + 25 = 0. 6-4
10. –b ± b2 – 4ac 2a x = Quadratic Formula = Substitute 4 for a, –10 for b, and 25 for c. –(–10) ± (–10)2 – 4(4)(25) 2(4) – (–10) ± –300 8 = Use the Order of Operations. 10 ± 10i 3 8 = –1 = 1 = Simplify. 5 2 5 ± 5i 3 4 5 ± 5i 3 4 The solutions are – and . Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (continued) Quick Check 6-4
11. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Factor x4 – 6x2 – 27. Step 1: Since x4 – 6x2 – 27 has the form of a quadratic expression, you can factor it like one. Make a temporary substitution of variables. x4 – 6x2 – 27 = (x2)2 – 6(x2) – 27 Rewrite in the form of a quadratic expression. = a2 – 6a – 27 Substitute a for x2. Step 2: Factor a2 – 6a – 27. a2 – 6a – 27 = (a + 3)(a – 9) Step 3: Substitute back to the original variables. (a + 3)(a – 9) = (x2 + 3)(x2 – 9) Substitute x2 for a. = (x2 + 3)(x + 3)(x – 3) Factor completely. Quick Check The factored form of x4 – 6x2 – 27 is (x2 + 3)(x + 3)(x – 3). 6-4
12. x = ± 3 or x = ± i 5 Solve for x, and simplify. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solve x4 – 4x2 – 45 = 0. x4 – 4x2 – 45 = 0 (x2)2 – 4(x2) – 45 = 0 Write in the form of a quadratic expression. Think of the expression as a2 – 4a – 45, which factors as (a – 9)(a + 5). (x2 – 9)(x2 + 5) = 0 (x – 3)(x + 3)(x2 + 5) = 0 x = 3 or x = –3 or x2 = –5 Use the factor theorem. Quick Check 6-4
13. ±2, ±i 7 5 ± 5i 3 4 –5, Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 1. Solve x3 – 2x2 – 3 = x – 4 by graphing. Where necessary, round to the nearest hundredth. Factor each expression. 2. 216x3 – 1 3. 8x3 + 125 4.x4 – 5x2 + 4 Solve each equation. 5.x3 + 125 = 0 6.x4 + 3x2 – 28 = 0 –0.80, 0.55, 2.25 (6x – 1)(36x2 + 6x + 1) (2x + 5)(4x2 – 10x + 25) (x + 1)(x – 1)(x + 2)(x – 2) 6-4
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# 5.3 The other trigonometric functions (Page 2/13)
Page 2 / 13
The point $\text{\hspace{0.17em}}\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\text{\hspace{0.17em}}$ is on the unit circle, as shown in [link] . Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t.$
$\mathrm{sin}\text{\hspace{0.17em}}t=-\frac{\sqrt{2}}{2},\mathrm{cos}\text{\hspace{0.17em}}t=\frac{\sqrt{2}}{2},\mathrm{tan}\text{\hspace{0.17em}}t=-1,\mathrm{sec}\text{\hspace{0.17em}}t=\sqrt{2},\mathrm{csc}\text{\hspace{0.17em}}t=-\sqrt{2},\mathrm{cot}\text{\hspace{0.17em}}t=-1$
## Finding the trigonometric functions of an angle
Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}t=\frac{\pi }{6}.$
We have previously used the properties of equilateral triangles to demonstrate that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{6}=\frac{1}{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{6}=\frac{\sqrt{3}}{2}.\text{\hspace{0.17em}}$ We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values.
$\begin{array}{l}\mathrm{sec}\frac{\pi }{6}=\frac{1}{\mathrm{cos}\frac{\pi }{6}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\hfill \end{array}$
$\mathrm{csc}\frac{\pi }{6}=\frac{1}{\mathrm{sin}\frac{\pi }{6}}=\frac{1}{\frac{1}{2}}=2$
$\begin{array}{l}\mathrm{cot}\frac{\pi }{6}=\frac{\mathrm{cos}\frac{\pi }{6}}{\mathrm{sin}\frac{\pi }{6}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\hfill \end{array}$
Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}t=\frac{\pi }{3}.$
$\begin{array}{l}\mathrm{sin}\frac{\pi }{3}=\frac{\sqrt{3}}{2}\\ \mathrm{cos}\frac{\pi }{3}=\frac{1}{2}\\ \mathrm{tan}\frac{\pi }{3}=\sqrt{3}\\ \mathrm{sec}\frac{\pi }{3}=2\\ \mathrm{csc}\frac{\pi }{3}=\frac{2\sqrt{3}}{3}\\ \mathrm{cot}\frac{\pi }{3}=\frac{\sqrt{3}}{3}\end{array}$
Because we know the sine and cosine values for the common first-quadrant angles, we can find the other function values for those angles as well by setting $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ equal to the cosine and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in [link] .
Angle $\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ Cosine 1 $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{1}{2}$ 0 Sine 0 $\frac{1}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{3}}{2}$ 1 Tangent 0 $\frac{\sqrt{3}}{3}$ 1 $\sqrt{3}$ Undefined Secant 1 $\frac{2\sqrt{3}}{3}$ $\sqrt{2}$ 2 Undefined Cosecant Undefined 2 $\sqrt{2}$ $\frac{2\sqrt{3}}{3}$ 1 Cotangent Undefined $\sqrt{3}$ 1 $\frac{\sqrt{3}}{3}$ 0
## Using reference angles to evaluate tangent, secant, cosecant, and cotangent
We can evaluate trigonometric functions of angles outside the first quadrant using reference angles as we have already done with the sine and cosine functions. The procedure is the same: Find the reference angle formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by x - and y -values in the original quadrant. [link] shows which functions are positive in which quadrant.
To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “A Smart Trig Class.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is “ A ,” a ll of the six trigonometric functions are positive. In quadrant II, “ S mart,” only s ine and its reciprocal function, cosecant, are positive. In quadrant III, “ T rig,” only t angent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “ C lass,” only c osine and its reciprocal function, secant, are positive.
Given an angle not in the first quadrant, use reference angles to find all six trigonometric functions.
1. Measure the angle formed by the terminal side of the given angle and the horizontal axis. This is the reference angle.
2. Evaluate the function at the reference angle.
3. Observe the quadrant where the terminal side of the original angle is located. Based on the quadrant, determine whether the output is positive or negative.
## Using reference angles to find trigonometric functions
Use reference angles to find all six trigonometric functions of $\text{\hspace{0.17em}}-\frac{5\pi }{6}.\text{\hspace{0.17em}}$
The angle between this angle’s terminal side and the x -axis is $\text{\hspace{0.17em}}\frac{\pi }{6},$ so that is the reference angle. Since $\text{\hspace{0.17em}}-\frac{5\pi }{6}\text{\hspace{0.17em}}$ is in the third quadrant, where both $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are negative, cosine, sine, secant, and cosecant will be negative, while tangent and cotangent will be positive.
$\begin{array}{l}\mathrm{cos}\left(-\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2},\mathrm{sin}\left(-\frac{5\pi }{6}\right)=-\frac{1}{2},\text{tan}\left(-\frac{5\pi }{6}\right)=\frac{\sqrt{3}}{3}\hfill \\ \text{sec}\left(-\frac{5\pi }{6}\right)=-\frac{2\sqrt{3}}{3},\text{csc}\left(-\frac{5\pi }{6}\right)=-2,\mathrm{cot}\left(-\frac{5\pi }{6}\right)=\sqrt{3}\hfill \end{array}$
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
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Probability of Simple Events Objective: Students will be able to find the probability of a simple event. Students will be able to understand the distinction
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• Probability
of Simple
Events
• Probability of Simple EventsObjective:Students will be able to find the probability of a simple event.Students will be able to understand the distinction between simple events and compound events.
Essential Question:(1) How do I find the probability of a simple event?(2) How can I distinguish between a simple and compound event?
• Probability of Simple EventsVocabulary:Outcome one possible result of a probability.Sample Space the list of possible outcomes for a probability event.Random outcomes that occur at random if each outcome is equally likely to occur.Simple Event a specific outcome or type of outcome. Complementary Events the events of one outcome happening and that outcomes not happening are complimentary; the sum of the probabilities of complementary events is 1.
• Real World Example:Best Buy is having an IPOD giveaway. They put all the IPOD Shuffles in a bag. Customers may choose an IPOD without looking at the color. Inside the bag are 4 orange, 5 blue, 6 green, and 5 pink IPODS. If Maria chooses one IPOD at random, what is the probability she will choose an orange IPOD?
Probability of Simple Events
• Real World Example:Best Buy is having an IPOD giveaway. They put all the IPOD Shuffles in a bag. Customers may choose an IPOD without looking at the color. Inside the bag are 4 orange, 5 blue, 6 green, and 5 pink IPODS. If Maria chooses one IPOD at random, what is the probability she will choose an orange IPOD?
P(orange) = 4/20 = 2/10 = 1/5 or 20%Probability of Simple Events
• What is a PROBABILITY?
- Probability is the chance that some event will happen
- It is the ratio of the number of ways a certain event can occur to the number of possible outcomesProbability of Simple Events
• What is a PROBABILITY?
number of favorable outcomes number of possible outcomes
Examples that use Probability:(1) Dice, (2) Spinners, (3) Coins, (4) Deck of Cards, (5) Evens/Odds, (6) Alphabet, etc.Probability of Simple EventsP(event) =
• What is a PROBABILITY?
0% 25% 50% 75% 100%
0 or .25 0r .5 or .75 1
Impossible Not Very Equally Likely Somewhat Certain Likely LikelyProbability of Simple Events
• Example 1: Roll a dice.What is the probability of rolling a 4?
# favorable outcomes # possible outcomes
1 6
The probability of rolling a 4 is 1 out of 6Probability of Simple EventsP(event) = P(rolling a 4) =
• Example 2: Roll a dice.What is the probability of rolling an even number?
# favorable outcomes # possible outcomes
3 1 6 2
The probability of rolling an even number is 3 out of 6 or .5 or 50%Probability of Simple EventsP(event) = P(even #) = =
• Example 3: Spinners.What is the probability of spinning green?
# favorable outcomes # possible outcomes
1 1 4 4
The probability of spinning green is 1 out of 4 or .25 or 25%Probability of Simple EventsP(event) = P(green) = =
• Example 4: Flip a coin.What is the probability of flipping a tail?
# favorable outcomes # possible outcomes
1 1 2 2
The probability of spinning green is 1 out of 2 or .5 or 50%Probability of Simple EventsP(event) = P(tail) = =
• Example 5: Deck of Cards.What is the probability of picking a heart?
# favorable outcomes 13 1
# possible outcomes 52 4The probability of picking a heart is1 out of 4 or .25 or 25%What is the probability of picking a non heart?
# favorable outcomes 39 3
# possible outcomes 52 4The probability of picking a heart is3 out of 4 or .75 or 75%
Probability of Simple EventsP(heart) = = =P(nonheart) = = =
• Key Concepts:
- Probability is the chance that some event will happen
- It is the ratio of the number of ways a certain even can occur to the total number of possible outcomesProbability of Simple Events
• Guided Practice: Calculate the probability of each independent event.
1) P(black) = 2) P(1) = 3) P(odd) = 4) P(prime) = Probability of Simple Events
• Guided Practice: Answers.
1) P(black) = 4/82) P(1) = 1/83) P(odd) = 1/24) P(prime) = 1/2Probability of Simple Events
• Independent Practice: Calculate the probability of each independent event.
1) P(red) = 2) P(2) = 3) P(not red) = 4) P(even) = Probability of Simple Events
• Independent Practice: Answers.
1) P(red) = 1/22) P(2) = 1/43) P(not red) = 1/24) P(even) = 1/2Probability of Simple Events
• Real World Example:A computer company manufactures 2,500 computers each day. An average of 100 of these computers are returned with defects. What is the probability that the computer you purchased is not defective?
P(not defective) = # not defective= 2,400 = 24 total # manufactured 2,500 25Probability of Simple Events
• Homework:- Workbook Practice 11-1
Probability of Simple Events
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