text:
## anonymous 5 years ago i have a couple of math problems to solve anyone care to help me one on one plz...i really appreciate it

1. anonymous

2. anonymous

$h(x)=\sqrt{4x+8}$ 4x+8≥0 4x≥-8 x≥-2 D:(x I x≥-2, x E R)

3. anonymous

I REALLY THANK U FOR HELPING ME OUT

4. anonymous

you're welcome

5. anonymous

I GET CONFUSED BECAUSE I HAVENT DONE THIS MATH IN A LONG WHILE..CAN U FINISH HELPING ME PLZ...ONLY IF U CAN

6. anonymous

ok number 11 a) f(x) - g(x) + h(x) just plug the functions f(x), g(x), and h(x) you'll get the result = -2x^2 + 27x -4

7. anonymous

b) g(x)/h(x) again, just plug g(x) and h(x) (10x^2 - 15x)/5x = 2x-3

8. anonymous

im confused on number 11a

9. anonymous

i inserted f g h

10. anonymous

but i got: $f(23x-4) - G(85) + h(5X)$

11. anonymous

f(x) - g(x) + h(x) = (8x^2 +7x-4) - (10x^2 -15x) + (5x) 8x^2 +7x-4 - 10x^2 + 15x + 5x

12. anonymous

where did you get that? you just need to substitute f(x) with 8x^2 +7x-4 substitute g(x) with 10x^2 -15x and substitute h(x) with 5x

13. anonymous

i got confused ...now i see it...ughhhhh *sign*

14. anonymous

15. anonymous

number 12 is about function composition g(f(3)) first, find f(3) f(x) = 8x^2 +7x-4 so f(3) = 8(3)^2 + 7(3) -4 f(3) = 8(9) + 21 -4 f(3) = 72+21-4 f(3) = 89 therefore g(f(3)) = g(89) g(x) = 10x^2 -15x g(89) = 10(89)^2 - 15(89)

16. anonymous

number 12 is confusing "O

17. anonymous

13) because M is the midpoint, AM=MB 2x^2 - 10x + 15 = x^2 - 2x + 3 x^2 - 8x + 12 = 0 (x-6)(x-2) x = 6 or x=2

18. anonymous

this is an article from wikipedia , related to number 12 http://en.wikipedia.org/wiki/Function_composition

19. anonymous

thanks

20. anonymous

21. anonymous

14) AC is an angle bisector, therefore angle BAD = angle CAD x^2 = 5x + 24 x^2 - 5x - 24=0 (x-8)(x+3) x=8 or x=-3

22. anonymous

15) BAD is a right angle 4x+8+7x+5 = 90 degree 11x + 13 = 90 11x = 77 x=7

23. anonymous

16) 8x+12 + 2x+8 = 180 degree 10x + 20 = 180 10x = 160 x= 16

24. anonymous

25. anonymous

17) x^2 = 6x-8

26. anonymous

for number 12 how did we get 8 if the problem is f(x) = 3x -4 and g(x) = x to the 2nd -5 you dont have to redo the problem just need to know how 8 got there

27. anonymous

oh, sorry, i did it wrong, i didnt see the f(x) and g(x) in number 11 and 12 are different, i'll do it again

28. anonymous

ok thanks

29. anonymous

f(x) = 3x -4 and g(x) = x^2 - 5 f(3) = 3(3) - 4 = 9-4 = 5 because f(3) = 5, g(f(3)) = g(5) g(5) = (5)^2 - 5 = 25-5 = 20

30. anonymous

thank you very much for correcting it ")

31. anonymous

32. anonymous

19, because it's an iscoceles triangle, angle C must be the same as angle A. a triangle has an angle of 180 degree, so : A + B + C = 180 degree 2x^2 + x^2 + 2x^2 = 180

33. anonymous

34. anonymous

god ur good at this

35. anonymous

can i borrow ur brains for my finals...lol ")

36. anonymous

this is for number 20

37. anonymous

lol, not really good, actually, this is kinda like SAT problems

38. anonymous

damn then i am dumb "(

39. anonymous

40. anonymous

you just need to keep practicing, you'll be good at this. sorry i can't continue this, i gotta go. bye

41. anonymous

ok thank u very much for all ur help


text:
Hey! I'm David, the author of the Real-World Cryptography book. I'm a crypto engineer at O(1) Labs on the Mina cryptocurrency, previously I was the security lead for Diem (formerly Libra) at Novi (Facebook), and a security consultant for the Cryptography Services of NCC Group. This is my blog about cryptography and security and other related topics that I find interesting.

# How to check if a binary contains the Dual EC backdoor for the NSA posted December 2015

tl;dr:

this is what you should type:

strings your_binary | grep -C5 -i "c97445f45cdef9f0d3e05e1e585fc297235b82b5be8ff3efca67c59852018192\|8e722de3125bddb05580164bfe20b8b432216a62926c57502ceede31c47816edd1e89769124179d0b695106428815065\|1b9fa3e518d683c6b65763694ac8efbaec6fab44f2276171a42726507dd08add4c3b3f4c1ebc5b1222ddba077f72943b24c3edfa0f85fe24d0c8c01591f0be6f63"

After all the Jupiner fiasco, I wondered how people could look if a binary contained an implementation of Dual EC, and worse, if it contained Dual EC with the NSA's values for P and Q.

The easier thing I could think of is the use of strings to check if the binary contains the hex values of some Dual EC parameters:

strings your_binary | grep -C5 -i python -c "print '%x' % 115792089210356248762697446949407573530086143415290314195533631308867097853951"

This is the value of the prime p of the curve P-256. Other curves can be used for Dual EC though, so you should also check for the curve P-384:

strings your_binary | grep -C5 -i python -c "print '%x' % 39402006196394479212279040100143613805079739270465446667948293404245721771496870329047266088258938001861606973112319"

and the curve P-521:

strings your_binary | grep -C5 -i python -c "print '%x' % 6864797660130609714981900799081393217269435300143305409394463459185543183397656052122559640661454554977296311391480858037121987999716643812574028291115057151 "

I checked the binaries of ScreenOS (taken from here) and they contained these three curves parameters. But this doesn't mean anything, just that these curves are stored, maybe used, maybe used for Dual EC...

To check if it uses the NSA's P and Q, you should grep for P and Q x coordinates from the same NIST paper.

This looks for all the x coordinates of the different P for each curves. This is not that informative since it is the standard point P as a generator of P-256

strings your_binary | grep -C5 -i "6b17d1f2e12c4247f8bce6e563a440f277037d812deb33a0f4a13945d898c296\|aa87ca22be8b05378eb1c71ef320ad746e1d3b628ba79b9859f741e082542a385502f25dbf55296c3a545e3872760ab7\|c6858e06b70404e9cd9e3ecb662395b4429c648139053fb521f828af606b4d3dbaa14b5e77efe75928fe1dc127a2ffa8de3348b3c1856a429bf97e7e31c2e5bd66"

Testing the ScreenOS binaries, I get all the matches. This means that the parameters for P-256 and maybe Dual EC are indeed stored in the binaries.

weirdly, testing for Qs I don't get any match. So Dual EC or not?

strings your_binary | grep -C5 -i "c97445f45cdef9f0d3e05e1e585fc297235b82b5be8ff3efca67c59852018192\|8e722de3125bddb05580164bfe20b8b432216a62926c57502ceede31c47816edd1e89769124179d0b695106428815065\|1b9fa3e518d683c6b65763694ac8efbaec6fab44f2276171a42726507dd08add4c3b3f4c1ebc5b1222ddba077f72943b24c3edfa0f85fe24d0c8c01591f0be6f63"

The Dual_EC_DRBG 'Q' parameter was replaced with 9585320EEAF81044F20D55030A035B11BECE81C785E6C933E4A8A131F6578107 and the secondary ANSI X.9.31 PRNG was broken, allowing raw Dual_EC output to be exposed to the network. Please see this blog post for more information.

Diffing the vulnerable (and patched binaries. I see that only the P-256 curve $Q$ was modified from Juniper's values, other curves were left intact. I guess this means that only the P-256 curve was being used in Dual EC.

If you know how Dual EC works (if you don't check my video), you know that to establish a backdoor into it you need to generate $P$ and $Q$ accordingly. So changing the value $Q$ with no correlation to $P$ is not going to work, worse it could be that $Q$ is too "close" to P and thus the secret $d$ linking them could be easily found ( $P = dQ$).

Now one clever way to generate a secure $Q$ with a strong value $d$ that only you would know is to choose a large and random $d$ and calculate its inverse $d^{-1} \pmod{ord_{E}}$. You have your $Q$ and your $d$!

$d^{-1} P = Q$

bonus: here's a small script that attempts to find $d$ in the hypothesis $d$ would be small (the fastest way to compute an elliptic curve discrete log is to use Pollard Rho's algorithm)

p256 = 115792089210356248762697446949407573530086143415290314195533631308867097853951
a256 = p256 - 3
b256 =  41058363725152142129326129780047268409114441015993725554835256314039467401291

## base point values
gx = 48439561293906451759052585252797914202762949526041747995844080717082404635286
gy = 36134250956749795798585127919587881956611106672985015071877198253568414405109

## order of the curve
qq = 115792089210356248762697446949407573529996955224135760342422259061068512044369

# init curve
FF = GF(p256)
EE = EllipticCurve([FF(a256), FF(b256)])

# define the base point
G = EE(FF(gx), FF(gy))

# P and Q
P = EE(FF(0x6b17d1f2e12c4247f8bce6e563a440f277037d812deb33a0f4a13945d898c296), FF(0x4fe342e2fe1a7f9b8ee7eb4a7c0f9e162bce33576b315ececbb6406837bf51f5))

# What is Q_y ?
fakeQ_x = FF(0x9585320EEAF81044F20D55030A035B11BECE81C785E6C933E4A8A131F6578107)
fakeQ = EE.lift_x(fakeQ_x)

print discrete_log(P, fakeQ, fakeQ.order(), operation='+')

The lift_x function allows me to get back the $y$ coordinate of the new $Q$:

EE.lift_x(fakeQ_x)
(67629950588023933528541229604710117449302072530149437760903126201748084457735 : 36302909024827150197051335911751453929694646558289630356143881318153389358554 : 1)
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# Analyze a TLS handshake posted December 2015

Short blogpost on a quick way to analyze a TLS handshake:

In one terminal, setup the server:

openssl req -x509 -newkey rsa:2048 -keyout key.pem -out cert.pem -nodes
openssl s_server -cert cert.pem -key key.pem

The first command use the req toolkit of openssl. It is usually used to create certificate request (and the request is then later approved by a CA), but since we're in a rush, let's just use it with the -x509 option to directly generate a certificate. rsa:2048 generates a key with the algorithm RSA and with a modulus of size 2048 bits. -nodes disable the use of a passphrase to protect the key (the default protect the key by encrypting it with DES).

In a second terminal, start capturing packets:

tcpdump -i lo0 -s 65535 -w exchange.cap

65535 is the maximum length of a packet.

Start a handshake in a third terminal:

openssl s_client -connect localhost:4433

Now open the .cap with Wireshark!

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# How to efficiently compute a batch GCD posted December 2015

Heard a bit late about the factorable research results and how they used batch gcd to recover a bunch of servers' private keys.

The question one could think of is how to efficiently do a batch gcd on a big set of public keys?

From this utility:

• Actual pairwise GCD
This performs n*(n-1)/2 GCD operations on the moduli. This is slow. Don't use this.
• Accumulating Product
This iterates over all input moduli, performing a GCD of each one against the product of all previous. Once it finds a candidate, it scans all previous moduli to find out which ones it shared a factor with (either GCD or division, depending on whether one or both were found). The main scan cannot be done in parallel, and even though it seems like this is O(n), the increasing size of the accumulated product results it lots of long multiplication and long divison so it's still painfully slow for large numbers of moduli.

Looks like the most efficient ways come from Dan Bernstein (again!), in a 7 pages paper

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# List of terms from "Threat Model for BGP Path Security" posted December 2015

the Threat Model for BGP Path Security document lists, as RFCs usually do, relevant terms with their respective definitions. It can be a quick way to get an understanding of these abbreviations you often come across but never dare to google:

• Autonomous System (AS): An AS is a set of one or more IP networks operated by a single administrative entity.

• AS Number (ASN): An ASN is a 2- or 4-byte number issued by a registry to identify an AS in BGP.

• Border Gateway Protocol (BGP): A path vector protocol used to convey "reachability" information among ASes in support of inter-domain routing.

• False (Route) Origination: If a network operator originates a route for a prefix that the operator does not hold (and that has not been authorized to originate by the prefix holder), this is termed false route origination.

• Internet Service Provider (ISP): An organization managing (and typically selling) Internet services to other organizations or individuals.

• Internet Number Resources (INRs): IPv4 or IPv6 address space and ASNs.

• Internet Registry: An organization that manages the allocation or distribution of INRs. This encompasses the Internet Assigned Number Authority (IANA), Regional Internet Registries (RIRs), National Internet Registries (NIRs), and Local Internet Registries (LIRs) (network operators).

• Network Operator: An entity that manages an AS and thus emits (E)BGP updates, e.g., an ISP.

• Network Operations Center (NOC): A network operator employs a set of equipment and a staff to manage a network, typically on a 24/7 basis. The equipment and staff are often referred to as the NOC for the network.

• Prefix: A prefix is an IP address and a mask used to specify a set of addresses that are grouped together for purposes of routing.

• Public Key Infrastructure (PKI): A PKI is a collection of hardware, software, people, policies, and procedures used to create, manage, distribute, store, and revoke digital certificates.

• Relying Parties (RPs): An RP is an entity that makes use of signed products from a PKI, i.e., it relies on signed data that is verified using certificates and Certificate Revocation Lists (CRLs) from a PKI.

• RPKI Repository System: The RPKI repository system consists of a distributed set of loosely synchronized databases.

• Resource PKI (RPKI): A PKI operated by the entities that manage INRs and that issue X.509 certificates (and CRLs) that attest to the holdings of INRs.

• RPKI Signed Object: An RPKI signed object is a data object encapsulated with Cryptographic Message Syntax (CMS) that complies with the format and semantics defined in [RFC6488].

• Route: In the Internet, a route is a prefix and an associated sequence of ASNs that indicates a path via which traffic destined for the prefix can be directed. (The route includes the origin AS.)

• Route Leak: A route leak is said to occur when AS-A advertises routes that it has received from AS-B to the neighbors of AS-A, but AS-A is not viewed as a transit provider for the prefixes in the route.
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# Zeroisation posted December 2015

In cryptography, zeroisation (also spelled zeroization) is the practice of erasing sensitive parameters (electronically stored data, cryptographic keys, and CSPs) from a cryptographic module to prevent their disclosure if the equipment is captured. This is generally accomplished by altering or deleting the contents to prevent recovery of the data. When encryption was performed by mechanical devices, this would often mean changing all the machine's settings to some fixed, meaningless value, such as zero. On machines with letter settings rather than numerals, the letter 'O' was often used instead. Some machines had a button or lever for performing this process in a single step. Zeroisation would typically be performed at the end of an encryption session to prevent accidental disclosure of the keys, or immediately when there was a risk of capture by an adversary.

from Wikipedia

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# How not to mess with implementing a stateful signature scheme posted December 2015

The Mcgrew draft (on the old stateful signature scheme LMS) has a section on "how not to mess with implementing a stateful scheme". It's pretty scary.

The LMS signature system, like all N-time signature systems, requires that the signer maintain state across different invocations of the signing algorithm, to ensure that none of the component one-time signature systems are used more than once. This section calls out some important practical considerations around this statefulness.

In a typical computing environment, a private key will be stored in non-volatile media such as on a hard drive. Before it is used to sign a message, it will be read into an application's Random Access Memory (RAM). After a signature is generated, the value of the private key will need to be updated by writing the new value of the private key into non-volatile storage. It is essential for security that the application ensure that this value is actually written into that storage, yet there may be one or more memory caches between it and the application. Memory caching is commonly done in the file system, and in a physical memory unit on the hard disk that is dedicated to that purpose. To ensure that the updated value is written to physical media, the application may need to take several special steps. In a POSIX environment, for instance,the O_SYNC flag (for the open() system call) will cause invocations of the write() system call to block the calling process until the data has been to the underlying hardware. However, if that hardware has its own memory cache, it must be separately dealt with using an operating system or device specific tool such as hdparm to flush the on-drive cache, or turn off write caching for that drive. Because these details vary across different operating systems and devices, this note does not attempt to provide complete guidance; instead, we call the implementer's attention to these issues.

When hierarchical signatures are used, an easy way to minimize the private key synchronization issues is to have the private key for the second level resident in RAM only, and never write that value into non-volatile memory. A new second level public/private key pair will be generated whenever the application (re)starts; thus, failures such as a power outage or application crash are automatically accommodated. Implementations SHOULD use this approach wherever possible.

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# Hash-Based Signatures Part IV: XMSS and SPHINCS posted December 2015

This post is the ending of a series of blogposts on hash-based signatures. You can find part I here

So now we're getting into the interesting part, the real signatures schemes.

PQCrypto has released an initial recommendations document a few months ago. The two post-quantum algorithms advised there were XMSS and SPHINCS:

This blogpost will be presenting XMSS, a stateful signature scheme, while the next will focus on SPHINCS, the first stateless signature scheme!

## XMSS

The eXtended Merkle Signature Scheme (XMSS) was introduced in 2011 and became an internet-draft in 2015.

The main construction looks like a Merkle tree, excepts a few things. The XMSS tree has a mask XORed to the child nodes before getting hashed in their parents node. It's a different mask for every node:

The second particularity is that a leaf of the XMSS tree is not a hash of a one-time signature public key, but the root of another tree called a L-tree.

A L-tree has the same idea of masks applied to its nodes hashes, different from the main XMSS Trees, but common to all the L-trees.

Inside the leaves of any L-tree are stored the elements of a WOTS+ public key. This scheme is explained at the end of the first article of this series.

If like me you're wondering why they store a WOTS+ public key in a tree, here's what Huelsing has to say about it:

The tree is not used to store a WOTS public key but to hash it in a way that we can prove that a second-preimage resistant hash function suffices (instead of a collision resistant one).

Also, the main public key is composed of the root node of the XMSS tree as well as the bit masks used in the XMSS tree and a L-tree.

## SPHINCS

SPHINCS is the more recent one, combining a good numbers of advances in the field and even more! Bringing the statelessness we were all waiting for.

Yup, this means that you don't have to keep the state anymore. But before explaining how they did that, let's see how SPHINCS works.

First, SPHINCS is made out of many trees.

Let's look at the first tree:

• Each node is the hash of the XOR of the concatenation of the previous nodes with a level bitmask.
• The public key is the root hash along with the bitmasks.
• The leaves of the tree are the compressed public keys of WOTS+ L-trees.

See the WOTS+ L-trees as the same XMSS L-tree we previously explained, except that the bitmask part looks more like a SPHINCS hash tree (a unique mask per level).

Each leaves, containing one Winternitz one-time signature, allow us to sign another tree. So that we know have a second layer of 4 SPHINCS trees, containing themselves WOTS+ public keys at their leaves.

This go on and on... according to your initial parameter. Finally when you reach the layer 0, the WOTS+ signatures won't sign other SPHINCS trees but HORS trees.

A HORST or HORS tree is the same as a L-tree but this time containing a HORS few-time signature instead of a Winternitz one-time signature. We will use them to sign our messages, and this will increase the security of the scheme since if we do sign a message with the same HORS key it won't be a disaster.

Here's a diagram taken out of the SPHINCS paper making abstraction of WOTS+ L-trees (displaying them as signature of the next SPHINCS tree) and showing only one path to a message.

When signing a message M you first create a "randomized" hash of M and a "random" index. I put random in quotes because everything in SPHINCS is deterministically computed with a PRF. The index now tells you what HORST to pick to sign the randomized hash of M. This is how you get rid of the state: by picking an index deterministically according to the message. Signing the same message again should use the same HORST, signing two different messages should make use of two different HORST with good probabilities.

And this is how this series end!

EDIT: here's another diagram from Armed SPHINCS, I find it pretty nice!

# Hash-Based Signatures Part III: Many-times Signatures posted December 2015

We saw previously what were one-time signatures (OTS), then what were few-time signatures (FTS). But now is time to see how to have practical signature schemes based on hash functions. Signature schemes that you can use many times, and ideally as many times as you'd want.

If you haven't read Part I and Part II it's not necessarily a bad thing since we will make abstraction of those. Just think about OTS as a public key/private key pair that you can only use once to sign a message.

## Dumb trees

The first idea that comes to mind could be to use a bunch of one-time signatures (use your OTS scheme of preference). The first time you would want to sign something you would use the first OTS keypair, and then never use it again. The second time you would want to sign something, you would use the second OTS keypair, and then never use it again. This can get repetitive and I'm sure you know where I'm going with this. This would also be pretty bad because your public key would consist of all the OTS public keys (and if you want to be able to use your signature scheme a lot, you will have a lot of OTS public keys).

One way of reducing the storage amount, secret-key wise, is to use a seed in a pseudo-random number generator to generate all the secret keys. This way you don't need to store any secret-key, only the seed.

But still, the public key is way too large to be practical.

## Merkle trees

To link all of these OTS public keys to one main public keys, there is one very easy way, it's to use a Merkle tree. A solution invented by Merkle in 1979 but published a decade later because of some uninteresting editorial problems.

Here's a very simple definition: a Merkle tree is a basic binary tree where every node is a hash of its childs, the root is our public key and the leaves are the hashes of our OTS public keys. Here's a drawing because one picture is clearer than a thousand words:

So the first time you would use this tree to sign something: you would use the first OTS public key (A), and then never use it again. Then you would use the B OTS public key, then the C one, and finally the D one. So you can sign 4 messages in total with our example tree. A bigger tree would allow you to sign more messages.

The attractive idea here is that your public key only consist of the root of the tree, and every time you sign something your signature consists of only a few hashes: the authentication path.

In our example, a signature with the first OTS key (A) would be: (1, signature, public key A, authentication path)

• 1 is the index of the signing leaf. You have to keep that in mind: you can't re-use that leaf's OTS. This makes our scheme a stateful scheme.

• The signature is our OTS published secret keys (see the previous parts of this series of articles).

• The public key is our OTS public key, to verify the signature.

• The authentication path is a list of nodes (so a list of hashes) that allows us to recompute the root (our main public key).

Let's understand the authentication path. Here's the previous example with the authentication path highlighted after signing something with the first OTS (A).

We can see that with our OTS public key, and our two hashes (the neighbor nodes of all the nodes in the path from our signing leaf to the root) we can compute the main public key. And thus we can verify that this was indeed a signature that originated from that main public key.

Thanks to this technique we don't to know all of the OTS public keys to verify that main public key. This saves space and computation.

And that's it, that's the simple concept of the Merkle's signature scheme. A many-times signature scheme based on hashes.

...part IV is here

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format: plain


text:
# 9.2.2A Represent, Solve & Translate

9-12
Subject:
Math
Strand:
Algebra
Standard 9.2.2

Recognize linear, quadratic, exponential and other common functions in real-world and mathematical situations; represent these functions with tables, verbal descriptions, symbols and graphs; solve problems involving these functions, and explain results in the original context.

Benchmark: 9.2.2.1 Represent & Solve: Linear & Quadratic

Represent and solve problems in various contexts using linear and quadratic functions.

For example: Write a function that represents the area of a rectangular garden that can be surrounded with 32 feet of fencing, and use the function to determine the possible dimensions of such a garden if the area must be at least 50 square feet.

Benchmark: 9.2.2.2 Represent & Solve: Exponential

Represent and solve problems in various contexts using exponential functions, such as investment growth, depreciation and population growth.

Benchmark: 9.2.2.3 Translate Between Representations

Sketch graphs of linear, quadratic and exponential functions, and translate between graphs, tables and symbolic representations. Know how to use graphing technology to graph these functions.

## Overview

Big Ideas and Essential Understandings

Standard 9.2.2 Essential Understandings

Mathematical modeling is a process that allows students to describe and make sense of the relationship between an independent variable and a dependent variable found in a real-world setting. This standard specifies that high school students should be able to model situations using a variety of functions that include linear, quadratic, exponential, absolute value, inverse, power, square root, and other common functions. Students need to be able to fluidly translate among tables, graphs, and equations of these functions in order to match key features of these functions to real-world situations. The key features that separate these function types include intercepts, asymptotes, restricted domain, and rate of change. As students learn to make connections between the representations of the models and the real-world situation they begin to understand how variables in their life are connected to the mathematics they study in school as well as extend their knowledge of function types themselves.

All Standard Benchmarks

9.2.2.1
Represent and solve problems in various contexts using linear and quadratic functions.
For example: Write a function that represents the area of a rectangular garden that can be surrounded with 32 feet of fencing, and use the function to determine the possible dimensions of such a garden if the area must be at least 50 square feet.

9.2.2.2
Represent and solve problems in various contexts using exponential functions, such as investment growth, depreciation and population growth.

9.2.2.3
Sketch graphs of linear, quadratic and exponential functions, and translate between graphs, tables and symbolic representations. Know how to use graphing technology to graph these functions.

9.2.2.4
Express the terms in a geometric sequence recursively and by giving an explicit (closed form) formula, and express the partial sums of a geometric series recursively.

For example: A closed form formula for the terms tn in the geometric sequence 3, 6, 12,24, ... is tn = 3(2)n-1, where n = 1, 2, 3, ... , and this sequence can be expressed recursively by writing t1 = 3 and  tn = 2tn-1, for n ≥ 2.

Another example: the partial sums sn of the series 3 + 6 + 12 + 24 + ... can be expressed recursively by writing s1 = 3 and sn = 3 + 2sn-1, for n ≥ 2.

9.2.2.5
Recognize and solve problems that can be modeled using finite geometric sequences and series, such as home mortgage and other compound interest examples. Know how to use spreadsheets and calculators to explore geometric sequences and series in various contexts.

9.2.2.6
Sketch the graphs of common non-linear functions such as $f\left ( x \right )=\sqrt{x}$, $f\left ( x \right )=\left |x\right |$, $f\left ( x \right )=\frac{1}{x}$,  $f\left ( x \right )=x^{3}$, and translations of these functions, such as $f\left ( x \right )=\sqrt{x-2}+4$. Know how to use graphing technology to graph these functions.

Benchmark Cluster

Benchmark Group A

9.2.2.1
Represent and solve problems in various contexts using linear and quadratic functions.

For example: Write a function that represents the area of a rectangular garden that can be surrounded with 32 feet of fencing, and use the function to determine the possible dimensions of such a garden if the area must be at least 50 square feet.

9.2.2.2
Represent and solve problems in various contexts using exponential functions, such as investment growth, depreciation and population growth.

9.2.2.3
Sketch graphs of linear, quadratic and exponential functions, and translate between graphs, tables and symbolic representations. Know how to use graphing technology to graph these functions.

What students should know and be able to do [at a mastery level] related to these benchmarks

• Representing and solving problems using mathematical models can be sorted into two categories. Some problem situations match mathematical models exactly while in other problem situations the model can only approximate the data. Most experiments that involve measurement or the use of real-world data will have models that approximate the relationship. Students should be able to select and find an exact model when appropriate and an approximate model when appropriate.
 Exact Model -  (equation not given to students)Toothpicks are used to make this picture of 5 overlapping triangles.a)      How many toothpicks are used to make the 5 overlapping triangles?b)      How many toothpicks would be used to make a picture with 13 overlapping triangles?c)      How many toothpicks are needed to make a picture with n overlapping triangles?d)     How many additional toothpicks are added each time you add a triangle?e)      How many overlapping triangles can be made using 62 toothpicks?
 Approximate Model -  (equation not given to students)Shopping carts in the parking lots of stores are put together to form trains. Susan English and Tim Erickson measured 12 different trains. They counted the number of carts in each train and measured the length of that train of carts in centimeters. a)      How long is one shopping cart?b)      How much does one shopping cart fit into the next one?c)      One set of carts was 486 cm long. How many carts?d)     Find an equation that models the length of a shopping cart train in centimeters given the number of shopping carts in the train.http://www.eeps.com/zoo/cages/ShoppingCarts.html
• Students should be able to identify the variables involved in a situation and write equations to show how they are related. These situations can involve linear, quadratic, or exponential functions. In benchmark 9.2.2.1 the example provided asks student to solve the following problem:

"Write a function that represents the area of a rectangular garden that can be surrounded with 32 feet of fencing, and use the function to determine the possible dimensions of such a garden if the area must be at least 50 square feet."

Students need to identify the independent and dependent variables in this situation. The dependent variable is the area of the rectangle and is clearly mentioned in the problem. The independent variable is harder to identify in this situation since it could be the length, width, or even both dimensions of the rectangle. Students need to recognize that the independent variable could be either the length or the width and further that since the perimeter of the fence is fixed then once the length of the rectangle is specified then so is the width. Therefore if students select the length of the rectangle as the independent variable then the area of the rectangle is dependent on that choice. The students then need to be able to combine the formulas, $A_{rectangle}=l\times w$ and $P_{rectangle}=2l+2w$, to get the function $A(l)=l(16-l)$.

• Students should know that exponential functions have a multiplicative rate of change and the factors that cause exponential growth or decay. Given specific instances of the equation $f(x)=a\cdot b^{c-x}$, they should be able to distinguish between functions that describe exponential growth and ones that describe decay.
• Once students are able to plot a variety of functions point by point they need to move past seeing a function as a collection of points to one object representing a relationship between infinitely many points. Students should be able to quickly sketch the graph of a line from the equation by identifying intercepts and slope. They should be able to quickly recognize the various symbolic forms of a quadratic function and sketch the graph based on the format presented. Students should be able to identify and use the y-intercept and horizontal asymptote to sketch the graph of an exponential function given its equation.
• Students should be able to identify key features of a set of bivariate data presented in graphical form so they can determine the type of function that would best model the relationship between the two variables and then relate these features to the original context of the problem. Examples are shown below.
 graph what they should see Students should be able to explain that there is a positive relationship between the variables (i.e. as the number of shopping carts increases the length of the train increases). They should be able to explain that the relationship between the variables looks linear and find a linear function that fits the data. They should further be able to explain how key features (i.e. slope and y-intercept) of the function relate to the original context. The graph shows the actual age of dogs and what veterinarians use for approximate human age of dogs. Students should be able to explain that the relationship between actual age and human equivalent is positive and why it makes sense in the original context. The students should be able to rule out an exponential model based on the y-intercept of (0,0) and the absence of a horizontal asymptote. They should be able to reason that a square root function or a quadratic function could model the data and why.
• Students should be familiar with the syntax required to graph various functions on a graphing utility and should be able to scale the window appropriately given the equation. For example, if a student was asked to graph the function y = (x + 5)(x - 23) they should be able to adjust the window to show the important features of the function including the vertex, y-intercept, and x-intercepts.
 not so good better best graph of y = (x + 5)(x - 23) using the "standard" window graph of y = (x + 5)(x - 23) showing the x-intercepts but not the vertex graph of y = (x + 5)(x - 23) showing vertex, y-intercept, and x-intercepts
• Students should be able to use a graphing utility to test and revise function models given a scatterplot of the data.
 The data in the graph show the population of the United States from 1800 to present. Students should be able to input the data into a graphing utility and reason that the relationship is exponential. The graph on the left shows an initial guess that should be adjusted further to fit the relationship better. Students should be able to enter the functions that they have studied and then make adjustments to find a model that captures the essence of the data well.
• One of the most important features of using a graphing utility is the ability to link representations. A change in the equation results in corresponding changes in the table and graph. Students need to be familiar enough with the graphing utility they are using to easily make connections between the table, graph, and equation and be able to make adjustments in the display of any of the representations to highlight important features.

Work from previous grades that supports this new learning includes:

• In middle school students learn to use variables as pattern generalizers instead of simply representing unknowns. Knowing how to describe and use variables in these situations is crucial for students entering high school as they expand their knowledge of different non-linear functions.
• Students in middle school know how to find exact or approximate rates of change for linear functions given a table, graph, or equation. They can also determine that a function is non-linear given a table or graph.
• Students should know how to graph linear functions on a graphing utility and be able to access the table, graph, and equation features. They should be able to change the window settings to show important points, change the table settings to access trends in the values of the table, and be able to change the settings on the how the graphs are displayed (colors, dotted, dashed, etc.) on the graph.
Correlations

NCTM Standards: (those that apply to 9.2.2 are bolded)

(p.296, PSSM) Instructional programs from Pre-K through grade 12 should enable all students to

Understand patterns, relations, and functions.  In grades 9-12 all students should

• generalize patterns using explicitly defined and recursively defined functions;
• understand relations and functions and select, convert flexibly among, and use various representations for them;
• analyze functions of one variable by investigating rates of change, intercepts, zeros, asymptotes, and local and global behavior;
• understand and perform transformations such as arithmetically combining, composing, and inverting commonly used functions, using technology to perform such operations on more-complicated symbolic expressions;
• understand and compare the properties of classes of functions, including exponential, polynomial, rational, logarithmic, and periodic functions;
• interpret representations of functions of two variables.

Represent and analyze mathematical situations and structures using algebraic symbols. In grades 9-12 all students should

• understand the meaning of equivalent forms of expressions, equations, inequalities, and relations;
• write equivalent forms of equations, inequalities, and systems of equations and solve them with fluency - mentally or with paper and pencil in simple cases and using technology in all cases;
• use symbolic algebra to represent and explain mathematical relationships;
• use a variety of symbolic representations, including recursive and parametric equations, for functions and relations;
• judge the meaning, utility, and reasonableness of the results of symbol manipulations, including those carried out by technology.

Use mathematical models to represent and understand quantitative relationships. In grades 9-12 all students should

• identify essential quantitative relationships in a situation and determine the class or classes of functions that might model the relationships;
• use symbolic expressions, including iterative and recursive forms, to represent relationships arising from various contexts;
• draw reasonable conclusions about a situation being modeled.

Analyze change in various contexts. In grades 9-12 all students should

• approximate and interpret rates of change from graphical and numerical data.

Common Core State Standards (CCSM)

### Interpret functions that arise in applications in terms of the context.

·         F-IF.4.      For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.

·         F-IF.5.      Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function.

·         F-IF.6.      Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.

## Analyze functions using different representations.

·         F-IF.7.      Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.

• Graph linear and quadratic functions and show intercepts, maxima, and minima.
• Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions.
• Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior.
• (+) Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior.
• Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude.

·         F-IF.8.      Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.

• Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context.
• Use the properties of exponents to interpret expressions for exponential functions. For example, identify percent rate of change in functions such as $y=(1.02)^{t}$, $y=(0.97)^{t}$, $y=(1.01)^{12t}$, $y=(1.2)^{\frac{t}{10}}$, and classify them as representing exponential growth or decay.

·         F-IF.9.      Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a graph of one quadratic function and an algebraic expression for another, say which has the larger maximum.

## Misconceptions

Student Misconceptions

Student Misconceptions and Common Errors

• Students sometimes only focus on the additive rate of change of the dependent variable to determine the rate of change of a function.

table

students say

students should say

 x y 0 3 1 5 2 7 3 9

"The slope is two since the values in the y-column are going up by two."

This response is correct but the rate of change of a function relies on coordinating the change in both variables. A statement that highlights this coordination would be: "The values of the y-variable are increasing by two for each change of one in the x-variable.

 x y 0 7 5 10 10 13 15 16

"The relationship is linear and it has a slope of three."

"The relationship is linear and the value of the y-variable increases by three for every increase of five of the x-variable. The rate of change is three-fifths."

 x y 0 3 3 9 -2 -1 11 25

"This relationship is not linear since the rate of change in the -variable is not constant."

"The relationship is linear with a rate of change of two." The correct rate of change could be determined symbolically using pairs of points. Some students may find the rate of change by making a graph.

 x y 0 10 1 15 4 20 9 25

"This relationship is linear because the y-variable goes up by 5 each time."

"The relationship is not linear. If you look at the graph of the function it looks like a curve. My guess is that the function is a translation of a square root function. I used my calculator to guess and check and came up with $f(x)=5\sqrt{x}+10$."

• The idea of slope as a rate of change is difficult for students. In the shopping cart example a model for the length of the shopping cart train given the number of carts in the trains (c) could be L(c) = 33c + 74. Students could misinterpret the slope of 33 centimeters as the length of each cart rather than thinking of the 33 centimeters as the additional length added to the train length with the addition of one more shopping cart.

## Vignette

In the Classroom

Students need to make sense of situations using tables, graphs, and equations. The example below shows a classroom scenario in which students are working with the Geometer's Sketchpad. A similar exploration can be done using Geogebra. The sketch shown below was created by the teacher and shows a rectangle that can be changed by moving point E around along segment AB. The rectangle was created so that the perimeter stays constant at 20 cm. The goal of the lesson is for students to study the relationship between the length of the rectangle (AE) and the area of the rectangle. The beginning of the lesson gets students to think about what the independent and dependent variables are and then moves to studying how they are related then eventually moves to how the table, graph, and equation help students make sense of the situation. Students come up with an argument for why the situation is quadratic and then move into questioning how the theoretical domain does not match the theoretical domain of the function.

 Whole Class DiscussionStudents are working in pairs and each have a computer with the teacher created document open.Teacher: Please spend two minutes exploring the rectangle. Write at least three things you notice about the rectangle as you move point E along segment AB.After 2 minutesTeacher: What did you notice, Elmer?Elmer: I noticed that the rectangle gets skinny and fat.Teacher: What do you mean by skinny?Elmer: When point E gets close to point B then the rectangle is skinny. It gets more fat when E is in the middle.Ella: The rectangle gets more fat when E is next to B, and more skinny when E is next to A.Teacher: Ella, what does fat and skinny mean?Ella: Fat is wide, and skinny is thin but tall.Elmer: I used skinny to mean not much area and fat to mean more area.Teacher: Let's use the terms more area and less area instead of fat and skinny. Does anyone notice anything else?Mia: I noticed that the rectangle disappears when E is on top of B.Teacher: What do you mean it disappears?Mia: It does not have any area.Teacher: So as you move point E around does the area of the rectangle change?Elmer: No, it stays the same.Mia: It is changing.Teacher: Measure the area and the perimeter of the rectangle and see how they change as you move point E around.Han: The perimeter stays the same but the area is changing.Elmer: I noticed that EB is the same length at DE so that is why the perimeter is staying the same.Ella: It is always going to stay at 20cm because AE and DE add up to 10 and that is half the rectangle.Teacher: We agree that the area is changing and that the perimeter stays the same. Furthermore, we know that the length of segment AE is between 0 cm and 10 cm. Spend some time working with your partner and determine how the area is changing as the length of AE changes.After 5 minutesElla: The area gets bigger when AE is about 5 cm. The area gets to almost 0 when AE is 10 cm.Han: The area is also 0 when AE is close to 0 cm.Elmer: How can the area of the rectangle be 25 cm2? If you multiply 5.4 and 4.6 you get 24.95.Teacher: The computer is rounding to the nearest tenth. Talk to your partner and try to determine the area of the rectangle if the length of AE is 6 cm.Mia: That is easy, it is 24 cm2. All you have to do is multiply 6 by 4.Teacher: On your own, sketch a graph of how the area of the rectangle changes with the length of AE.three different possible student sketches Teacher: Your computer will let you look at the graph of the relationship between the length of AE and the area. What does the graph look like?Elmer: It looks like an upside down parabola. This makes sense since 25 cm2 is the largest the area could be. Also the area is 0 when AE is 10 cm or 0 cm. I am not sure why there is a curve on the graph.Teacher: With your partner, find the equation for this parabola and explain why it matches this situation.After 10 minutesMia: I found the equation to be $A=l(10-l)$ because area is length times width and the width is always $10$ minus the length. This proves that it is a parabola because $l(10-l)=10l-l^{2}$ and $A=10l-l^{2}$ is the equation of a parabola.Han: I knew the vertex of the parabola was at (5,25) and it was pointing down so I guessed $A=-1(x-5)^{2}+25$ and it worked. When I enter the equation in the computer the graph matches but goes further down on both ends. My equation must be wrong since it says I can have negative area when the length is bigger than 10 cm.Teacher: Han, what does the $x$ represent in your equation?Han: I'm not sure, oh it represents the length of AE.Teacher: The length of AE has to be larger than 0 cm and less than 10 cm. Your equation is correct but it only works in this context for values of $x$ between 0 and 10 cm. The domain of your function is limited to these values. Mia's equation also has this restriction on the domain.

## Resources

Instructional Notes

Teacher Notes

• Students are provided many examples that include tables that are well-ordered and struggle making sense of tables that are not well-ordered. For example, in the Leaky Faucet problem below, students will often state that the faucet is not leaking at a constant rate since the amount of water presented in the table is not increasing at a constant rate.

Leaky Faucet

Cassandra decided to see how fast her bathtub faucet was leaking. She got a large container and put it under her faucet when she got up in the morning, and then checked periodically during the day to see how much water was in the container. She wrote the times and the amounts down. They are in the table below.

Did the faucet leak at a constant rate?

 Time Amount of Water 7:00 a.m. 2 ounces 8:15 a.m. 12 ounces 9:45 a.m. 24 ounces 2:30 p.m. 62 ounces 5:15 p.m. 84 ounces 6:00 p.m. 90 ounces 9:30 p.m. 118 ounces

Source: Lobato

To solve the Leaky Faucet problem students first need to identify the independent variable. The time that the measurement occurs helps inform this relationship but a better independent variable to test is Elapsed Time in minutes. Some student may select 7:00 a.m. as time t = 0 minutes. Others may choose 6:45 a.m. as time t = 0 since that is the time when there is 0 ounces of water in the container. The relationship between elapsed time and amount of water in the container will be linear regardless of when t = 0 is selected because the faucet drips two ounces of water every 15 minutes. Instruction should focus on helping students see how the change in the elapsed time is related to the change in the number of ounces of water regardless of how the intervals of measurement are chosen.

Instructional Resources
• Square Graphs
Interactive graph that shows relationships among characteristics of a square
• Barbie Bungee
This lesson involves students doing an experiment to study the relationship between weight and length of stretch of a bungee apparatus made of rubber bands.
• Egg Launch Contest
Students will represent quadratic functions as a table, with a graph, and with an equation. They will compare data and move between representations.

• GeoGebra
Geogebra is an interactive mathematics tool that can be used to investigate algebraic functions and geometry. The web site includes the program and sample lessons.
New Vocabulary

rate of change:

The change in the value of a quantity divided by the elapsed time.  Commonly used examples include miles per gallon (miles/gallon), feet per second (feet/second),gallons per minute (gallons/minute), etc.

asymptote:

independent variable:

dependent variable:

mathematical model:

An equation or a system of equations representing real-world phenomena.  Models also represent patterns found in graphs and/or data.  Usually models are not exact matches the objects or behavior they represent.  A good model should capture the essential character of whatever is being modeled.

Professional Learning Communities

Reflection - Critical Questions regarding the teaching and learning of these benchmarks

• Does my instruction of functions focus too much on one representation?
• How do real-world examples help my students understand the differences and similarities between linear and exponential functions?
• Do my students relate their final answer to the original context consistently?
• How do my students organize data using table that are not well ordered?
• Which representations do my students prefer most?

Materials - suggested articles and books

• Usiskin, Zalman. "Conceptions of school algebra and uses of variables." In The Ideas of Algebra, K-12, 1988 Yearbook of the National Council of Teachers of Mathematics, pp. 8-19. Reston, VA: 1988.
• Gay, A. Susan & Jones, Alyson. "uncovering variables in the context of modeling activities." In Algebra and Algebraic Thinking in School Mathematics, 70th Yearbook of the National Council of Teachers of Mathematics, pp. 211-221. Reston, VA: 2008.
• Anderson, Nancy. "Walk the line: Making sense of." In Algebra and Algebraic Thinking in School Mathematics, 70th Yearbook of the National Council of Teachers of Mathematics, pp. 233-246. Reston, VA: 2008.
References

Common Core State Standards (http://www.corestandards.org/the-standards/mathematics)

Ruddell, M.R., & Shearer, B.A. (2002). "Extraordinary," "tremendous," "exhilarating, "magnificent": Middle school at-risk students become avid word learners with the vocabulary self-collection strategy (VSS). Journal of Adolescent & Adult Literacy, 45, 352-363.

Lobato - Leaky Faucet

## Assessment

 Example Item 1Source: MCAII Item Sampler 11th Grade Mathematics (http://education.state.mn.us/MDE/Accountability_Programs/Assessment_and_Testing/Assessments/MCA/Samplers/index.html)Correct Answer: DCognitive Level: Understanding
 Example Item 2The graph above shows distance versus time for a race between runners A and B. The race is already in progress, and the graph shows only the portion of the race that occurred after 11 A.M.The table on the next page lists several characteristics of the graph. Interpret these characteristics in terms of what happened during this portion of the race. Include times and distances to support your interpretation.(A sample interpretation of the y-intercepts is given in the table.)Source: NAEP Grade 12, 2009, Question M2-9 (http://nces.ed.gov/nationsreportcard/itmrlsx/detail.aspx?subject=mathematics)Correct Answer:Cognitive Level: Hard (28.28% correct)
 Example Item 310.    The following chart indicates the maximum number of connecting line segments y that can be drawn connecting x points, where no three points lie on the same line.         (a)     The relationship between x and y is represented by the equation y = kx(x - 1) for any positive number of points x. Use the information in the table to determine the value of the real number k.                  Answer: k = ____________________         (b)     Use the equation from part (a) to determine the maximum number of line segments that can be drawn connecting 100 points, no three of which lie on the same line.                  Answer: ____________________Source: NAEP Grade 12, 2009, Question M7-10 (http://nces.ed.gov/nationsreportcard/itmrlsx/detail.aspx?subject=mathematics)Correct Answer:            (a) Answer: $k=\frac{1}{2}$            (b) Answer: 4,950Cognitive Level: Hard (28.69% correct)

Example Item 4

 Exercise Time  (minutes) Total  Calories  Burned 101214161820 374249586470

5.      Which of the following graphs best illustrates the relationship between exercise time and total calories burned, as shown in the table above?

Source: NAEP Grade 12, 2005, Question M3-5 (http://nces.ed.gov/nationsreportcard/itmrlsx/detail.aspx?subject=mathematics)

Cognitive Level: Easy (28.69% correct)

 Example Item 5

## Differentiation

Struggling Learners

It takes time for students to learn the key features of linear, quadratic, and exponential functions. Extra time may be needed for students to identify these key features in tables, graphs, equations, and real-world situations. Teachers should observe the representations that a struggling student prefers and help them make connections with the other ones. Identifying individual data points for each representation is a good start before looking at the entire function that can describe the relationship between infinitely many pairs of points.

English Language Learners

English Learners (EL) need to taught strategies for identifying key content vocabulary. One strategy is the Vocabulary Self-Collection Strategy (VSS) (Ruddell & Shearer, 2002). Students first read a problem then work individually or in groups to identify key content vocabulary. For example, students might read the Shopping Cart Problem individually then before they attempt to solve the problem they work in pairs to identify key vocabulary that they need to know in order to make sense of and solve the problem.

 Shopping Cart ProblemShopping carts in the parking lots of stores are put together to form trains. Susan English and Tim Erickson measured 12 different trains. They counted the number of carts in each train and measured the length of that train of carts in centimeters.a)      How long is one shopping cart?b)      How much does one shopping cart fit into the next one?c)      One set of carts was 486 cm long. How many carts?d)    Find an equation that models the length of a shopping cart train in centimeters given the number of shopping carts in the train.

The words in the problem highlighted in red above are possible words that may be selected by a classroom of students. A classroom discussion then takes place about the meanings of these words in the context of this problem. Ruddell and Shearer (2002) found that if students are taught to use the VSS and given opportunities to practice they will begin to self-select words they need to know independently and their reading comprehension will improve.

Extending the Learning

As students develop in their modeling abilities they begin to notice that function rules have limitations for many situations. For example, in the shopping cart example the students might come up with the rule L(c) = 33c + 74 for the length of a cart train with c carts. Students should realize that the domain should be limited to whole numbers greater than 0. Some students may argue that since 0 carts is 0 centimeters long then the relationship is not linear. It is true that the equation L(c) = 33c + 74 does not contain the point (0,0) as shown in the graph on the right below students should be taught that rules can be written for situations in which there may be multiple formulas for different values of the domain.

Students who study beyond the standard should learn how to develop rules to match situations and define the rules piecewise.

\begin{align}L(c)=\begin{cases}
0 &c=0 \\
33c+74 & c>0
\end{cases}\end{align}

Furthermore they should understand that this notation defines only one rule that has multiple parts.


text:
In order to solve a radical inequality, we will first consider any domain restrictions. Then we will convert our inequality into an equality and solve the resulting equation. The solution will give us a boundary or critical value. From there, we will test values on each side of the boundary. We must restrict any value from our solution that violates the domain.

Test Objectives
• Demonstrate the ability to solve a radical equation
• Demonstrate the ability to solve a radical inequality
• Demonstrate the ability to write an inequality solution in interval notation
• Demonstrate the ability to graph an interval on the number line
Solving Radical Equations Part 1 Practice Test:

#1:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\sqrt{-1 - 6x}> \sqrt{3x + 8}$$

#2:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\sqrt{3x + 22}+ 2 > 4$$

#3:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\sqrt{10 - x}≤ \sqrt{3x + 7}- 3$$

#4:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\sqrt{8 - x}≥ \sqrt{2x - 7}- 3$$

#5:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}1 + \sqrt{6x - 2}≤ \sqrt{9x - 2}$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}-\frac{8}{3}< x < -1$$ $$\left(-\frac{8}{3}, -1 \right)$$

#2:

Solutions:

$$a)\hspace{.2em}x > -6$$ $$(-6, \infty)$$

#3:

Solutions:

$$a)\hspace{.2em}6 ≤ x ≤ 10$$ $$[6, 10]$$

#4:

Solutions:

$$a)\hspace{.2em}\frac{7}{2}≤ x ≤ 8$$ $$\left[\frac{7}{2}, 8\right]$$

#5:

Solutions:

$$a)\hspace{.2em}x=\frac{1}{3}\hspace{.2em}or \hspace{.2em}x ≥ 3$$ $$\left[\frac{1}{3}, \frac{1}{3}\right] ∪ [3, \infty)$$


text:
# Data Science Posts and Resources

Articles on Data Science

## Classification

Classifcation models a function to predict a specific outcome out of mulitple possible outcomes.

Laxmi K Soni

## Classification

With regression we now predicted specific output-values for certain given input-values. Sometimes, however, we are not trying to predict outputs but to categorize or classify our elements. For this, we use classification algorithms

In the figure above, we see one specific kind of classification algorithm, namely the K-Nearest-Neighbor classifier. Here we already have a decent amount of classified elements. We then add a new one (represented by the stars) and try to predict its class by looking at its nearest neighbors.

## CLASSIFICATION ALGORITHMS

There are various different classification algorithms and they are often used for predicting medical data or other real life use-cases. For example, by providing a large amount of tumor samples, we can classify if a tumor is benign or malignant with a pretty high certainty.

## K-NEAREST-NEIGHBORS

As already mentioned, by using the K-Nearest-Neighbors classifier, we assign the class of the new object, based on its nearest neighbors. The K specifies the amount of neighbors to look at. For example, we could say that we only want to look at the one neighbor who is nearest but we could also say that we want to factor in 100 neighbors. Notice that K shouldn’t be a multiple of the number of classes since it might cause conflicts when we have an equal amount of elements from one class as from the other.

## NAIVE-BAYES

The Naive Bayes algorithm might be a bit confusing when you encounter it the first time. However, we are only going to discuss the basics and focus more on the implementation in Python later on.

Outlook Temperture Humidity Windy Play
Sunny Hot High FALSE No
Sunny Hot High TRUE No
Rainy Mild High FALSE No
Rainy Hot High TRUE No
Overcast Hot Normal TRUE Yes
Sunny Hot Normal TRUE Yes
Sunny Mild High TRUE Yes
Overcast Cold Normal TRUE No

Imagine that we have a table like the one above. We have four input values (which we would have to make numerical of course) and one label or output. The two classes are Yes and No and they indicate if we are going to play outside or not.

What Naive Bayes now does is to write down all the probabilities for the individual scenarios. So we would start by writing the general probability of playing and not playing. In this case, we only play three out of eight times and thus our probability of playing will be 3/8 and the probability of not playing will be 5/8.

Also, out of the five times we had a high humidity we only played once, whereas out of the three times it was normal, we played twice. So our probability for playing when we have a high humidity is 1/5 and for playing when we have a medium humidity is 2/3. We go on like that and note all the probabilities we have in our table. To then get the classification for a new entry, we multiply the probabilities together and end up with a prediction.

## LOGISTIC REGRESSION

Another popular classification algorithm is called logistic regression . Even though the name says regression , this is actually a classification algorithm. It looks at probabilities and determines how likely it is that a certain event happens (or a certain class is the right one), given the input data. This is done by plotting something similar to a logistic growth curve and splitting the data into two.

## DECISION TREES

With decision tree classifiers, we construct a decision tree out of our training data and use it to predict the classes of new elements.

Since we are not using a line (and thus our model is not linear), we are also preventing mistakes caused by outliers.

This classification algorithm requires very little data preparation and it is also very easy to understand and visualize. On the other hand, it is very easy to be overfitting the model. Here, the model is very closely matched to the training data and thus has worse chances to make a correct prediction on new data.

## RANDOM FOREST

Rndom forest classifier is based on decision trees. What it does is creating a forest of multiple decision trees. To classify a new object, all the various trees determine a class and the most frequent result gets chosen. This makes the result more accurate and it also prevents overfitting. It is also more suited to handle data sets with higher dimensions. On the other hand, since the generation of the forest is random , you have very little control over your model.

Now let us get into the code. In this example, we will get our data directly from the sklearn module. For the program we need the following imports:

import numpy as np
import pandas as pd
from sklearn.model_selection import train_test_split
from sklearn.neighbors import KNeighborsClassifier
from sklearn.datasets import load_breast_cancer

At the last import, we import a dataset containing data on breast cancer. Also notice that we are only importing the KNeighborsClassifier for now.

data = load_breast_cancer()
print (data.feature_names)
## ['mean radius' 'mean texture' 'mean perimeter' 'mean area'
##  'mean smoothness' 'mean compactness' 'mean concavity'
##  'mean concave points' 'mean symmetry' 'mean fractal dimension'
##  'radius error' 'texture error' 'perimeter error' 'area error'
##  'smoothness error' 'compactness error' 'concavity error'
##  'concave points error' 'symmetry error' 'fractal dimension error'
##  'worst radius' 'worst texture' 'worst perimeter' 'worst area'
##  'worst smoothness' 'worst compactness' 'worst concavity'
##  'worst concave points' 'worst symmetry' 'worst fractal dimension']
print (data.target_names)
## ['malignant' 'benign']

targets, we have two options in this dataset: malignant and benign .

## PREPARING DATA

Again, we convert our data back into NumPy arrays and split them into training and test data.

X = np.array(data.data)
Y = np.array(data.target)
X_train, X_test, Y_train, Y_test = train_test_split(X,Y, test_size = 0.1 )

The data attribute refers to our features and the target attribute points to the classes or labels. We again choose a test size of ten percent.

## TRAINING AND TESTING

We start by first defining our K-Nearest-Neighbors classifier and then training it.

knn = KNeighborsClassifier( n_neighbors = 5 )
knn.fit(X_train, Y_train)
## KNeighborsClassifier()

The n_neighbors parameter specifies how many neighbor points we want to consider. In this case, we take five. Then we test our model again for its accuracy.

accuracy = knn.score(X_test, Y_test)
print (accuracy)
## 0.9473684210526315

We get a pretty decent accuracy for such a complex task.

0.9649122807017544

### THE BEST ALGORITHM

Now let’s put all the classification algorithms that we’ve discussed up until now to use and see which one performs best.

from sklearn.neighbors import KNeighborsClassifier
from sklearn.naive_bayes import GaussianNB
from sklearn.linear_model import LogisticRegression
from sklearn.tree import DecisionTreeClassifier
from sklearn.ensemble import RandomForestClassifier
clf1 = KNeighborsClassifier( n_neighbors = 5 )
clf2 = GaussianNB()
clf3 = LogisticRegression()
clf4 = DecisionTreeClassifier()
clf5 = RandomForestClassifier()

clf1.fit(X_train, Y_train)
## KNeighborsClassifier()
clf2.fit(X_train, Y_train)
## GaussianNB()
clf3.fit(X_train, Y_train)
## LogisticRegression()
##
## C:\Users\slaxm\AppData\Local\R-MINI~1\envs\R-RETI~1\lib\site-packages\sklearn\linear_model\_logistic.py:765: ConvergenceWarning: lbfgs failed to converge (status=1):
## STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.
##
## Increase the number of iterations (max_iter) or scale the data as shown in:
##     https://scikit-learn.org/stable/modules/preprocessing.html
## Please also refer to the documentation for alternative solver options:
##     https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
##   extra_warning_msg=_LOGISTIC_SOLVER_CONVERGENCE_MSG)
clf4.fit(X_train, Y_train)
## DecisionTreeClassifier()
clf5.fit(X_train, Y_train)
## RandomForestClassifier()
print (clf1.score(X_test, Y_test))
## 0.9473684210526315
print (clf2.score(X_test, Y_test))
## 0.9649122807017544
print (clf3.score(X_test, Y_test))
## 0.9649122807017544
print (clf4.score(X_test, Y_test))
## 0.9473684210526315
print (clf5.score(X_test, Y_test))
## 0.9824561403508771

When you run this program a couple of times, you will notice that we can’t really say which algorithm is the best. Every time we run this script, we will see different results, at least for this specific data set.

### PREDICTING LABELS

Again, we can again make predictions for new, unknown data. The chance of success in the classification is even very high. We just need to pass an array of input values and use the predict function .

X_new = np.array([[...]])
Y_new = clf.predict(X_new)

Nothing yet.


text:
# Vertex Poisson algebras associated with Courant algebroids and their deformations; I

Research paper by Gaywalee Yamskulna

Indexed on: 06 Sep '05Published on: 06 Sep '05Published in: Mathematics - Quantum Algebra

#### Abstract

This is the first of two papers on vertex Poisson algebras associated with Courant algebroids, and their deformations. In this work, we study relationships between vertex Poisson algebras and Courant algebroids. For any $\N$-graded vertex Poisson algebra $A=\coprod_{n\in\N} A_{(n)}$, we show that $A_{(1)}$ is a Courant $A_{(0)}$-algebroid. On the other hand, for any Courant $\mathcal{A}$-algebroid $\mathcal{B}$, we construct an $\N$-graded vertex Poisson algebra $A=\coprod_{n\in\N}A_{(n)}$ such that $A_{(0)}$ is $\mathcal{A}$ and the Courant $\mathcal{A}$-algebroid $A_{(1)}$ is isomorphic to $\mathcal{B}$ as a Courant $\mathcal{A}$-algebroid.


text:
# Heat Exchanger

 Question 1
A shell and tube heat exchanger is used as a steam condenser. Coolant water enters the tube at 300 K at a rate of 100 kg/s. The overall heat transfer coefficient is 1500 $W/m^2.K$, and total heat transfer area is 400 $m^2$. Steam condenses at a saturation temperature of 350 K. Assume that the specific heat of coolant water is 4000 J/kg.K. The temperature of the coolant water coming out of the condenser is _______K (round off to the nearest integer).
 A 254 B 339 C 254 D 654
GATE ME 2021 SET-2   Heat Transfer
Question 1 Explanation:

\begin{aligned} \mathrm{NTU}&=\frac{U A}{\left(\dot{m} c_{p}\right)_{\text {small }}}=\frac{1500 \times 400}{100 \times 4000}=1.5\\ \epsilon_{H E}&=1-e^{-N T U}=\frac{T_{c e}-T_{c i}}{T_{h i}-T_{c i}}=\frac{T_{C e}-300}{350-300}=0.7768\\ T_{c e}&=338.84 \mathrm{~K} \simeq 339 \mathrm{~K} \end{aligned}
 Question 2
In a concentric tube counter-flow heat exchanger, hot oil enters at $102^{\circ}C$ and leaves at $65^{\circ}C$. Cold water enters at $25^{\circ}C$ and leaves at $42^{\circ}C$. The log mean temperature difference (LMTD) is _______$^{\circ}C$ (round off to one decimal place)
 A 40 B 43.25 C 56.5 D 49.3
GATE ME 2020 SET-1   Heat Transfer
Question 2 Explanation:
\begin{aligned} T_{h i}&= 102^{\circ} \mathrm{C}, \quad T_{h e}=65^{\circ} \mathrm{C} \\ T_{c, i}&=25^{\circ} \mathrm{C}, \quad T_{c e}=42^{\circ} \mathrm{C} \\ \mathrm{LMTD}&=\frac{\Delta T_{1}-\Delta T_{2}}{\ln \left(\frac{\Delta T_{1}}{\Delta T_{2}}\right)} \\ &\Delta T_{1}=102^{\circ} \mathrm{C}-42^{\circ} \mathrm{C}=60^{\circ} \mathrm{C} ; \\ & \Delta T_{2}=65^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}=40^{\circ} \mathrm{C}_{7} \\ \mathrm{LMTD}&=\frac{60-40}{\ln \left(\frac{60}{40}\right)}=\frac{20}{0.4054}=49.3^{\circ} \mathrm{C} \end{aligned}
 Question 3
Hot and cold fluids enter a parallel flow double tube heat exchanger at $100^{\circ}C \; and \; 15^{\circ}C$, respectively. The heat capacity rates of hot and cold fluids are $C_h=2000W/K$ and $C_c=1200W/K$, respectively. If the outlet temperature of the cold fluid is $45^{\circ}C$, the log mean temperature difference (LMTD) of the heat exchanger is _________K (round off to two decimal places).
 A 21.45 B 45.25 C 57.71 D 98.65
GATE ME 2019 SET-2   Heat Transfer
Question 3 Explanation:

Given data:
C_{h} = 2000 W/K Energy balance:
Energy released by hot fluid = Energy received by cold fluid
\begin{aligned} \mathrm{C}_{\mathrm{h}}\left(\mathrm{T}_{\mathrm{hi}}-\mathrm{T}_{\mathrm{he}}\right) &=\mathrm{C}_{\mathrm{c}}\left(\mathrm{T}_{\mathrm{cc}}-\mathrm{T}_{\mathrm{ci}}\right) \\ 2000\left(100-\mathrm{T}_{\mathrm{hc}}\right)&=1200(45-15) \\ 2000 \times\left(100-\mathrm{T}_{\mathrm{he}}\right)& =1200 \times 30 \mathrm{T}_{\mathrm{he}}=82^{\circ} \mathrm{C} \\ \mathrm{LMTD}&=\frac{\theta_{1}-\theta_{2}}{\ln \left(\frac{\theta_{1}}{\theta_{2}}\right)} \\ &=\frac{85-37}{\ln \left(\frac{85}{37}\right)}=\frac{48}{\ln \left(\frac{85}{37}\right)} \\ \mathrm{LMTD}&=57.71^{\circ} \mathrm{C} \text{ or } \mathrm{K}\end{aligned}
This is logarithmic mean temperature difference either we take in $^{o}C$ or we take in K the final answer will be 57.71 K (or)$^{o}C$ because it is a difference.
 Question 4
Steam in the condenser of a thermal power plant is to be condensed at a temperature of $30^{\circ}C$ with cooling water which enters the tubes of the condenser at $14^{\circ}C$ and exits at $22^{\circ}C$. The total surface area of the tubes is 50 $m^{2}$, and the overall heat transfer coefficient is 2000 W/$m^{2}$ K. The heat transfer (in MW) to the condenser is ______ (correct to two decimal places).
 A 0.5 B 1.16 C 2.25 D 3.45
GATE ME 2018 SET-2   Heat Transfer
Question 4 Explanation:
Since steam is condensing, the temperature of hot fluid remains constant.
$\therefore$ Temperature profiles of hot and cold fluids are

$\therefore (\mathrm{LMTD}) of H E=\frac{\Delta T_{i}-\Delta T_{e}}{\ln \frac{\Delta T_{i}}{\Delta T_{e}}}=\frac{16-8}{\ln \left(\frac{16}{8}\right)}=11.54^{\circ} \mathrm{C}$
Total heat transfer rate between steam
\begin{aligned} \mathrm{C.W.} &=Q=U A \mathrm{LMTD} \mathrm{W} \\ &=2000 \times 50 \times 11.54 \\ &=11.54 \times 10^{5} \mathrm{W} \\ &=1.154 \mathrm{MW} \end{aligned}
 Question 5
In a counter-flow heat exchanger, water is heated at the rate of 1.5kg/s from $40\,^{\circ} C$ to $80\,^{\circ} C$ by an oil entering at $120\,^{\circ} C$ and leaving at $60\,^{\circ} C$. The specific heat of water and oil are $4.2\:kJ/kg \! \cdot\!K$ and $2\:kJ/kg \! \cdot\!K$, respectively. The overall heat transfer coefficient is $400\, W/m^{2} \cdot K$ The required heat transfer surface area (in $m^{2}$ ) is
 A 0.104 B 0.022 C 10.4 D 21.84
GATE ME 2017 SET-2   Heat Transfer
Question 5 Explanation:

\begin{aligned} \Delta T_{i} &=120-80=40^{\circ} \mathrm{C} \\ \Delta T_{e} &=60-40=20^{\circ} \mathrm{C} \\ (L M T D) \text { Counterflow }&=\frac{\Delta T_{i}-\Delta T_{e}}{\ln \frac{\Delta T_{i}}{\Delta T_{e}}} =28.86^{\circ} \mathrm{C} \\ &=\frac{40-20}{\ln \frac{40}{20}} = 28.86^{o}\mathrm{C}\\ Q &=\dot{m}_{C} c_{p_{c}}\left(T_{c e}-T_{c i}\right) \\ &=1.5 \times 4.2 \times 10^{3}(80-40) \mathrm{watt} \\ Q &=252000 \mathrm{W} \\ Q &=U A \Delta T_{m} \\ A &=\frac{Q}{U \Delta T_{m}}=\frac{1.5 \times 4.2 \times 10^{3} \times 40}{400 \times 28.86} \\ &=21.83 \mathrm{m}^{2} \end{aligned}
 Question 6
Saturated steam at $100^{\circ}$C condenses on the outside of a tube.cold fluid enters the tube at $20^{\circ}$C and exits at $50^{\circ}$C the value of the log mean Temperature Difference (LMTD) is _________$^{\circ}$C.
 A 30 B 80 C 64 D 84
GATE ME 2017 SET-1   Heat Transfer
Question 6 Explanation:
Temperature profiles are

\begin{aligned} \Delta T_{i}=100-20=80^{\circ} \mathrm{C} \\ \Delta T_{e}=100-50=50^{\circ} \mathrm{C} \end{aligned} (LMTD) (parallel or counter HE)
\begin{aligned} =\frac{\Delta T_{i}-\Delta T_{e}}{\ln \frac{\Delta T_{i}}{\Delta T_{e}}}=\frac{80-50}{\ln \frac{80}{50}} \\ =63.82^{\circ} \mathrm{C} \end{aligned}
 Question 7
For a heat exchanger, $\Delta T_{max}$ is the maximum temperature difference and $\Delta T_{min}$ is the minimum temperature difference between the two fluids. LMTD is the log mean temperature difference. $C_{min}$ and$C_{max}$ are the minimum and the maximum heat capacity rates. The maximum possible heat transfer ($Q_{max}$) between the two fluids is
 A $C_{min}$ LMTD B $C_{min}$ $\Delta T_{max}$ C $C_{max}$ $\Delta T_{max}$ D $C_{max}$ $\Delta T_{min}$
GATE ME 2016 SET-3   Heat Transfer
Question 7 Explanation:
Maximum heat transfer possible
$=C_{\min } \times(\Delta T)_{\max }$
 Question 8
Consider a parallel-flow heat exchanger with area $A_{p}$ and a counter-flow heat exchanger with area $A_{c}$. In both the heat exchangers, the hot stream flowing at 1 kg/s cools from $80^{\circ}$C to $50^{\circ}$C . For the cold stream in both the heat exchangers, the flow rate and the inlet temperature are 2 kg/s and $10^{\circ}$C , respectively. The hot and cold streams in both the heat exchangers are of the same fluid. Also, both the heat exchangers have the same overall heat transfer coefficient. The ratio $A_{c}$/ $A_{p}$ is _________
 A 0.258 B 0.684 C 0.489 D 0.927
GATE ME 2016 SET-2   Heat Transfer
Question 8 Explanation:
Parallel flow

$A_{p} =$ Area of parallel flow heat exchanger
Counter flow

$A_{c}=$ Area of counter flow heat exchanger
Heat lost by hot stream in both heat exchanger
\begin{aligned} &=m c\left[T_{h i}-T_{h o}\right] \\ &=1 \times c \times\left[T_{h i}-T_{h o}\right] \\ &=1 \times c \times[80-50]=30 \mathrm{ckW} \end{aligned}
Heat gained by cold stream in both exchange
\begin{aligned} &=m c\left[T_{co}-T_{c i}\right] \\ &=2 \times c\left[T_{c o}-10\right] \end{aligned}
Heat lost by hot stream = Heat gain by cool stream
\begin{aligned} 30c &=2c(T_\infty -10) \\ T_\infty &= \frac{30}{2}+10\\ &=15+10=25^{\circ} C\\ A_P, LMTD&= \frac{\Delta T_i-\Delta T_e}{ln(\Delta T_i/\Delta T_e)}\\ &= \frac{70-25}{ln(70/25)}=43.705^{\circ} C\\ Q_H&=UA_P 43.705 \\ A_C, LMTD&= \frac{\Delta T_i-\Delta T_e}{ln(\Delta T_i/\Delta T_e)}\\ &= \frac{55-40}{ln(55/40)}=47.103^{\circ} C\\ Q_H&=UA_C 47.103 \\ \therefore \; UA_P 43.705 &=UA_C 47.103\\ \frac{43.705}{47.103}&=\frac{A_C}{A_P}=0.927 \end{aligned}
 Question 9
Saturated vapor is condensed to saturated liquid in a condenser. The heat capacity ratio is $C_{r}=\frac{C_{min}}{C_{max}}$. The effectiveness $(\varepsilon )$ of the condenser is
 A $\frac{1-exp[-NTU(1+C_{r})]}{1+C_{r}}$ B $\frac{1-exp[-NTU(1-C_{r})]}{1-C_{r} exp[-NTU(1-C_{r})]}$ C $\frac{NTU}{1+NTU}$ D $1-exp(-NTU)$
GATE ME 2015 SET-3   Heat Transfer
Question 9 Explanation:
For parallel
$\varepsilon=\frac{1-\exp \left[-N T U\left(1+C_{f}\right)\right]}{1+C_{r}}$
For condenser,
\begin{aligned} C_{\max } &=\infty \\ C_{r} &=0 \\ \varepsilon &=\frac{1-\exp [-\mathrm{NTU}]}{1} \end{aligned}
Therefore (D) is correct option.
Considering counter-flow heat exchanger
$\frac{1-\exp \left[-N T U\left(1+C_{r}\right)\right]}{1-C_{f} \exp \left[-N T U\left(1+C_{r}\right)\right]}$
$\therefore \quad c_{r}=0$
$1-\exp (-N T U)$
 Question 10
A balanced counterflow heat exchanger has a surface area of 20 $m^{2}$ and overall heat transfer coefficient of 20 W/$m^{2}$ -K. Air ($C_{p}$ =1000 J/kg-K) entering at 0.4 kg/s and 280 K is to be preheated by the air leaving the system at 0.4 kg/s and 300 K. The outlet temperature (in K) of the preheated air is
 A 290 B 300 C 320 D 350
GATE ME 2015 SET-2   Heat Transfer
Question 10 Explanation:
Given data:
\begin{aligned} A&=20 \mathrm{m}^{3} ; \quad U=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} ; \quad c_{p h}=1000 \mathrm{J} / \mathrm{kg} \mathrm{K} \\ m_{c}&=0.4 \mathrm{kg} / \mathrm{s} ; \quad T_{c 1}=280 \mathrm{K} ; \quad m_{h}=0.4 \mathrm{kg} / \mathrm{s} \\ T_{h 2}&=300 \mathrm{K} ; \quad T_{c 2}=? \end{aligned}
By energy balance,
Heat lost by hot fluid = Heat gained by cold fluid
$m_{h} c_{p h}\left(T_{h 1}-T_{h 2}\right)=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right)$
$\text{or }\quad T_{h 1}-T_{h 2}=T_{c 2}-T_{c 1} \quad \because m_{h} c_{p h}=m_{c} c_{p c}$
\begin{aligned} T_{h 1}-T_{c 2} &=T_{h 2}-T_{c 1} \\ \theta_{1} &=\theta_{2} \\ \text { If } \theta_{1}&=\theta_{2} \text { , then } L M T D=\theta_{1}=\theta_{2} \\ Q &=A U L M T D \\ Q &=A U \times \theta_{1} \\ &=20 \times 20 \times\left(T_{h 2}-T_{c 1}\right) \\ &=400(300-280)=8000 \mathrm{kW} \\ \text{also }\quad Q&=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right) \\ \therefore \quad 8000 &=0.4 \times 1000\left(T_{c 2}-280\right) \\ 20&=T_{c 2}-280 \\ \text{or}\quad T_{c 2}&=300^{\circ} \mathrm{C} \end{aligned}
Alternatively:
Given data:
$A=20 \mathrm{m}^{3} ; U=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} ; \quad$
$c_{\mathrm{ph}}=1000 \mathrm{J} / \mathrm{kgK}$
$m_{c}=0.4 \mathrm{kg} / \mathrm{s} ; \quad T_{c 1}=280 \mathrm{K} ; \quad m_{h}=0.4 \mathrm{kg} / \mathrm{s}$
$T_{h 2}=300 \mathrm{K} ; \quad T_{c 2}=?$

By energy balance,
Heat lost by hot fluid = Heat gained by cold fluid
$m_{h} c_{p h}\left(T_{h 1}-T_{h 2}\right)=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right)$
\begin{aligned} \text { or } T_{h 1}-T_{h 2} &=T_{c 2}-T_{c 1} \qquad \because m_{h} c_{p h}=m_{c} c_{p c} \\ T_{h 1}-300 &=T_{c 2}-280 \\ T_{h 1}-T_{c 2} &=300-280 \qquad \ldots(1) \\ T_{h 1}-T_{c 2} &=20 \\ \epsilon&=\frac{N T U}{N T U+1} \quad[ \text{For balancing heat exchanger}]\\ \frac{T_{h 1}-T_{h 2}}{T_{h 1}-T_{c 1}} &=\frac{1}{1+1} \\ \frac{T_{h 1}-300}{T_{h 1}-280} &=\frac{1}{2} \\ 2 T_{h 1}-600 &=T_{h 1}-280 \\ T_{h 1} &=600-280=320 \mathrm{K} \end{aligned}
Substitute the value of $T_{h 1}=320 \mathrm{K}$ in Eq. (1), we get
$320-T_{c 2}=20$
$\text{or }\quad T_{c 2}=300 \mathrm{K}$
There are 10 questions to complete.


text:
# 3. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

P Pankaj Sanodiya

Length =825cm=3 $\times$ 5 $\times$ 5 $\times$11

Breadth =675cm=3 $\times$$\times$ 3 $\times$ 5 $\times$ 5

Height =450cm=2 $\times$3 $\times$ 3 $\times$ 5 $\times$

Longest tape = HCF of 825,675, and 450=  3 $\times$ 5 $\times$ 5

=75cm

Therefore, the longest tape is 75cm.

Exams
Articles
Questions


text:
# Complex Conditionals (and, or, and not)¶

We can also use and and or to join several logical expressions together as shown in the table below. An or joining two expressions means that if either of the expressions is true, the whole expression is true. An and used to join two expressions is only true if both expressions are true. A not negates the logical value that follows it. If it was true, then a not changes the result to false. If it was false, the not changes the result to true.

Expression

Meaning

(a < b) or (c < d)

The whole expression is true if a is less than b or c is less than d.

(a < b) and (c < d)

The whole expression is true if a is less than b AND ALSO c is less than d.

not a < b

Only true if a is actually greater than or equal to b. The logical expression not a < b is the same as a >= b.

A common use of and is to check that a value is in a range between a minimum value and a maximum value. For example, if you have asked a person to pick a number between 1 and 10 you can check for this using the following.

A common use of or is to check if either one of two conditions are true. For example, a parent has told a teen that she can go out if she has cleaned her room or finished her homework. If either of these is true she can go out. In Python a value of 0 means false and any non-zero value is true, but 1 is often used for true.

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186 Answered Questions for the topic Percents

09/14/19

#### Multiplying percents?

If there are 39900000 flowers, how many are blue at 1.8% and how many are dotted at 11.5%, how many are blue and dotted?

09/10/19

A friend has a 80% average before the final exam for a course. That score includes everything but final, which counts for 15% of the grade. What is the best course grade for your friend to earn?... more

08/31/19

#### 7 is 25% of what number?

how do you solve this problume

08/09/19

#### Math problems need help

Jerome bought a pair of shoes for $61.49 including tax. The shoes had a sticker price of$57.20 what percent sales tax did he pay  Can you tell me the answer

01/18/19

#### Percents Study question

Tom has already read 30% of the book. he is considering two options to finish the book. option 1- read 15% of the remaining pages each night. Or... option 2- read 20 pages each night. which option... more

11/29/18

09/09/18

#### Percent question

How many gallons of 5% ethanol gasoline must be mixed with how many gallons of 12% ethanol gasoline to get 700 gallons of 10% ethanol?

08/28/18

#### Step by step process on how to set this up, then I'm sure I can handle the rest.

Four exams contribute to 75% of your final grade. Assume you score a 64% on the first exam, a 72% on the second, and 84% on the third. What would you have to score on your fourth exam to guarantee... more

06/05/18

#### 8 pounds plus 30 percent

I have a math test and it said that a baby was born and was eight pounds and now it increased by thirty percent. What is it’s current weight?

05/24/18

James, priya, and siobhan work in a grocery store. James makes $7.00 per hour priya makes 40% more than James and siobhan makes 5% less then priya. How much does siobhan make per hour? 05/24/18 #### Mae ling earns a weekly salary of$360 plus a 5.5% commission on sales at a gift shop. How much would she make in...

Mae ling earns a weekly salary of \$360 plus a 5.5% commission on sales at a gift shop. How much would she make in a work week if she sold 4,200 worth of merchandise?

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text:
Question 9

# A and B can complete a task together in 35 days. If A works alone and completes 5/7th of the task and then leaves the rest for B to complete by herself, it will take a total of 90 days to complete the task. How many days would it take by A, the more efficient among the duo, to complete the entire work by herself?

Solution

Fraction of work completed by A is not mentioned.

• 45+ RRB previous papers with solutions PDF

1 year, 4 months ago

Hsvsg

##### Sudhansu Patra

11 months, 4 weeks ago

Critical question .....
A 's efficiency here , 5/6 = 0.8333
B's efficiency = 1/6 = 0.16666
A + B = 7 unit

OR


text:
G02 Chapter Contents
G02 Chapter Introduction
NAG Library Manual

# NAG Library Routine DocumentG02FAF

Note:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.

## 1  Purpose

G02FAF calculates two types of standardized residuals and two measures of influence for a linear regression.

## 2  Specification

 SUBROUTINE G02FAF ( N, IP, NRES, RES, H, RMS, SRES, LDSRES, IFAIL)
 INTEGER N, IP, NRES, LDSRES, IFAIL REAL (KIND=nag_wp) RES(NRES), H(NRES), RMS, SRES(LDSRES,4)

## 3  Description

For the general linear regression model
 $y=Xβ+ε,$
 where $y$ is a vector of length $n$ of the dependent variable, $X$ is an $n$ by $p$ matrix of the independent variables, $\beta$ is a vector of length $p$ of unknown parameters, and $\epsilon$ is a vector of length $n$ of unknown random errors such that $\mathrm{var}\epsilon ={\sigma }^{2}I$.
The residuals are given by
 $r=y-y^=y-Xβ^$
and the fitted values, $\stackrel{^}{y}=X\stackrel{^}{\beta }$, can be written as $Hy$ for an $n$ by $n$ matrix $H$. The $i$th diagonal elements of $H$, ${h}_{i}$, give a measure of the influence of the $i$th values of the independent variables on the fitted regression model. The values of $r$ and the ${h}_{i}$ are returned by G02DAF.
G02FAF calculates statistics which help to indicate if an observation is extreme and having an undue influence on the fit of the regression model. Two types of standardized residual are calculated:
(i) The $i$th residual is standardized by its variance when the estimate of ${\sigma }^{2}$, ${s}^{2}$, is calculated from all the data; this is known as internal Studentization.
 $RIi=ris1-hi .$
(ii) The $i$th residual is standardized by its variance when the estimate of ${\sigma }^{2}$, ${s}_{-i}^{2}$ is calculated from the data excluding the $i$th observation; this is known as external Studentization.
 $REi=ris-i1-hi =rin-p-1 n-p-RIi2 .$
The two measures of influence are:
(i) Cook's $D$
 $Di=1pREi2hi1-hi .$
(ii) Atkinson's $T$
 $Ti=REi n-pp hi1-hi .$

## 4  References

Atkinson A C (1981) Two graphical displays for outlying and influential observations in regression Biometrika 68 13–20
Cook R D and Weisberg S (1982) Residuals and Influence in Regression Chapman and Hall

## 5  Parameters

1:     N – INTEGERInput
On entry: $n$, the number of observations included in the regression.
Constraint: ${\mathbf{N}}>{\mathbf{IP}}+1$.
2:     IP – INTEGERInput
On entry: $p$, the number of linear parameters estimated in the regression model.
Constraint: ${\mathbf{IP}}\ge 1$.
3:     NRES – INTEGERInput
On entry: the number of residuals.
Constraint: $1\le {\mathbf{NRES}}\le {\mathbf{N}}$.
4:     RES(NRES) – REAL (KIND=nag_wp) arrayInput
On entry: the residuals, ${r}_{i}$.
5:     H(NRES) – REAL (KIND=nag_wp) arrayInput
On entry: the diagonal elements of $H$, ${h}_{i}$, corresponding to the residuals in RES.
Constraint: $0.0<{\mathbf{H}}\left(\mathit{i}\right)<1.0$, for $\mathit{i}=1,2,\dots ,{\mathbf{NRES}}$.
6:     RMS – REAL (KIND=nag_wp)Input
On entry: the estimate of ${\sigma }^{2}$ based on all $n$ observations, ${s}^{2}$, i.e., the residual mean square.
Constraint: ${\mathbf{RMS}}>0.0$.
7:     SRES(LDSRES,$4$) – REAL (KIND=nag_wp) arrayOutput
On exit: the standardized residuals and influence statistics.
For the observation with residual, ${r}_{i}$, given in ${\mathbf{RES}}\left(i\right)$.
${\mathbf{SRES}}\left(i,1\right)$
Is the internally standardized residual, ${\mathrm{RI}}_{i}$.
${\mathbf{SRES}}\left(i,2\right)$
Is the externally standardized residual, ${\mathrm{RE}}_{i}$.
${\mathbf{SRES}}\left(i,3\right)$
Is Cook's $D$ statistic, ${D}_{i}$.
${\mathbf{SRES}}\left(i,4\right)$
Is Atkinson's $T$ statistic, ${T}_{i}$.
8:     LDSRES – INTEGERInput
On entry: the first dimension of the array SRES as declared in the (sub)program from which G02FAF is called.
Constraint: ${\mathbf{LDSRES}}\ge {\mathbf{NRES}}$.
9:     IFAIL – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).

## 6  Error Indicators and Warnings

If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
 On entry, ${\mathbf{IP}}<1$, or ${\mathbf{N}}\le {\mathbf{IP}}+1$, or ${\mathbf{NRES}}<1$, or ${\mathbf{NRES}}>{\mathbf{N}}$, or ${\mathbf{LDSRES}}<{\mathbf{NRES}}$, or ${\mathbf{RMS}}\le 0.0$.
${\mathbf{IFAIL}}=2$
 On entry, ${\mathbf{H}}\left(i\right)\le 0.0$ or $\text{}\ge 1.0$, for some $i=1,2,\dots ,{\mathbf{NRES}}$.
${\mathbf{IFAIL}}=3$
 On entry, the value of a residual is too large for the given value of RMS.

## 7  Accuracy

Accuracy is sufficient for all practical purposes.

None.

## 9  Example

A set of $24$ residuals and ${h}_{i}$ values from a $11$ parameter model fitted to the cloud seeding data considered in Cook and Weisberg (1982) are input and the standardized residuals etc calculated and printed for the first $10$ observations.

### 9.1  Program Text

Program Text (g02fafe.f90)

### 9.2  Program Data

Program Data (g02fafe.d)

### 9.3  Program Results

Program Results (g02fafe.r)


text:
2019
Том 71
№ 11

# Strongly Nonlinear Differential Equations with Carlitz Derivatives over a Function Field

Kochubei A. N.

Abstract

In earlier papers the author studied some classes of equations with Carlitz derivatives for $\mathbb{F}_q$ -linear functions, which are the natural function field counterparts of linear ordinary differential equations. Here we consider equations containing self-compositions $u \circ u ... \circ u$ of the unknown function. As an algebraic background, imbeddings of the composition ring of $\mathbb{F}_q$ -linear holomorphic functions into skew fields are considered.

English version (Springer): Ukrainian Mathematical Journal 57 (2005), no. 5, pp 794-805.

Citation Example: Kochubei A. N. Strongly Nonlinear Differential Equations with Carlitz Derivatives over a Function Field // Ukr. Mat. Zh. - 2005. - 57, № 5. - pp. 669–678.

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