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# Multiplication of Rational Numbers

## Multiply fractions: multiply straight across

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Multiplication of Real Numbers

Jacob received tips of $4.00 each from three of his paper route customers. How much did he receive in total? ### Watch This ### Guidance Multiplication of two integers with the same signs produces a positive result and multiplication of two integers with unlike signs results in a negative answer. These rules can be applied to the multiplication of all real numbers. To multiply fractions, you multiply the numerators and then you multiply the denominators. The product of the numerators over the product of the denominators is the answer to the problem. Sometimes the answer can be expressed as an equivalent fraction. The rules for multiplying integers also apply to multiplying decimals. The sum of the number of digits after the decimal points determines the placement of the decimal point in the answer. #### Example A Sam spent$2.00 for a bottle of chocolate milk at the school cafeteria every school day. At the end of the week, how does this affect his net worth?

Solution: The result of (+5)×(2)\begin{align*}(+5) \times (-2)\end{align*} is –10. The product of a positive integer and a negative integer is always negative.

#### Example B

What is (2)×(3)\begin{align*}(-2) \times (-3)\end{align*}?

Solution: The result of (2)×(3)\begin{align*}(-2) \times (-3)\end{align*} is +6. The product of two negative integers is always positive.

#### Example C

i) (23)×(57)\begin{align*}\left(\frac{2}{3}\right) \times \left(\frac{5}{7}\right)\end{align*}

ii) (78)×(334)\begin{align*}\left(\frac{7}{8}\right) \times \left(3 \frac{3}{4}\right)\end{align*}

iii) (534)×(235)\begin{align*}\left(5 \frac{3}{4} \right) \times \left(2 \frac{3}{5}\right)\end{align*}

Solution: Remember, there are three simple steps to follow to multiply fractions:

1. Multiply the numerators of the fractions
2. Multiply the denominators of the fractions.
3. Simplify the fraction if necessary.

i) (23)×(57)\begin{align*}\left(\frac{2}{3}\right) \times \left(\frac{5}{7}\right)\end{align*}

=2×53×7=1021

ii) (78)×(334)\begin{align*}\left(\frac{7}{8}\right) \times \left(3 \frac{3}{4}\right)\end{align*} Express the mixed number as an improper fraction.

=(78)×(154)(4×3)+34=7×158×4=10532=3932

iii) (534)×(235)\begin{align*}\left(5 \frac{3}{4}\right) \times \left(2 \frac{3}{5}\right)\end{align*} Express the mixed numbers as improper fractions.

=(234)×(135)(4×5)+34 and (5×2)+35=23×134×5=29920=141920

#### Example D

(14.65)×(2.7)\begin{align*}(14.65) \times (2.7)\end{align*}

Solution: Multiply the numbers as you would whole numbers. To place the decimal point in the answer, count the number of digits after the decimal points in the problem. There are two digits after the decimal point in 14.65 and one digit after the decimal point in 2.7. This is a total of three digits after the decimal points. From the right of the answer, count three places to the left and insert the decimal point.

14.65×2.7  10255+29300  39.555

Jacob received tips of 4.00 each from three of his paper route customers. How much did he receive in total? The result of (+3)×(+4)\begin{align*}(+3) \times (+4)\end{align*} is +12. The product of two positive integers is always positive. ### Vocabulary Denominator The denominator of a fraction is the number on the bottom that indicates the total number of equal parts in the whole or the group. 58\begin{align*}\frac{5}{8}\end{align*} has denominator 8. Fraction A fraction is any rational number that is not an integer. Improper Fraction An improper fraction is a fraction in which the numerator is larger than the denominator. 83\begin{align*}\frac{8}{3}\end{align*} is an improper fraction. Integer All natural numbers, their opposites, and zero are integers. A number in the list ..., –3, –2, –1, 0, 1, 2, 3... Mixed Number A mixed number is a number made up of a whole number and a fraction such as 435\begin{align*}4 \frac{3}{5}\end{align*}. Numerator The numerator of a fraction is the number on top that is the number of equal parts being considered in the whole or the group. 58\begin{align*}\frac{5}{8}\end{align*} has numerator 5. ### Guided Practice Multiply the following fractions: 1. (59)×(47)\begin{align*}\left(\frac{5}{9}\right) \times \left(\frac{-4}{7}\right)\end{align*} 2. (323)×(415)\begin{align*}\left(3\frac{2}{3}\right) \times \left(4 \frac{1}{5}\right)\end{align*} 3. Determine the answer to (135.697)×(34.32)\begin{align*}(-135.697) \times (-34.32)\end{align*} Answers: 1. Multiply the numerators. Multiply the denominators. Simplify the fraction. (59)×(47)=5×(4)9×7=2063 The answer can be written as 2063\begin{align*}\frac{-20}{63}\end{align*} or 2063\begin{align*}-\frac{20}{63}\end{align*}. 2. Write the two mixed numbers as improper fractions. Multiply the denominator and the whole number. Add the numerator to this product. Write the answer over the denominator. Follow the steps for multiplying fractions. Simplify the fraction if necessary. (323)×(415)(113)×(215)(113)×(215)=23115=1525 3. Multiply the numbers as you would whole numbers. Remember the rule for multiplying integers. When you multiply two integers that have the same sign, the product will always be positive. 135.697×34.32 271394 4070910 542788004070910004657.12104 There are three digits after the decimal point in 135.697 and two digits after the decimal point in 34.32. Beginning at the right of the product, count five places to the left and insert the decimal point. ### Practice Multiply. 1. (7)×(2)\begin{align*}(-7) \times (-2)\end{align*} 2. (+3)×(+4)\begin{align*}(+3) \times (+4)\end{align*} 3. (5)×(+3)\begin{align*}(-5) \times (+3)\end{align*} 4. (+2)×(4)\begin{align*}(+2) \times (-4)\end{align*} 5. (+4)×(1)\begin{align*}(+4) \times (-1)\end{align*} Match each given phrase with the correct multiplication statement. Then, determine each product. 1. take away six groups of 3 balls 2. net worth after losing seven5 bills
3. take away nine sets of 8 forks
4. take away four sets of four plates
5. receive eight groups of 4 glasses
6. buy seven sets of 12 placemats
1. (+8)×(+4)\begin{align*}(+8) \times (+4)\end{align*}
2. (+7)×(5)\begin{align*}(+7) \times (-5)\end{align*}
3. (4)×(+4)\begin{align*}(-4) \times (+4)\end{align*}
4. (9)×(+8)\begin{align*}(-9) \times (+8)\end{align*}
5. (+7)×(+12)\begin{align*}(+7) \times (+12)\end{align*}
6. (6)×(+3)\begin{align*}(-6) \times (+3)\end{align*}

Use the rules that you have learned for multiplying real numbers to answer the following problems.

1. (13)×(9)\begin{align*}(-13) \times (-9)\end{align*}
2. (3.68)×(82.4)\begin{align*}(-3.68) \times (82.4)\end{align*}
3. (49)×(57)\begin{align*}\left(\frac{4}{9}\right) \times \left(\frac{5}{7}\right)\end{align*}
4. (723)×(612)\begin{align*}\left(7 \frac{2}{3} \right) \times \left(6 \frac{1}{2}\right)\end{align*}
5. (15.734)×(8.1)\begin{align*}(15.734) \times (-8.1)\end{align*}

### Vocabulary Language: English

Associative Property

Associative Property

The associative property states that you can change the groupings of numbers being added or multiplied without changing the sum. For example: (2+3) + 4 = 2 + (3+4), and (2 X 3) X 4 = 2 X (3 X 4).
Commutative Property

Commutative Property

The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example $a+b=b+a \text{ and\,} (a)(b)=(b)(a)$.
Denominator

Denominator

The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
distributive property

distributive property

The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
fraction

fraction

A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
improper fraction

improper fraction

An improper fraction is a fraction in which the absolute value of the numerator is greater than the absolute value of the denominator.
Integer

Integer

The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
Mixed Number

Mixed Number

A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
multiplicative identity property

multiplicative identity property

The product of any number and one is the number itself.
Numerator

Numerator

The numerator is the number above the fraction bar in a fraction.


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## Changelog for QuickCheck-2.12.5

QuickCheck 2.12.3, 2.12.4 (released 2018-09-12) * Shrinking for Float and Decimal now works by reducing the number of digits in the number. The new function shrinkDecimal implements this shrinking behaviour. * Shrinking for Rational now tries to make the numerator and denominator of the number smaller. Previously it tried to reduce the magnitude of the number. QuickCheck 2.12.2 (released 2018-09-10) * Fix infinite shrinking loop for fractional types. * Add SortedList modifier. QuickCheck 2.12.1 (released 2018-09-06) * Fix bug in 'classify'. QuickCheck 2.12 (released 2018-09-03) * Silently breaking changes! - The Arbitrary instance for Word now generates only small values, the same as Int - cover no longer causes a property failure if coverage is insufficient. It just prints a warning. (But see next item!) * Overhaul of label/cover family of combinators: - New property combinator checkCoverage, which checks coverage requirements in a statistically sound way, and *does* fail if they are not met. - Order of arguments to cover swapped, to make it easier to switch between classify and cover - New combinators tabulate and coverTable, for reporting test case distribution more flexibly than label. - When label is called multiple times in a property, each call produces a separate table of frequencies. * New functions: - (=/=): like (/=), but prints a counterexample (thanks to tom-bop) - forAllShow/forAllShrinkShow: quantification using an explicit show function (thanks to Stevan Andjelkovic) - forAllBlind/forAllShrinkBlind: quantification without printing anything - verboseShrinking: see how a counterexample is shrunk - labelledExamples: given a property which uses label, generate an example test case for each label - idempotentIOProperty: a variant of ioProperty which shrinks better but only works for idempotent I/O actions * Other improvements: - MonadFix Gen instance (thanks to Jon Fowler) - Rational numbers shrink using continued fractions (thanks to Justus Sagemüller) - Function instances for Const, Identity, and the types in Data.Monoid; instance Functor Fun (thanks to Erik Schnetter and Xia Li-yao) - More of Test.QuickCheck.Function is exported from Test.QuickCheck - Semantics of .||. changed to improve short-circuiting: if the left argument's precondition is false, the right argument is not evaluated and the whole disjunction is considered to have a false precondition - Bug fix: suchThatMaybe always increased size to at least 1 * Miscellaneous API changes: - Result type has changed a bit: - InsufficientCovered constructor is gone - Type of labels has changed - New fields classes, tables QuickCheck 2.11.1 - 2.11.3 (released 2018-01-12) * Cosmetic fixes. QuickCheck 2.11 (released 2018-01-12) * New features: - InfiniteList modifier generates infinite lists and shows only the relevant part. - applyArbitrary2/3/4 for applying a function to random arguments. - Template Haskell function allProperties returns all properties in a module. * Applicative Gen instances do less splitting. * Property now has a Typeable instance. * (===) now prints correct output when the property is true. * Test.QuickCheck now exports Fun constructor. * verboseCheck output is now slightly less confusing. QuickCheck 2.10.1 (released 2017-10-06) * Arbitrary instances for Foreign.C.Types are available in more GHC versions. * Fixed a bug where withMaxSuccess didn't adjust the allowed number of discarded tests. * Remove quadratic behaviour in terminal output. QuickCheck 2.10 (released 2017-06-15) * New combinators: - withMaxSuccess sets the maximum number of test cases for a property. - shrinkMap/shrinkMapBy are helpers for defining shrink functions. - total checks that a value is non-crashing. - suchThatMap is similar to 'suchThat' but takes a Maybe-returning function instead of a predicate. - getSize returns the current test case size. * Random strings and characters now include Unicode characters by default. To generate only ASCII characters, use the new ASCIIString modifier or arbitraryASCIIChar generator. The following modifiers and generators also control the kind of strings generated: UnicodeString, PrintableString, arbitraryUnicodeChar, arbitraryPrintableChar. * QuickCheck no longer catches asynchronous exceptions, which means that pressing ctrl-C will now cancel testing without printing a counterexample. If you are debugging an infinite loop, please use the 'within' combinator or 'verboseCheck' instead. ('within' is better as it allows the counterexample to be shrunk.) * Much of Test.QuickCheck.Function (showable random functions) is now exported from Test.QuickCheck. - Test.QuickCheck.Function now defines functions and pattern synonyms which simplify testing functions of more than one argument: apply2, apply3, Fn2, Fn3. * New typeclasses Arbitrary1 and Arbitrary2 which lift Arbitrary to unary/binary type constructors, like in Data.Functor.Classes. * Some Arbitrary instances have been removed: NonEmpty, Natural. This is because they resulted in a lot of extra dependencies. You can now find them in the quickcheck-instances package. Alternatively, use the NonEmptyList and NonNegative modifiers. * New Arbitrary instances for the following types: Proxy, ExitCode, WrappedMonad, WrappedArrow, QCGen, and the types in Foreign.C.Types and Data.Functor.{Product,Compose}. Also a Function instance for Word. * The functions in Test.QuickCheck.Monadic which take an argument of type PropertyM m a now expect that 'a' to be Testable, and test it. To reduce breakage from this, () is now an instance of Testable which always succeeds. - PropertyM now has a MonadFail instance on recent GHCs. Furthermore, the constraints on some instances were loosened. * Miscellaneous API changes: - Result now returns the counterexample as a list of strings. See the "failingTestCase" field. - Args now has a maxShrinks argument, the maximum number of shrinks to try before giving up shrinking. - The 'labels' field of Result now encodes frequencies as Doubles rather than Ints. * Bugfixes: - 'Test.QuickCheck.Function', 'Test.QuickCheck.Poly', and 'Test.QuickCheck.Monadic' are now Safe modules. - Result.theException and Result.reason were taken from the pre-shrunk counterexample, not the shrunk one. - The Testable Property instance improperly used 'again'. - Gen.>>= is strict in the result of split, fixing a space leak. - within now gives a better error message on timeout * Some more examples and links have been added to the documentation. QuickCheck 2.9.2 (released 2016-09-15) * Fix a bug where some properties were only being tested once * Make shrinking of floating-point values less aggressive * Add function chooseAny :: Random a => Gen a QuickCheck 2.9.1 (released 2016-07-11) * 'again' was only used in forAllShrink, not forAll QuickCheck 2.9 (released 2016-07-10) * Arbitrary, CoArbitrary and Function instances for more types * Generics for automatic Function instances * A new combinator "again" which undoes the effect of "once" * Remove "exhaustive" from Testable typeclass; instead, combinators which are nonexhaustive (such as forAll) call "again", which should be more robust * Drop support for GHC 6.x * Fixed bugs: * arbitrarySizedBoundedIntegral wasn't generating huge integers * verboseCheck failed with Test.QuickCheck.Function * label had a space leak QuickCheck 2.8.2 (released 2016-01-15) * GHC 8 support * Add Arbitrary and CoArbitrary instances for types in containers package * Improve speed of shuffle combinator * Only print to stderr if it's a terminal. * Small changes: slightly improve documentation, remove redundant constraints from some functions' types, small improvements to Test.QuickCheck.All. QuickCheck 2.8.1 (released 2015-04-03) * Fix bug where exceptions thrown printing counterexamples weren't being caught when terminal output was disabled * Don't export Test.QuickCheck.Property.result QuickCheck 2.8 (released 2015-03-18) * New features: * Support for GHC 7.10 * Arbitrary instance for Natural * New generators shuffle and sublistOf * Support for generic coarbitrary * When using the cover combinator, insufficient coverage now causes the property to fail * API changes: * Test.QuickCheck.Function: new pattern synonym Fn * genericShrink no longer requires Typeable * Result has a new constructor InsufficientCoverage * resize throws an error if the size is negative * Bug fixes: * Fix memory leaks * Exceptions thrown by callbacks now cause the test to fail * Fixed a bug where the cover combinator wouldn't give a warning if coverage was 0% QuickCheck 2.7.3 (released 2014-03-24) * Add annotations for Safe Haskell. QuickCheck 2.7.2 (released 2014-03-22) * Fix bug in cabal file which broke cabal test QuickCheck 2.7.1 (released 2014-03-20) * Fixed bug - the Small modifier didn't work on unsigned types * Changed arbitrarySizedIntegral to have an Integral constraint instead of just Num QuickCheck 2.7 (released 2014-03-19) * New features: * New genericShrink function provides generic shrinking with GHC. * New combinator x === y: fails if x /= y, but also prints their values * New function generate :: Gen a -> IO a for running a generator. * New combinators infiniteList and infiniteListOf for generating infinite lists. * Several combinators added to the main Test.QuickCheck module which were previously languishing in other modules. Of particular interest: quickCheckAll, ioProperty. * New combinators delay and capture which can be used (unsafely!) to reuse the random number seed. Useful for generating polymorphic (rank-2) values. * A new Discard data type and a Testable instance for discarding test cases. * All modifiers now have Functor instances and accessor functions. * Pressing ctrl-C during shrinking now shows the last failed test case, rather than the current shrinking candidate. * Experimental support for UHC. You will need the latest version of Cabal from git. * Better distribution of test data: * The Int generator now only generates fairly small numbers. * The new Small and Large modifiers control the distribution of integers (Small generates small numbers, Large from the whole range). * Floating-point numbers shrink better. * Improved random number generation: * QuickCheck now uses TFGen rather than StdGen on newer versions of GHC, because StdGen's random numbers aren't always random. * 'variant' now uses a prefix code. This should prevent some potential bananaskins with coarbitrary. * API changes: * The Gen monad now uses an abstract type QCGen rather than StdGen. * The Result type now returns the thrown exception and number of failed shrink attempts. * Property is now a newtype rather than Gen Prop as it was before. * promote is moved into the new module Test.QuickCheck.Gen.Unsafe. * 'printTestCase' is deprecated - its new name is 'counterexample' * 'morallyDubiousIOProperty' is deprecated - its new name is 'ioProperty', no moral judgement involved :) QuickCheck 2.6, released 2013-03-07 * Add convenience Function instances for up to 7-tuples * Make stderr line buffered to reduce console I/O. * Return a flag to say whether the test case was interrupted. QuickCheck 2.5, released 2012-06-18 * Replace maxDiscard with maxDiscardRatio * Remove Testable () instance. * Added a 'discard' exception that discards the current test case * Add accessors for modifiers (where it makes sense) * Rename 'stop' to 'abort' to avoid a name clash * Added a 'once' combinator * If a property is of type Bool, only run it once * Add coarbitraryEnum to Test.QuickCheck module. * Add 'coarbitrary' helper for Enums. * Rejiggled the formatting code to support multi-line error messages * Add instances for Ordering and Fixed. * Added arbitraryBoundedEnum generator (thanks to Antoine Latter). * Add verboseCheckAll and polyverboseCheck function for usability.

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# Laplace to Fourier conversion + plotting

#### AlexMak

Joined Jan 2, 2018
32
Hello, so I have a signal that I need to convert to frequency. I have the signal in laplace transform now and I heard that I can convert it to fourier transform by s = jw. The problem is I can't plot the signal that way. Is there something I'm missing / doing wrong ?

#### MrAl

Joined Jun 17, 2014
6,838
Hello,

If i understand you right then you have to take your expression and plot the magnitude and phase shift for a frequency of choice 'w' where w=2*pi*f.

#### AlexMak

Joined Jan 2, 2018
32
Hello,

If i understand you right then you have to take your expression and plot the magnitude and phase shift for a frequency of choice 'w' where w=2*pi*f.
I must turn the signal to frequency and plot it. Im not aware of the magnitude and phase shift as they might have a different meaning in my language. I understood that you can link the laplace to the fourier transform and obtain the frequency signal by replacing s=iw but i still dont know how to plot it.

#### simozz

Joined Jul 23, 2017
104
When you plot the power spectrum in the frecueny domain, you plot $$|X(\text{i}\omega)|^2$$.

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# Diagonalization of a Toeplitz matrix

Let $0<\lambda\leq1$ so that the $n \times n$ matrix $$\Sigma = \begin{pmatrix} 1&1-\lambda& \cdots &1-\lambda\\ 1-\lambda&\ddots&\ddots& \vdots\\ \vdots &\ddots&\ddots&1-\lambda\\ 1-\lambda&\cdots&1-\lambda&1\\ \end{pmatrix}$$ is positive definite. I believe we can orthogonally diagonalize $\Sigma$ as $$\Sigma = VDV^T$$ where $$V = \begin{pmatrix} \frac{-1}{\sqrt{2 \cdot 1}} & \frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\cdots&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ 0&\cdots&\cdots&0&\sqrt{\frac{n-1}{n}}&\frac{1}{\sqrt{n}}\\ 0&\cdots&0&\sqrt{\frac{n-2}{n-1}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ \vdots&\iddots&\iddots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\vdots&\vdots\\ 0&\iddots&\iddots&\vdots&\vdots&\vdots\\ \sqrt{\frac{1}{2}}&\frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\ \end{pmatrix}$$

$$D = \begin{pmatrix} \lambda&0& \cdots &0\\ 0&\ddots&\ddots& \vdots\\ \vdots &\ddots&\lambda&0\\ 0&\cdots&0&n-(n-1)\lambda\\ \end{pmatrix}$$

I am having some trouble showing this result, can someone offer a suggestion for the proof?

## migrated from mathoverflow.netJun 25 '14 at 1:15

This question came from our site for professional mathematicians.

The matrix $\Sigma$ can be written as $$\Sigma=\lambda I+(1-\lambda)E,$$ where $E=ee^T$, $e=[1,1,\ldots,1]^T$. If $V$ is an orthogonal matrix such that $V^TEV=D$ is diagonal, then $$\tag{1} V^T\Sigma V=\lambda I+(1-\lambda)D$$ is the diagonalization of $\Sigma$.
The only thing which remains to find are the matrices $D$ and $V$. Since $E$ has rank one, it has one nonzero eigenvalue and $n-1$ zero eigenvalues and we can chose an orthonormal set of eigenvectors since $E$ is symmetric.
We have $$Ee=ne,$$ hence $n$ is the only nonzero eigenvalue of $E$ with the constant eigenvector $v_1=e/\sqrt{n}$ (normalized to $\|v_1\|=1$). The other eigenvectors $v_2,\ldots,v_n$ of $E$ corresponding to the zero eigenvalues can be any set of orthonormal eigenvectors orthogonal to $e$. Indeed, if $0\neq v\perp e$, then $$Ev=(ee^T)v=e(e^Tv)=0.$$ Since the eigenvalues of $\Sigma$ are $\lambda$ plus $(1-\lambda)$ times the eigenvalues of $E$, the eigenvalues of $\Sigma$ are $\lambda$ with the multiplicity $n-1$ and the simple eigenvalue $\lambda+(1-\lambda)n$.
Note that there are infinitely many bases of the eigenspace of $E$ corresponding to the zero eigenvalues. You can chose any $n-1$ orthonormal vectors orthogonal to a constant vector.


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• The SHiP experiment is designed to search for very weakly interacting particles beyond the Standard Model which are produced in a 400 GeV/c proton beam dump at the CERN SPS. An essential task for the experiment is to keep the Standard Model background level to less than 0.1 event after $2\times 10^{20}$ protons on target. In the beam dump, around $10^{11}$ muons will be produced per second. The muon rate in the spectrometer has to be reduced by at least four orders of magnitude to avoid muon-induced combinatorial background. A novel active muon shield is used to magnetically deflect the muons out of the acceptance of the spectrometer. This paper describes the basic principle of such a shield, its optimization and its performance.
• A new general purpose fixed target facility is proposed at the CERN SPS accelerator which is aimed at exploring the domain of hidden particles and make measurements with tau neutrinos. Hidden particles are predicted by a large number of models beyond the Standard Model. The high intensity of the SPS 400~GeV beam allows probing a wide variety of models containing light long-lived exotic particles with masses below ${\cal O}$(10)~GeV/c$^2$, including very weakly interacting low-energy SUSY states. The experimental programme of the proposed facility is capable of being extended in the future, e.g. to include direct searches for Dark Matter and Lepton Flavour Violation.


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## Railways Quant Test 99

Instructions

For the following questions answer them individually

Q 1

The quotient in a division is 403. The divisor is 100 and the remainder is 58, the dividend is-

Q 2

Which of the fractions given below is not equal to $$\frac{5}{9}$$ ?

Q 3

Marked price of an article is Rs. 500. The shopkeeper sells the article at successive discounts of 15% and 10%. Another shopkeeper sells the same article and gives the successive discounts of 9% and 16%. From which shopkeeper the customer will get more benefit ?

Q 4

Ram spends 20% of his monthly income on his household expenditure, 15% of the rest on books. 30% of rest on clothes and saves the rest. On counting, he comes to know he has finally saved Rs. 9,520. Find out his monthly income?

Q 5

When product of two numbers is 768 and LCM is 96 what is the HCF of these numbers ?


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Subjects -> TRANSPORTATION (Total: 214 journals)     - AIR TRANSPORT (9 journals)    - AUTOMOBILES (26 journals)    - RAILROADS (10 journals)    - ROADS AND TRAFFIC (9 journals)    - SHIPS AND SHIPPING (43 journals)    - TRANSPORTATION (117 journals) TRANSPORTATION (117 journals)
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Similar Journals
 Journal on Vehicle Routing AlgorithmsNumber of Followers: 0      Hybrid journal (It can contain Open Access articles) ISSN (Print) 2367-3591 - ISSN (Online) 2367-3605 Published by Springer-Verlag  [2469 journals]
• Allocating limited deicing resources in winter snow events

Abstract: Abstract As the number and severity of snowfall events continue to grow, the need to direct road maintenance intelligently during these snowfall events will also grow. In several locations, local governments lack the resources to treat all roadways completely during snow events. Furthermore, some governments utilize only traffic data to determine which roads should be treated. As a result, many schools, businesses, and government offices must be unnecessarily closed, which directly impacts the social, educational, and economic well-being of citizens and institutions. In this work, we propose a mixed integer programming (MIP) formulation to allocate resources to manage snowfall on roads using meteorological, geographical, and environmental parameters. Given that MIPs often scale poorly with problem size, we also present a constructive heuristic designed for the problem. Additionally, we evaluate the impacts of an increase in budget for winter road maintenance on snow control resources.
PubDate: 2019-12-01
DOI: 10.1007/s41604-019-00015-4

• On a multi-trip vehicle routing problem with time windows integrating
European and French driver regulations

Abstract: Abstract In the European Union, truck drivers must comply with Regulation (EC) No 561/2006, which provides a set of rules on the daily or weekly working time and on break or rest periods after specific working times. Truck drivers can be back to the depot much earlier than the end of their working period, and be able to perform other trips. The resulting problem is a Multi-Trip Vehicle Routing Problem with Time Windows, where truck drivers have the possibility to accomplish, during their time horizon, more than one trip, combined with a Truck Driver Scheduling Problem. It aimed to find and assign a set of routes to a fleet of vehicles allowing each customer to be visited within its given time window, and the accumulated load to be delivered (or collected) not exceeding the capacity of the vehicle. Each truck driver must comply with applicable hours of service regulations, and the objective is to minimize total transportation cost. To solve this problem, we have developed a hybrid method composed of a genetic algorithm and a local neighborhood search. Experiments have been conducted on instances from the literature and real-life cases from SYSTRANS, a specialized French software company.
PubDate: 2019-12-01
DOI: 10.1007/s41604-019-00016-3

• A dial-a-ride problem using private vehicles and alternative nodes

Abstract: Abstract This paper addresses the dial-a-ride problem (DARP) using private vehicles and alternative nodes (DARP-PV-AN). The DARP consists of creating vehicle routes in order to ensure a set of users’ transportation requests. Each request corresponds to a client needing to be transported from his/her origin to his/her destination. Routing costs have to be minimized while respecting a set of constraints like time windows and maximum route length. In the classical DARP, vehicles have to start from a depot and come back to it at the end of their route. In the DARP-PV-AN, the on-demand transportation service can be done either by a public fleet or by clients that use their private vehicles. The use of these vehicles adds more flexibility and unclog the public transportation fleet by allowing clients to organize their own transportation. However, it also raises some privacy concerns. The DARP-PV-AN addresses these concerns and focuses on location privacy, i.e., the ability to prevent the third parties from learning clients’ locations, by keeping both original and final locations private. This is addressed by setting several pickup/delivery nodes for the transportation requests, thus masking the private address. A compact mixed integer linear model is presented, and an evolutionary local search (ELS) is proposed to compute solutions of good quality for the problem. These methods are benchmarked on a modified set of benchmark instances. A new set of realistic instances is also provided to test the ELS in a more realistic way.
PubDate: 2019-12-01
DOI: 10.1007/s41604-019-00014-5

• Solving a real-life roll-on–roll-off waste collection problem with
column generation

Abstract: Abstract In this paper, we develop an intelligent solution to a complex, real-life vehicle routing problem in the waste management sector. Waste management companies deliver various types of empty waste containers to industrial customers, to be filled with different types of waste, after which the containers are picked up again. The full containers are transported to a so-called waste management depot or waste handling depot, where they are emptied, after which they are reused. Empty containers are stocked in various stock depots. Time windows in which containers may be picked up and delivered at the customers and opening hours of the different depots have to be taken into account additionally. Very importantly, there are two types of waste collection trucks that can, respectively, carry one or two containers. The resulting vehicle routing problem belongs to a class of so-called roll-on–roll-off problems, characterized by unit demand. Additionally, the problem discussed in this paper has several characteristics (multiple waste types, multiple container types, multiple depots, a heterogeneous fleet, pickup and delivery, time window constraints, and other) that make solving it a difficult task. To solve this problem, we develop a novel column generation scheme that incorporates a heuristic approach to generate new columns. The master problem is solved by linear relaxation of a set partitioning problem. New routes (columns) are generated by a constructive heuristic loosely based on Solomon’s insertion heuristic for the vehicle routing problem with time windows. The construction heuristic operates on a reformulated version of the roll-on–roll-off problem, i.e., a generalized vehicle routing problem with time windows. The algorithm—called wmpopt—is submitted to an extensive sensitivity analysis to determine its response to different parameter settings. After this, we test it on four real-life problem instances and compare its results to those obtained by a commercial solver. We show that wmpopt achieves much better solutions than the commercial solver in similar computing times.
PubDate: 2019-05-10
DOI: 10.1007/s41604-019-00013-6

• Emergent intelligence technique-based transport depot resource management
in a metropolitan area

Abstract: Abstract A metropolitan area experiences transport resource scarcities and dynamic resource allocation problems due to huge number of commuters’ population, dynamic commuters’ arrival rates, etc. The progress of these problems results into economic loss, traffic congestion, inefficient utilization of resources and increase in indefinite waiting time of commuters. Hence, we propose a dynamic transport resource allocation scheme. The scheme efficiently manages number of vehicles over time and space using emergent intelligence (EI) technique. Based on the historical information, commuters’ arrival rates, resource availability, deficit resources, surplus resources of neighbor depots, etc., the EI technique with agents is used for collecting, analyzing, sharing and allocating resources, for optimal utilization of resources in a metropolitan area. Analytical model of EI technique-based transport resource allocation scheme is developed using Markov chain, and derived the closed-form expression of optimal utilization of resources. Performance evaluation is carried out with varying number of commuters’ arrival rates, vehicles, requests and number of neighborhood depots. We observe that results obtained in the simulation and analysis reflect the effectiveness of the proposed scheme in a metropolitan area.
PubDate: 2019-03-18
DOI: 10.1007/s41604-019-00012-7

• The VeRoLog Solver Challenge 2019

Abstract: Abstract The VeRoLog Solver Challenge 2019 is the $${4{\mathrm{th}}}$$ solver challenge facilitated by VeRoLog, the EURO Working Group on Vehicle Routing and Logistics. The challenge is organized in cooperation with ORTEC. On behalf of the organizing committee, the authors report on the problem and organization of the challenge.
PubDate: 2019-03-13
DOI: 10.1007/s41604-019-00011-8

• On automatic algorithm configuration of vehicle routing problem solvers

Abstract: Abstract Many of the algorithms for solving vehicle routing problems expose parameters that strongly influence the quality of obtained solutions and the performance of the algorithm. Finding good values for these parameters is a tedious task that requires experimentation and experience. Therefore, methods that automate the process of algorithm configuration have received growing attention. In this paper, we present a comprehensive study to critically evaluate and compare the capabilities and suitability of seven state-of-the-art methods in configuring vehicle routing metaheuristics. The configuration target is the solution quality of eight metaheuristics solving two vehicle routing problem variants. We show that the automatic algorithm configuration methods find good parameters for the vehicle route optimization metaheuristics and clearly improve the solutions obtained over default parameters. Our comparison shows that despite some observable differences in configured performance there is no single configuration method that always outperforms the others. However, largest gains in performance can be made by carefully selecting the right configurator. The findings of this paper may give insights on how to effectively choose and extend automatic parameter configuration methods and how to use them to improve vehicle routing solver performance.
PubDate: 2019-02-22
DOI: 10.1007/s41604-019-00010-9

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How many terms of the A.P. -6, $$- \dfrac{11}{2}$$, -5, …. are needed to give the sum –25?

Asked by Pragya Singh | 1 year ago |  104

##### Solution :-

-25 is the sum of n terms of the given A.P.

The common difference of the A.P. is $$d=-\dfrac{11}{2}+6$$

$$\dfrac{-11+12}{2}= \dfrac{1}{2}$$

The sum of first n terms of an arithmetic progression is given by the equation

$$s_n= \dfrac{n}{2}[2a+(n-1)d]$$

Substitute Sn=-25, a=-6 and $$d= \dfrac{1}{2}$$ in the equation

$$-25= \dfrac{n}{2}[2(-6)+(n-1)\dfrac{1}{2}]$$

$$-50= n[-12+\dfrac{n}{2}-\dfrac{1}{2}]$$

$$-50= n[\dfrac{25}{2}+\dfrac{1}{2}]$$

-100 = n [-25 + n]

= n2-25n+100=0

Factorize the equation.

= n2-5n+20n+100=0

= n(n-5)-20(n-5)=0

= n=20 or 5

Therefore, 5 or 20 terms of the A.P. are needed to give the sum -25  .

Answered by Abhisek | 1 year ago

### Related Questions

#### Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

#### Find the two numbers whose A.M. is 25 and GM is 20.

Find the two numbers whose A.M. is 25 and GM is 20.

#### If a is the G.M. of 2 and 1/4 find a.

If a is the G.M. of 2 and $$\dfrac{1}{4}$$ find a.

#### Find the geometric means of the following pairs of numbers

Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a3b and ab3

(iii) –8 and –2

Insert 5 geometric means between $$\dfrac{32}{9}$$ and $$\dfrac{81}{2}$$.


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# What is the ionic compound name for k2s?

##### 1 Answer
May 22, 2018

$\text{Potassium sulfide....}$

#### Explanation:

${K}_{2} S$ is composed of $2 \times {K}^{+}$ and $1 \times {S}^{2 -}$ ions...


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## Promiscuidade Os Pivetes De Katia Tvrip Avi 34

Promiscuidade Os Pivetes De Katia Tvrip Avi 34

Promiscuidade Os Pivetes De Katia Tvrip Avi 34

Format: avi Quality: TVRip Starring: Frank Birch, Paul Neil, Annie Hulme, Sheila Poole.. Çalışırken Geceler (1991) Le crime de fuite 9. Guest. Sign up. ; W. Ware.9 (1961) Kaur-Saneyja, P.9 (1961) Kaur-Saneyja, P.9: Sleepless 9. Now that was a night to remember. By [C1] | 11-08-2014, 08:51 PM. 16:36. promiscuidade os pivetes de katia tvrip avi 34 Cracked 2022 Latest Version Porra Urrutia asi. – 1893. 67. promiscuidade os pivetes de katia tvrip avi 34 Cracked Version Porra Urrutia asi. – 1893. 67. Now that was a night to remember. By [C1] | 11-08-2014, 08:51 PM. 16:36. Promiscuidade, os Pivetes de Katia TVrip Avi 34 – Porra Urrutia Asi. Serhebti ) 93 67.Q: How can I see my CPU usage in my MYSQL database? On my db server, I don’t have a lot of GUI programs running (except for MYSQL). I’ve noticed some time is being spent with the CPU and I’m not sure why. Is there a way to view my CPU usage in my MYSQL database? Thanks. A: You can use the SHOW STATUS command. See the manual for details on which variables can be retrieved. Since you are using MySQL 5.0 or higher you can use the redesigned View system tables feature introduced in 5.0, which lets you see information about all the tables on your server. Q: How to show these two sequences converge to $0$ in $l^2$ norm. I have seen these questions on MSE, but I don’t think this is really homework, so I was wondering how to approach it? I understand that since they are in $l^2$, the sequence has to satisfy \$||a_n-0||=|\int_0^1a_n(x)dx|


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# Inorganic Chemistry/Chemical Bonding/Orbital hybridization

Four sp3 orbitals.
Three sp2 orbitals.

In chemistry, hybridisation (or hybridization) is the concept of mixing atomic orbitals into new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory. Hybrid orbitals are very useful in the explanation of molecular geometry and atomic bonding properties.[1]

## Historical developmentEdit

Chemist Linus Pauling first developed hybridisation theory[2] in order to explain the structure of molecules such as methane (CH4). Historically, this concept was developed for such simple chemical systems but the approach was later applied more widely, and today it is considered an effective heuristic for rationalizing the structures of organic compounds.

Orbitals are a model representation of the behavior of electrons within molecules. In the case of simple hybridisation, this approximation is based on atomic orbitals, similar to those obtained for the hydrogen atom, the only atom for which an exact analytic solution to its Schrödinger equation is known. In heavier atoms, like carbon, nitrogen, and oxygen, the atomic orbitals used are the 2s and 2p orbitals, similar to excited state orbitals for hydrogen. Hybrid orbitals are assumed to be mixtures of these atomic orbitals, superimposed on each other in various proportions. It provides a quantum mechanical insight to Lewis structures. Hybridisation theory finds its use mainly in organic chemistry.

## spx and sdx terminologyEdit

This terminology describes the weight of the respective components of a hybrid orbital. For example, in methane, the C hybrid orbital which forms each C-H bond consists of 25% s character and 75% p character and is thus described as sp3 (read as s-p-three) hybridised. Quantum mechanics describes this hybrid as an sp3 wavefunction of the form N[s + (√3)pσ], where N is a normalization constant (here 1/2) and pσ is a p orbital directed along the C-H axis to form a sigma bond. The p-to-s ratio (denoted λ in general) is √3 in this example, and N2λ2 = 3/4 is the p character or the weight of the p component.

For atoms forming equivalent hybrids with no lone pairs, there is a correspondence to the number and type of orbitals used. For example, sp3 hybrids are formed from one s and three p orbitals. However, in all other cases, there is no such correspondence. The two bond-forming hybrid orbitals of oxygen in water can be described as sp4, which means that they have 20% s character and 80% p character, but does not imply that they are formed from one s and four p orbitals. As a result, the amount of p-character is not restricted to integer values; i.e., hybridisations like sp2.5 are also readily described.

In general, any two hybrid orbitals on the same atom must be mutually orthogonal. For an atom with s and p orbitals forming hybrids hi and hj with included angle $\theta$, the orthogonality condition implies the relation: 1 + $\lambda$i$\lambda$j cos($\theta$) = 0. The p-to-s ratio for hybrid i is $\lambda$i2, and for hybrid j it is $\lambda$j2. The bond directed towards a more electronegative substituent tends to have higher p-character as stated in Bent's rule. In the special case of equivalent hybrids on the same atom, again with included angle $\theta$, the equation reduces to just 1 + $\lambda$2 cos($\theta$) = 0. For example, BH3 has a trigonal planar geometry, three 120° bond angles, three equivalent hybrids about the boron atom, and thus 1 + $\lambda$2 cos($\theta$) = 0 becomes 1 + $\lambda$2 cos(120°) = 0, giving $\lambda$2 = 2 for the p-to-s ratio. In other words, sp2 hybrids.

An analogous notation is used to describe sdx hybrids. For example, the permanganate ion (MnO4-) has sd3 hybridisation with orbitals that are 25% s and 75% d.

## Types of hybridisationEdit

### sp3 hybridsEdit

Hybridisation describes the bonding atoms from an atom's point of view. That is, for a tetrahedrally coordinated carbon (e.g., methane CH4), the carbon should have 4 orbitals with the correct symmetry to bond to the 4 hydrogen atoms.

Carbon's ground state configuration is 1s2 2s2 2px1 2py1 or more easily read:

$C\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\downarrow}{2s}\; \frac{\uparrow\,}{2p_x}\; \frac{\uparrow\,}{2p_y}\; \frac{\,\,}{2p_z}$

(Note: The 1s orbital is lower in energy than the 2s orbital, and the 2s orbital is lower in energy than the 2p orbitals)

The carbon atom can utilize its two singly occupied p-type orbitals (the designations px py or pz are meaningless at this point, as they do not fill in any particular order), to form two covalent bonds with two hydrogen atoms, yielding the "free radical" methylene CH2, the simplest of the carbenes. The carbon atom can also bond to four hydrogen atoms by an excitation of an electron from the doubly occupied 2s orbital to the empty 2p orbital, so that there are four singly occupied orbitals.

$C^{*}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\,}{2s}\; \frac{\uparrow\,}{2p_x} \frac{\uparrow\,}{2p_y} \frac{\uparrow\,}{2p_z}$

As the additional bond energy more than compensates for the excitation, the formation of four C-H bonds is energetically favoured.

Quantum mechanically, the lowest energy is obtained if the four bonds are equivalent which requires that they be formed from equivalent orbitals on the carbon. To achieve this equivalence, the angular distributions of the orbitals change via a linear combination of the valence-shell (Core orbitals are almost never involved in bonding) s and p wave functions[3] to form four sp3 hybrids.

$C^{*}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\,}{sp^3}\; \frac{\uparrow\,}{sp^3} \frac{\uparrow\,}{sp^3} \frac{\uparrow\,}{sp^3}$

In CH4, four sp3 hybrid orbitals are overlapped by hydrogen's 1s orbital, yielding four σ (sigma) bonds (that is, four single covalent bonds) of the same length and strength.

translates into

### sp2 hybridsEdit

Other carbon based compounds and other molecules may be explained in a similar way as methane. Take, for example, ethene (C2H4). Ethene has a double bond between the carbons. The Lewis structure looks like this:

Ethene Lewis Structure. Each C bonded to two hydrogen atoms and one double bond between them.

Carbon will sp2 hybridise, because hybrid orbitals will form only σ bonds and one π (pi) bond is required for the double bond between the carbons. The hydrogen-carbon bonds are all of equal strength and length, which agrees with experimental data.

In sp2 hybridisation the 2s orbital is mixed with only two of the three available 2p orbitals:

$C^{*}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\,}{sp^2}\; \frac{\uparrow\,}{sp^2} \frac{\uparrow\,}{sp^2} \frac{\uparrow\,}{p}$

forming a total of 3 sp2 orbitals with one p-orbital remaining. In ethylene the two carbon atoms form a σ bond by overlapping two sp2 orbitals and each carbon atom forms two covalent bonds with hydrogen by ssp2 overlap all with 120° angles. The π bond between the carbon atoms perpendicular to the molecular plane is formed by 2p–2p overlap.

### sp hybridsEdit

The chemical bonding in compounds such as alkynes with triple bonds is explained by sp hybridization.

$C^{*}\quad \frac{\uparrow\downarrow}{1s}\; \frac{\uparrow\,}{sp}\; \frac{\uparrow\,}{sp} \frac{\uparrow\,}{p} \frac{\uparrow\,}{p}$

In this model the 2s orbital mixes with only one of the three p-orbitals resulting in two sp orbitals and two remaining unchanged p orbitals. The chemical bonding in acetylene (C2H2) consists of spsp overlap between the two carbon atoms forming a σ bond and two additional π bonds formed by pp overlap. Each carbon also bonds to hydrogen in a sigma ssp overlap at 180° angles.

## Hybridisation and molecule shapeEdit

Hybridisation helps to explain molecule shape:

Classification Main group Transition metal[4]
AX2
• Linear (180°)
• sp hybridisation
• E.g., CO2
• Bent (90°)
• sd hybridisation
• E.g., VO2+
AX3
• Trigonal planar (120°)
• sp2 hybridisation
• E.g., BCl3
• Trigonal pyramidal (90°)
• sd2 hybridisation
• E.g., CrO3
AX4
• Tetrahedral (109.5°)
• sp3 hybridisation
• E.g., CCl4
• Tetrahedral (109.5°)
• sd3 hybridisation
• E.g., MnO4
AX5 -
• Square pyramidal (66°, 114°)[5]
• sd4 hybridisation
• E.g., Ta(CH3)5
AX6 -
• Trigonal prismatic (63°, 117°)[5]
• sd5 hybridisation
• E.g., W(CH3)6

### Main group compounds with lone electron pairsEdit

For main group compounds with lone electron pairs, the s-orbital lone pair (analogous to s-p mixing in molecular orbital theory) can be hybridised to a certain extent with the bond pairs[6] to maximize energetic stability according to its Walsh diagram. This rationalisation is applied to explain deviations in ideal bond angles (i.e. only p-orbitals used for bonding), most commonly in second and third period elements.

• Trigonal pyramidal (AX3E1)
• s-orbital can be hybridised with the three p-orbital bonds to give bond angles greater than 90°.
• E.g., NH3
• Bent (AX2E1-2)
• s-orbital lone pair can be hybridised with the two p-orbital bonds to give bond angles greater than 90°. The out-of-plane p-orbital can either be a lone pair or pi bond. If it is a lone pair, it results in inequivalent lone pairs contrary to the common picture depicted by VSEPR theory (see below).
• E.g., SO2, H2O
• Monocoordinate (AX1E1-3)
• s-orbital lone pair can be hybridised with the p-orbital bond. The two out-of-line p-orbitals can either be lone pairs or pi bonds. The p-orbital lone pairs are inequivalent from the s-rich lone pair.
• E.g., CO, SO, HF

## Hybridisation of hypervalent moleculesEdit

In general chemistry courses and mainstream textbooks, hybridisation is often presented for main group AX5 and above, as well as for transition metal complexes, using the hybridisation scheme first proposed by Pauling.

Classification Main group Transition metal
AX2 -
• Linear (180°)
• sp hybridisation
• E.g., Ag(NH3)2+
AX3 -
• Trigonal planar (120°)
• sp2 hybridisation
• E.g., Cu(CN)32−
AX4 -
• Square planar (90°)
• dsp2 hybridisation
• E.g., PtCl42−
AX5
• Trigonal bipyramidal (90°, 120°)
• sp3d hybridisation
• E.g., PCl5
• Trigonal bipyramidal (90°, 120°)
• dsp3 hybridisation
• E.g., Fe(CO)5
AX6
• Octahedral (90°)
• sp3d2 hybridisation
• E.g., SF6
• Octahedral (90°)
• d2sp3 hybridisation
• E.g., Mo(CO)6
AX7
• Pentagonal bipyramidal (90°, 72°)
• sp3d3 hybridisation
• E.g., IF7
• Pentagonal bipyramidal (90°, 72°)
• d3sp3 hybridisation
• E.g., V(CN)74−
AX8
• Square antiprismatic
• sp3d4 hybridisation
• E.g., IF8
• Square antiprismatic
• d4sp3 hybridisation
• E.g., Re(CN)83−
AX9 -
• Tricapped trigonal prismatic
• d5sp3 hybridisation
• E.g., ReH92−

However, such a scheme is now superseded as more recent calculations based on molecular orbital theory have shown that in main-group molecules the d component is insignificant, while in transition metal complexes the p component is insignificant (see below).

### Resonance descriptionEdit

As shown by computational chemistry, hypervalent molecules can only be stable given strongly polar (and weakened) bonds with electronegative ligands such as fluorine or oxygen to reduce the valence electron occupancy of the central atom to below 8[7] (or 12 for transition metals). This requires an explanation that invokes sigma resonance in addition to hybridisation, which implies that each resonance structure has its own hybridisation scheme. As a guideline, all resonance structures have to obey the octet rule for main group compounds and the dodectet (12) rule for transition metal complexes.

Classification Main group Transition metal
AX2 - Linear (180°)
AX3 - Trigonal planar (120°)
AX4 - Square planar (90°)
AX5 Trigonal bipyramidal (90°, 120°) Trigonal bipyramidal (90°, 120°)
• Fractional hybridisation (s and d orbitals)
• E.g., Fe(CO)5
AX6 Octahedral (90°) Octahedral (90°)
AX7 Pentagonal bipyramidal (90°, 72°) Pentagonal bipyramidal (90°, 72°)
• Fractional hybridisation (s and three d orbitals)
• E.g., V(CN)74−
AX8 Square antiprismatic Square antiprismatic
• Fractional hybridisation (s and three p orbitals)
• E.g., IF8
• Fractional hybridisation (s and four d orbitals)
• E.g., Re(CN)83−
AX9 - Tricapped trigonal prismatic
• Fractional hybridisation (s and five d orbitals)
• E.g., ReH92−

#### Main group compounds with lone pairsEdit

For hypervalent main group compounds with lone electron pairs, the bonding scheme can be split into two components: the "resonant bonding" component and the "regular bonding" component. The "regular bonding" component has the same description (see above), while the "resonant bonding" component consists of resonating bonds utilizing p orbitals. The table below shows how each shape is related to the two components and their respective descriptions.

Regular bonding component (marked in red) Bent Monocoordinate Seesaw (AX4E1) (90°, 180°, >90°) T-shaped (AX3E2) (90°, 180°) Linear (AX2E3) (180°) - Square pyramidal (AX5E1) (90°, 90°) Square planar (AX4E2) (90°) - Pentagonal pyramidal (AX6E1) (90°, 72°) Pentagonal planar (AX5E2) (72°)

## Clarifying misconceptionsEdit

### VSEPR electron domains and hybrid orbitals are differentEdit

The simplistic picture of hybridisation taught in conjunction with VSEPR theory does not agree with high-level theoretical calculations[6] despite its widespread usage in many textbooks. For example, following the guidelines of VSEPR, the hybridization of the oxygen in water is described with two equivalent lone electron-pairs.[8] However, molecular orbital calculations give orbitals that reflect the C2v symmetry of the molecule.[9] One of the two lone pairs is in a pure p-type orbital, with its electron density perpendicular to the H–O–H framework.[10] The other lone pair is in an approximately sp0.8 orbital that is in the same plane as the H–O–H bonding.[10] Photoelectron spectra confirm the presence of two different energies for the nonbonded electrons.[11]

### Non-inclusion of d-orbitals in main group compoundsEdit

In 1990, Magnusson published a seminal work definitively excluding the role of d-orbital hybridization in bonding in hypervalent compounds of second-row elements. This had long been a point of contention and confusion in describing these molecules using molecular orbital theory. Part of the confusion here originates from the fact that one must include d-functions in the basis sets used to describe these compounds (or else unreasonably high energies and distorted geometries result), and the contribution of the d-function to the molecular wavefunction is large. These facts were historically interpreted to mean that d-orbitals must be involved in bonding. However, Magnusson concludes in his work that d-orbital involvement is not implicated in hypervalency.[12]

### Non-inclusion of p-orbitals in transition metal complexesEdit

Similarly, p-orbitals have long been thought to be utilized by transition metal centers in bonding with ligands, hence the 18-electron description; however, recent molecular orbital calculations have found that such p-orbital participation in bonding is insignificant,[13] even though the contribution of the p-function to the molecular wavefunction is calculated to be somewhat larger than that of the d-function in main group compounds.

## Hybridisation theory vs. MO theoryEdit

Hybridisation theory is an integral part of organic chemistry and in general discussed together with molecular orbital theory in advanced organic chemistry textbooks although for different reasons. One textbook notes that for drawing reaction mechanisms sometimes a classical bonding picture is needed with 2 atoms sharing two electrons [14]. It also comments that predicting bond angles in methane with MO theory is not straightforward. Another textbook treats hybridisation theory when explaining bonding in alkenes [15] and a third [16] uses MO theory to explain bonding in hydrogen but hybridisation theory for methane.

Bonding orbitals formed from hybrid atomic orbitals may be considered as localized molecular orbitals, which can be formed from the delocalized orbitals of molecular orbital theory by an appropriate mathematical transformation. For molecules with a closed electron shell in the ground state, this transformation of the orbitals leaves the total many-electron wave function unchanged. The hybrid orbital description of the ground state is therefore equivalent to the delocalized orbital description for explaining the ground state total energy and electron density, as well as the molecular geometry which corresponds to the minimum value of the total energy.

There is no such equivalence, however, for ionized or excited states with open electron shells. Hybrid orbitals cannot therefore be used to interpret photoelectron spectra, which measure the energies of ionized states, identified with delocalized orbital energies using Koopmans' theorem. Nor can they be used to interpret UV-visible spectra which correspond to electronic transitions between delocalized orbitals. From a pedagogical perspective, hybridisation approach tends to over-emphasize localisation of bonding electrons and does not effectively embrace molecular symmetry as does MO theory.

## ReferencesEdit

1. "It is important to recognize that the VSEPR model provides an approach to bonding and geometry based on the Pauli principle that is completely independent of the valence bond (VB) theory or of any orbital description of bonding." Gillespie, R. J. J. Chem. Educ. 2004, 81, 298-304.
2. L. Pauling, J. Am. Chem. Soc. 53 (1931), 1367
3. McMurray, J. (1995). Chemistry Annotated Instructors Edition (4th ed.). Prentice Hall. p. 272. ISBN 0-13-140221-8
4. Weinhold, Frank; Landis, Clark R. (2005). Valency and bonding: A Natural Bond Orbital Donor-Acceptor Perspective. Cambridge: Cambridge University Press. pp. 381–383. ISBN 0-521-83128-8.
5. a b Martin Kaupp Prof. Dr. (2001). ""Non-VSEPR" Structures and Bonding in d(0) Systems.". Angew Chem Int Ed Engl. 40 (1): 3534-3565. doi:10.1002/1521-3773(20011001)40:19<3534::AID-ANIE3534>3.0.CO;2-#.
6. a b Weinhold, Frank. "Rabbit Ears Hybrids, VSEPR Sterics, and Other Orbital Absurdities". University of Wisconsin. Retrieved 2012-11-11.
7. David L. Cooper , Terry P. Cunningham , Joseph Gerratt , Peter B. Karadakov , Mario Raimondi (1994). "Chemical Bonding to Hypercoordinate Second-Row Atoms: d Orbital Participation versus Democracy". Journal of the American Chemical Society 116 (10): 4414–4426. doi:10.1021/ja00089a033.
8. Petrucci R.H., Harwood W.S. and Herring F.G. "General Chemistry. Principles and Modern Applications" (Prentice-Hall 8th edn 2002) p. 441
9. Levine I.N. “Quantum chemistry” (4th edn, Prentice-Hall) p. 470–2
10. a b Laing, Michael J. Chem. Educ. (1987) 64, 124–128 "No rabbit ears on water. The structure of the water molecule: What should we tell the students?"
11. Levine p. 475
12. E. Magnusson. Hypercoordinate molecules of second-row elements: d functions or d orbitals? J. Am. Chem. Soc. 1990, 112, 7940-7951. doi:10.1021/ja00178a014
13. O’Donnell, Mark (2012). "Investigating P-Orbital Character In Transition Metal-to-Ligand Bonding". Brunswick, ME: Bowdoin College. Retrieved 2012-09-16.
14. Organic Chemistry. Jonathan Clayden, Nick Greeves, Stuart Warren, and Peter Wothers 2001 ISBN 0-19-850346-6
15. Organic Chemistry, Third Edition Marye Anne Fox James K. Whitesell 2003 ISBN 978-0-7637-3586-9
16. Organic Chemistry 3rd Ed. 2001 Paula Yurkanis Bruice ISBN 0-13-017858-6


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# Find the force needed to accelerate body to a certain velocity for a certain time with respect to drag force

So, the problem is straightworward when we suggest that air resistance force is constant: $$\vec F = \frac {\vec V_1 - \vec V_0} {t} m / b$$ $$\vec V_0, \vec V_1 - \text {initial and final velocities respectively},\\ t - \text {time, during which velocity will become its final value}, \\b - \text {some constant for drag force}$$

The problem arises when using quadratic air resistance: $$m a = -c v^2$$ I don't have any clue for solving this and asking for help.

What I must do is to accelerate plane to the certain velocity in a certain period of time. So, I must find needed acceleration or force, itsn't matter, because can be easily recalculated.

Thanks in advance!

P.S. It's sad, but I'm not familiar with calculus, I'm more programmer, not mathematician. P.P.S. I know what differentiating and integration is, what derivative and displacement is, but really not familiar with differential equations, but guess it is here.

• Hi just letting you know your post is not rendering properly on my tablet (might be a temporary issue) and you have a better chance of an answer if you show the details of any attempts you have tried yourself. Best of luck with it.
– user81619
Jun 21, 2015 at 22:15
• I really don't understand this sentence: "The problem arises when using quadratically drag force. I don't have any clue for solving this and asking for help." Jun 21, 2015 at 22:30
• Hm, I don't know why it's rendering not properly... gonenc, sorry, I wrote this incorrect, was meant: "quadratic air resistance". Corrected now Jun 21, 2015 at 22:35
• Your description of $t$ is ... odd. So is the phrase "for a certain time". I don't know what you mean. It seems the problem is over specified, but I may not understand what you are saying. Is $t$ the duration that the force is acting? What do you know about the drag force? May 30, 2016 at 18:56

## 1 Answer

If your object has mass $M$ and you want to accelerate it with acceleration $a$ to a specific end-velocity $v$ you have to keep in mind that the energy

$$e = \frac{M\cdot v^2}{2}$$

and also

$$e = F\cdot x$$

where $x$ is the distance over which the force $F$ which equals $M\cdot a$ is applied. Knowing that you can solve for the distance over which you have to accelerate:

$$\frac{M\cdot v^2}{2} = M\cdot a\cdot x \to$$

$$\to x = \frac{v^2}{2\cdot a}$$

Now you have the distance and can easily solve for the time you need to reach your velocity $v$:

$$t = \sqrt{\frac{2\cdot x}{a}}$$

Example: If you have a drag force of $Fd = -4 \frac{kg \cdot m}{s^2}$, your object has a mass of $M = 3 kg$ and you want to reach $v = 10 m/s$ in \$t = 5 s:

$$a = \frac{v}{t} \to a = \frac{10 m/s}{5 s} = 2 \frac{m}{s^2}$$

You multiply that with yor mass and get a force of

$$F = 6 \frac{kg \cdot m}{s^2}$$

This is the force you would need without drag force. Now you simply substract the drag:

$$F_{total} = F-Fd = (6+4)N = 10 N$$

• Thanks, but you didn't understand me correctly. Because t is known. Maybe I should give concrete example. What I must do is to accelerate plane to the certain velocity in the some period of time. So, I must find needed acceleration and start simulation. Maybe it became clear. Jun 21, 2015 at 23:11
• In the example you find the way to calculate the force you have to put in when the time in which you have to reach your velocity is known. Jun 21, 2015 at 23:21
• Ha, yes, it is, but there is a problem, because Fd will vary on speed. This is actual problem. The main problem that further body's speed is unknown at start. Jun 21, 2015 at 23:22
• Drag will be ma=−cv^2, v is unknown. How can we solve this? Jun 21, 2015 at 23:24
• There is no analytical solution for this problem, if you want to have real air resistance you have to use the ballisic differential equation which is a litte more complicated and can only be solved numerically, see yukterez.net/ballistik/#plot Jun 21, 2015 at 23:40


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Definition:Smooth Real Function

Definition

A real function is smooth if and only if it is of differentiability class $C^\infty$.

That is, if and only if it admits of continuous derivatives of all orders.

Also defined as

Some sources define a smooth function as a real function which has a continuous first derivative everywhere.

Also see

My brother Esau is an hairy man, but I am a smooth man. -- Take A Pew


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# IRBIS: a systematic search for conserved complementarity.

 Title IRBIS: a systematic search for conserved complementarity. Publication Type Journal Article Year of Publication 2014 Authors Pervouchine DD Journal RNA (New York, N.Y.) Volume 20 Pagination 1519–1531 Date Published Oct ISSN 1469-9001 Abstract IRBIS is a computational pipeline for detecting conserved complementary regions in unaligned orthologous sequences. Unlike other methods, it follows the first-fold-then-align'' principle in which all possible combinations of complementary k-mers are searched for simultaneous conservation. The novel trimming procedure reduces the size of the search space and improves the performance to the point where large-scale analyses of intra- and intermolecular RNA-RNA interactions become possible. In this article, I provide a rigorous description of the method, benchmarking on simulated and real data, and a set of stringent predictions of intramolecular RNA structure in placental mammals, drosophilids, and nematodes. I discuss two particular cases of long-range RNA structures that are likely to have a causal effect on single- and multiple-exon skipping, one in the mammalian gene Dystonin and the other in the insect gene Ca-$\alpha$1D. In Dystonin, one of the two complementary boxes contains a binding site of Rbfox protein similar to one recently described in Enah gene. I also report that snoRNAs and long noncoding RNAs (lncRNAs) have a high capacity of base-pairing to introns of protein-coding genes, suggesting possible involvement of these transcripts in splicing regulation. I also find that conserved sequences that occur equally likely on both strands of DNA (e.g., transcription factor binding sites) contribute strongly to the false-discovery rate and, therefore, would confound every such analysis. IRBIS is an open-source software that is available at http://genome.crg.es/{\textasciitilde}dmitri/irbis/. URL http://www.ncbi.nlm.nih.gov/pubmed/25142064 DOI 10.1261/rna.045088.114


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DI Management Home > Cryptography > Public key cryptography using discrete logarithms > Part 1: Diffie-Hellman key exchange

# Public key cryptography using discrete logarithms. Part 1: Diffie-Hellman key exchange

In the introduction to the basic Diffie-Hellman key exchange method we noted that it is possible to fool users into using a generator $g$ which only generates a small subgroup of $\mathbb{Z}_p^*$. This page describes the standard solution.

 << previous: Introduction next: MQV Key Agreement >>

## Diffie-Hellman with a large cyclical subgroup

The usual solution to prevent the "small-group" problem is to make sure that the order of the group, $p-1$, has a large prime divisor $q$. This means there is a unique cyclic subgroup of order $q$ mod $p$, and we make sure that $g$ is a generator of this subgroup. In practice $q$ should be a prime at least 160 bits long, preferably 256 bits. This is too large for a feasible brute-force search or any of the other techniques available. The modulus $p$ should be at least 1024 bits.

The public domain parameters are now $(p,q,g)$, and parties Alice and Bob must select private keys in the range $[2,q-2]$ (see note **). A user being given new parameters generated by others can carry out basic checks to make sure they are not being fooled into using a small subgroup.

This method is described in ANSI X9.42 [X9-42] and RFC 2631 [RFC2631]. The technique of using a prime order $q$ subgroup was first identified by Claus-Peter Schnorr [SCH91].

Algorithm: Generate Domain Parameters.
INPUT: Required bit lengths for modulus $p$ and prime divisor $q$.
OUTPUT: Parameters $(p,q,g)$.
1. Generate a random prime $q$ of the required bit length.
2. Choose an even random number $j$ of bit length equal to bitlen($p$) $-$ bitlen($q$)
3. Compute $p = jq+1$. If $p$ is not prime, then go to step 2.
4. Choose a random number $h$ in the range $1 \lt h \lt p-1$.
5. Compute $g=h^j\mod p$. If $g = 1$ then go to step 4.
6. Return $(p,q,g)$.

Algorithm: Verify Domain Parameters.
INPUT: Parameters $(p,q,g)$.
OUTPUT: "Accept parameters" or "Reject parameters".
1. Check that $1 \lt g \lt p-1$. If not, then return "Reject parameters" and stop
2. Test $q$ for primality. If $q$ is not prime then return "Reject parameters" and stop
3. Test $p$ for primality. If $p$ is not prime then return "Reject parameters" and stop
4. Compute $(p-1) \mod q$. If this is not equal to 0 then return "Reject parameters" and stop
5. Compute $g^q \mod p$. If this is not equal to 1 then return "Reject parameters" and stop
6. Return "Accept parameters".

### ▷ Why does this work?

The group $\mathbb{Z}_p^*$ has order $p-1$. Any subgroup of $\mathbb{Z}_p^*$ will have order equal to a divisor of $p-1$ (by Lagrange's theorem). The divisors of $p-1$ are $q$ and $j$: $q$ is large and prime but $j$ is small and composite. We want to ensure that our generator $g$ does not have order equal to $j$ or one of its divisors. In fact, the method always gives a generator of order $q$ as we show below.

If, for any $h \in [2,p-2]$, we have $h^j = 1$ then $h$ has order $j$ or less, so we reject it. Otherwise $g=h^j \gt 1$ has the property that $g^q = \left(h^{(p-1)/q}\right)^q = h^{p-1} = 1$ (by Fermat's little theorem) and so $\text{order}(g)\leq q$. Since $q$ is prime and therefore has no proper divisors then the order of $g$ cannot be less than $q$, so $\text{order}(g)\geq q$. It follows that $g$ has order exactly $q$. You usually find a suitable $g$ on the first go.

### Example of parameter generation

This is an illustrative example with trivially small numbers.

• 1. "Generate" random prime $q=47$.
• 2a. Try $j=2$:
• 3a. Compute $p = jq + 1 = 2\cdot 47 + 1 = 95$, which is not prime, so go to step 2.
• 2b. Try $j=6$:
• 3b. Compute $p = jq + 1 = 6\cdot 47 + 1 = 283$, which is prime, so continue.
• 4. Try $h=5$:
• 5. Compute $g = h^j\mod p = 5^6 \mod 283 = 60 \neq 1$, so continue.
• 6. Publish $(p,q,g) = (283,47,60)$.

To verify these, we carry out the following checks:

1. $1 \lt g=60 \lt p-1 = 282$, so OK.
2. $q=47$ is prime, so OK.
3. $p=283$ is prime, so OK.
4. $(p-1)\mod q = (283-1)\mod 47 = 0$, so OK.
5. $g^q\mod p = 60^{47}\mod 283 = 1$, so OK.
6. All OK, so "Accept parameters".

Algorithm: Key Pair Generation.
INPUT: Parameters $(p,q,g)$.
OUTPUT: Party A's private/public key pair $(a, A)$.
1. Party A chooses a number $a$ in the range $[2,q-2]$.
2. Compute $A = g^a \mod p$.
3. Return $(a, A)$. Keep $a$ secret.

Think of $g^a\mod p$ as the action of "throwing" the secret key $a$ somewhere in the large range $[1,p-1]$. Because it is done modulo $p$ there is no efficient way to go backwards and find $a$ for a large enough $p$.

### Example of key pair generation

• A.1. Alice chooses $a=24$ and keeps it secret.
• A.2. Compute $A = g^a\mod p = 60^{24}\mod 283 = 158$.
• A.3. Alice sends $A=158$ to Bob.
• B.1. Bob chooses $b=7$ and keeps it secret.
• B.2. Compute $B = g^b\mod p = 60^7\mod 283 = 216$.
• B.3. Bob sends $B=216$ to Alice.

Algorithm: Computation of shared secret by Party A.
INPUT: Parameters $(p,q,g)$, $B$, $a$.
OUTPUT: Shared secret $Z$.
1. Check that $1 \lt B \lt p$ and that $B^q\mod p = 1$. If not then return "Failure" and stop.
2. Compute $Z = B^a \mod p$.
3. Return $Z$.

### ▷ Why do we do the checks in step 1?

The public keys $B=0$ and $B=1$ give no security at all, so we make sure $B\gt 1$. You have the same effect if you can trick someone into using $B=kp$ or $B=1+kp$, so we limit $B\lt p$.

The check $B^q\mod p = 1$ only holds if $g^q\mod p = 1$, which ensures that $g$ is a generator for the subgroup of order $q\mod p$. We have $B^q\equiv \left(g^b\right)^q \equiv \left(g^q\right)^b\equiv 1^b\equiv 1\pmod{p}$.

### Example of computation of shared secret

• COMMON INPUT: $(p=283, q=47, g=60)$
•
• INPUT FOR Alice: $B=216$, $a=24$.
• A.1. Check: $1 \lt B = 216 \lt p = 283$ and $B^q\mod p = 216^{47} \mod 283 = 1$ $\Rightarrow$ OK.
• A.2. Compute $Z = B^a\mod p = 216^{24}\mod 283 = 181$.
•
• INPUT FOR Bob: $A=158$, $b=7$.
• B.1. Check: $1 \lt A = 158 \lt p = 283$ and $A^q\mod p= 158^{47} \mod 283 = 1$ $\Rightarrow$ OK.
• B.2. Compute $Z = A^b\mod p = 158^{7}\mod 283 = 181$.

## Remarks

• In practice $q$ is at least 160 bits and $p$ at least 1024 bits. A real example with numbers in decimal is
q = 983633858469108611936846792207646525014934079943
j = 19648785705229832512739279235719869326994578857372093636756123297040568424679124
64540061234193169057945703804694798339619887840647564390726543645590797170304574
76946989704881415213211796495985450103517961540422421579976333688259982527955809
16414671992092563312
p = 19327210897467885519624495407304217845488409100133554803661172025039322784872775
17278952189544417869074042858818503169545381538675666261955584944665679490522111
57880020162452917682834724804605237775109730850324717111878065901859872191793450
22033106753600355795626394426859896564719805266547324204357196851217
g = 20088512678116493013820556973260022253215016292246160430979593078444726373397837
79480891271906681929732776937543331689329117914118665148580824850572191418544875
10980215434186216265442406596314406393660737560679656370638936273176777219436857
6684632589065496658911743756860379357301492526015846031839304359976
a = 443154410456340133792289316319263982636340525614
A = 19214527844626903057876819500198193169641589347178408096123216510760347079712191
31303229964898309632972369585510990825414095578611652854141854136282707158639017
76349124806492330431394779299032591894335300024413826806576151022179543880341303
35905388651243538583579147122642435281102052849841738314682543989601
b = 708552627548105121354432083524985769585714694203
B = 39370801881546774932592357170039231704520402513227223839981915392904026892875070
04151691833350902076026964761378570941292367186512432962797136953267476511207359
49561972379770636776112864650706982533730860846695796684838147638233797651141503
6913259368572438791510731359101113653373912723872286563003569148155
Z = 3319378743040852066281248525554807677790302702508026945509475209822628115979919
7208611387091356050777485694513339213618300369077644017879638157922142896658726
2535373398289694901821015274930091093239014497316409413910122183039716148353078
3129064601312291695971038344663201164529792929266992327467270782207

• The ordinary Diffie-Hellman method above is still susceptible to a man-in-the-middle attack because the key exchange messages cannot be authenticated.
• The parameter generation method described in X9.42 and RFC 2631 [X9-42, RFC2631] uses a random seed and counter to derive $p$ and $q$ using the SHA-1 hash algorithm. These extra two parameters are supplied along with $(p,q,g)$ and a recipient can use them to check that the parameters were indeed generated correctly from a random seed and are therefore unlikely to be vulnerable to a small subgroup attack or other trickery.
• The shared secret $Z$ is usually hashed along with mutually agreed salting material and a counter to generate a session key to be used to encrypt material to be passed between the parties.
Session Key K_1 = Hash(Z || salt || 0001)
Session Key K_2 = Hash(Z || salt || 0002)
...etc...

where the hashes are either concatenated or truncated to form a symmetrical encryption key of the required length (e.g. 128 bits for AES-128). Incidentally, the X9.42 test vectors are wrong.

## Notes

** Selecting private keys in the range $[2,q-2]$, or is it $[1,q-1]$?
Depending on which references you read, some will say select the private key $a$ in the range $[1,q-1]$ and others will say use $[2,q-2]$. In practice, we are looking at picking a random number in a range of billions, so it is extremely unlikely we'll end up with a value at the limits. However, in the pedantically unlikely event we do generate a private key $a=1$, we will get a public key $A=g^1$ which equals, er, $g$ and may give a clue to an adversary as to the value of your private key. Similarly if we generate $a=q-1$ we also get $A=g^{q-1}=1$ by Fermat's Little Theorem. So why take the risk?

## References

• [RFC2631] RFC 2631 Diffie-Hellman Key Agreement Method, E. Rescoria, June 1999. <http://www.ietf.org/rfc/rfc2631.txt>
• [SCH91] Schnorr, C. P. Efficient signature generation by smart cards. Journal of Cryptology, 4(3):161-174, 1991. <pdf-link>
• [X9-42] ANSI X9.42-2003 Agreement of Symmetric Keys Using Discrete Logarithm Cryptography, American National Standards Institute, November 2003.


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# Let f(x) = \ln(7x-2) and g(x)=e^{7x-2} a. Find domain of f(x) b. Find domain of g(x) c. Find...

## Question:

Let {eq}f(x) = \ln(7x-2) {/eq} and {eq}g(x)=e^{7x-2} {/eq}

a. Find domain of {eq}f(x) {/eq}

b. Find domain of {eq}g(x) {/eq}

c. Find {eq}f(1)+g(0) {/eq}

## Domain of Exponential and Logarithmic Functions

Since exponential and logarithmic functions are inverses of one another, their domains are related. Specifically, the domain of a standard exponential function is all real numbers, but the range is all positive numbers. A logarithmic function has these two reversed, so the domain is all positive numbers and the range is all real numbers.

a) We can only take a logarithm of a positive number. Thus, the domain of this function is whatever values of x allow the expression within the logarithm is positive. We can find this as follows.

{eq}7x-2 > 0\\ 7x > -2\\ x > -\frac{7}{2} {/eq}

Thus, the domain of this function is all real numbers larger than -3.5.

b) An exponential function doesn't have the same restrictions as a logarithm. We can take the exponential function of any real number, provided that the expression in the exponent is defined. As this is a linear expression, it is defined everywhere. Thus, the domain of this function is all real numbers.

c) Lastly, we've been asked to calculate the sum of two function values. We can find them by evaluating the individual functions at these values first.

{eq}f(1) = \ln (7(1) - 2) = \ln 5\\ g(0) = e^{7(0)-2} = e^{-2} {/eq}

We acn now add these two values together.

{eq}f(1) + g(0) = \ln 5 + e^{-2} \approx 1.7447732 {/eq}


text:
Geometry

# What is the volume of a truncated octagonal based pyramid?

###### 08/23/2015

The volume of a pyramidal frustum whose bases are n-sided regular polygons is

where a1 and a2 are the sides of the two bases.

Note: cot(x) = 1 / tan(x)

For example, I was not provided my a's to calculate the volume my pyramidal octagonal frustrum. I was provided only an overal width (let overal width = W) so I had to calculate a1 with this equation:

a1 = W * (√2 / (2 + √2))

Then with the given h I solved for the overal width for the top octagon using trigometry;

Used the same a1 equation but for a2;

Then used this equation to find my volume:

It works, I tried it and proved it with AutoCAD using the command MASSPROP


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TheInfoList

Fluid mechanics is the branch of physics concerned with the mechanics of fluids (liquids, gases, and plasmas) and the forces on them.[1]:3 It has applications in a wide range of disciplines, including mechanical, civil, chemical and biomedical engineering, geophysics, oceanography, meteorology, astrophysics, and biology.

It can be divided into fluid statics, the study of fluids at rest; and fluid dynamics, the study of the effect of forces on fluid motion.[1]:3 It is a branch of continuum mechanics, a subject which models matter without using the information that it is made out of atoms; that is, it models matter from a macroscopic viewpoint rather than from microscopic. Fluid mechanics, especially fluid dynamics, is an active field of research, typically mathematically complex. Many problems are partly or wholly unsolved and are best addressed by numerical methods, typically using computers. A modern discipline, called computational fluid dynamics (CFD), is devoted to this approach.[2] Particle image velocimetry, an experimental method for visualizing and analyzing fluid flow, also takes advantage of the highly visual nature of fluid flow.

• Aerodynamics
• Applied mechanics
• Bernoulli's principle
• Communicating vessels
• In some applications, another rough broad division among fluids is made: ideal and non-ideal fluids. An ideal fluid is non-viscous and offers no resistance whatsoever to a shearing force. An ideal fluid really does not exist, but in some calculations, the assumption is justifiable. One example of this is the flow far from solid surfaces. In many cases, the viscous effects are concentrated near the solid boundaries (such as in boundary layers) while in regions of the flow field far away from the boundaries the viscous effects can be neglected and the fluid there is treated as it were inviscid (ideal flow). When the viscosity is neglected, the term containing the viscous stress tensor ${\displaystyle \mathbf {\tau } }$ in the Navier–Stokes equation vanishes. The equation reduced in this form is called the Euler equation.


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# What is the enthalpy of formation of 1 mol of sulfuric acid?

In the contact process, the commercial process of producing $\ce{H2SO4}$, several reactions occur before the final product is produced:

\begin{align} \ce{S8(s) + 8O2(g) &-> 8SO2(g)} &\Delta H_\mathrm{f} &= \pu{-70.9 kcal mol^{-1}}\tag{1} \\ \ce{2SO2(g) + O2(g) &->[V2O5] 2SO3(g)} &\Delta H_\mathrm{f} &= \pu{-94.5 kcal mol^{-1}}\tag{2} \\ \ce{SO3(g) + H2O(g) &-> H2SO4(g)} &\Delta H_\mathrm{f} &= \pu{-193.9 kcal mol^{-1}}\tag{3} \end{align}

Given that $\Delta H_\mathrm{f}$ for $\ce{H2O(g)}$ is $\pu{-68.3 kcal mol^{-1}}$, what is the enthalpy of formation of $\pu{1 mol}$ of $\ce{H2SO4}$?

What I've tried to do:

1. Summing all the enthalpies, I arrived at to $\Delta H_\mathrm{f} = \pu{-427.6 kcal mol^{-1}}$.
2. Using the data from equation $(3)$, I conclude that the answer is $\Delta H_\mathrm{f} = \pu{-193.9 kcal}$.
• The second point in what you have tried that is not $d$H$_f but only dH – DSinghvi May 11 '15 at 15:51 • @DSinghvi actually the question stated like this (given that the standard formation enthalpy of SO2, SO3, H2O, and H2SO4 consecutively are -70,9 kkal/mol, -94,5 kkal/mol, -68,3 kkal/mol, and -193,9 kkal/mol) I moved the enthalpy to the reaction. – Prajogo Atmaja May 11 '15 at 15:55 • @DSinghvi By the way, from your response, that means I have to conclude every steps In order for me to have the ΔHf of H2SO4? – Prajogo Atmaja May 11 '15 at 16:03 • what do you understand by$/Delta H_f/$– DSinghvi May 12 '15 at 7:08 ## 1 Answer Since enthalpy is a state function, all that is needed is the$\Delta H_{\mathrm{f}}$of each reactants that form$\ce{H2SO4}\$. Multiply the equations through by the correct fractions (including the enthalpies) to result in the correct coefficients and enthalpy for the desired reaction:

$$\ce{SO3{(g)} +H2O{(g)} -> H2SO4{(g)}}~~~~~\Delta H_{\mathrm{f}}=-193.9~\mathrm{kcal~mol^{-1}}$$

$$\ce{H2{(g)} + \frac{1}{2}O2{(g)}->H2O{(g)}}~~~~~\Delta H_{\mathrm{f}}=-68.3~\mathrm{kcal~mol^{-1}}$$

$$\ce{SO2{(g)} + \frac{1}{2}O2{(g)} \xrightarrow{\ce{V2O5}} SO3{(g)}}~~~~~\Delta H_{\mathrm{f}}=-47.25~\mathrm{kcal~mol^{-1}}$$

$$\ce{\frac{1}{8}S8{(s)} +O2{(g)} -> SO2{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-8.8625~\mathrm{kcal~mol^{-1}}$$

Combining these equations yields:

$$\ce{\frac{1}{8}S8{(s)} +2O2{(g)} +H2{(g)} -> H2SO4{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-318~\mathrm{kcal~mol^{-1}}$$

(according to the equations and numerical values provided)


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Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279

Justify each step.$\frac{z+7}{3}=-11$ $z+7=-33$ $z=-40$

$\frac{z+7}{3}=-11$ Given

$z+7=-33$ Multiplication property of equality

$z=-40$ Subtraction property of equality

See the step by step solution

Step 1. Apply the multiplication property of equality.

If $a=b$, then $ca=cb$.

Step 2. Description of step.

Multiply each side of $\frac{z+7}{3}=-11$ by 3 and simplify.

$\begin{array}{c}3×\frac{z+7}{3}=3×\left(-11\right)\\ z+7=-33\end{array}$

Step 3. Apply the subtraction property of equality.

If $a=b$ and $c=d$, then $a-c=b-d$.

Step 4. Description of step.

Subtract 7 from each side of the equation obtained in step 2 to find the value of z.

$\begin{array}{c}z+7-7=-33-7\\ z=-40\end{array}$

Therefore, the value of z is $-40$.


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## Ch. 14: Cost of Capital

Cost of Capital

• The return earned on assets depend on the riskiness of those assets.
• The return to an investor is the same as the cost to the company.
• As a manager, cost of capital provides us with an indication of how the market views the riskiness of our assets.
• This estimate provides a natural starting point for calculating the required rate of return for capital budgeting projects.
• Why is cost of capital important?
• Allows us to implement good capital budgeting decisions.
• Informs us how to structure our capital structure.

Cost of Equity

• Cost of equity is the return that shareholders require for their investment in the firm.
• Shareholders bear two types of risk:
• Financial risk
• We have already discussed two methods for determining the cost of equity:
• Dividend Growth Model
• $R_E = \frac{D_1}{P_0} + g$
• Cost of equity is a function of dividend yield and capital gains yield
• How do we implement?
• Only applicable to companies currently paying dividends
• Requires a reasonably constant growth of dividends
• Sensitive to estimated growth rate (1 for 1)
• Avoids taking an explicit stand on risk
• CAPM
• $R_E = R_f + \beta_E (E(R_m)-R_f)$
• Cost of equity is a function of:
• Pure TVM
• The Reward for Bearing Systematic Risk (Market Risk Premium)
• The Amount of Systematic Risk
• How do we implement?
• Disadvantage: reliance on past returns to predict the future
• Example
• Our company has a beta of 1.5. The market risk premium is expected to be 9%, and the current risk-free rate is 6%. We have used analysts’ estimates to determine that the market believes our dividends will grow at 6% per year and our last dividend was $\$$2. Our stock is currently selling for \$$15.65. • What is our cost of equity using the SML? • What is our cost of equity using the DGM? Cost of Debt • Cost of debt is the return that bondholders require for their investment in the firm. • Required return is best estimated by computing the yield-to-maturity (YTM) on the existing long-term debt. • Note: We could also use the CAPM. • Example • Suppose we have a corporate bond issue currently outstanding that has 25 years left to maturity. The coupon rate is 9%, and coupons are paid semiannually. The bond is currently selling for$\$$908.72 per \$$1,000 bond.
• What is the cost of debt?

Cost of Preferred Stock

• Preferred stock is a claim on a fixed dividend paid forever.
• $R_{ps} = \frac{D}{P_0}$
• Example
• Your company has preferred stock that has an annual dividend of $\$$3.00. The current price is \$$25. • What is the cost of preferred stock? Weighted-average Cost of Capital (WACC) • Know we know the cost of capital for each source of funds raised by the firm. What is the riskiness of the firm as a whole? • We need only take the average cost of capital across the sources of funds weighted by their contribution to the assets of the firm. • Recall$V = D + E$. • Market Value of Debt:$D = $# of outstanding bonds$ \times $market price of one bond • Market Value of Equity:$E = $# of outstanding shares$ \times $market price of one share • Thus,$R_{WACC} = \frac{D}{D+E} R_D + \frac{E}{D+E} R_E$. • What about taxes? • With taxes, we get to deduct the interest expense of debt ($T_c R_D D$). • This tax shield of debt is like having the government pay for a portion of our interest expense ($T_c R_D$). • Example • Suppose a firm borrows$\$$1 million at 9 percent interest. The corporate tax rate is 34%. • What is the total interest bill for the firm? • What is the aftertax interest bill for the firm? • What is the aftertax interest rate on this loan (cost of debt)? • Thus with taxes, R_{WACC} = \frac{D}{D+E} (1-T_c) R_D + \frac{E}{D+E} R_E • Example • Equity Information: 50 million shares; \$$80 per share; $\beta$ = 1.15; Market risk premium = 9%; Risk-free rate = 5%
• Debt Information: $1 billion in outstanding debt (face value); Current quote = 110; Coupon rate = 9%; semiannual coupons; 15 years to maturity • Tax rate: 40 percent • What is the cost of debt? • What is the cost of equity? • What is the weighted-average cost of capital? Divisional and Project-level Costs of Capital • The firm-level WACC is only valid for projects of the same riskiness as the entire firm. • If we are looking at a project that does NOT have the same risk as the firm, then we need to determine the appropriate discount rate for that project. • If the firm’s WACC was used for every division, then the riskier divisions would get more investment capital and the less risky divisions would lose the opportunity to invest in positive NPV projects. • Two tools: • Pure-Play Approach • Find one or more companies that specialize in the product or service that we are considering. • Compute (or research) the beta for each company. • Take an average of the betas. • Use that beta along with the CAPM to find the appropriate return for a project of that identical risk. • Use this computed cost of capital for capital budgeting computations. • Subjective Approach • If the project has more risk than the firm, use a discount rate greater than the WACC. • If the project has less risk than the firm, use a discount rate less than the WACC. • You may still accept projects that you shouldn’t and reject projects you should accept, but your error rate should be lower than not considering differential risk at all. Flotation Costs • Flotation costs are the fees paid to issue stocks or bonds. • While the required return for a project depends on the risk, it should not depend upon how the money is raised. • However, the cost of issuing new securities should not just be ignored either. • Basic Approach: • Compute the weighted-average flotation cost. • Use the target weights because the firm will issue securities in these percentages over the long term. • Example • Your company is considering a project that will cost$\$$1 million. The project will generate after-tax cash flows of \$$250,000 per year for 7 years. The WACC is 15%, and the firm’s target D/E ratio is .6. The flotation cost for equity is 5%, and the flotation cost for debt is 3%.
• What is the NPV for the project before adjusting for flotation costs?
• What is the NPV for the project after adjusting for flotation costs?

Example

• A corporation has 10,000 bonds outstanding with a 6% annual coupon rate, 8 years to maturity, a $\$$1,000 face value, and a \$$1,100 market price. • The company’s 100,000 shares of preferred stock pay a$\$$3 annual dividend, and sell for \$$30 per share.
• The company’s 500,000 shares of common stock sell for $\$\$25 per share and have a beta of 1.5. The risk free rate is 4%, and the market return is 12%.
• Assuming a 40% tax rate, what is the company’s WACC?