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Like terms

Combining these expressions with equation 117 and collecting like terms give equation 120 ... [Pg.491]

Addition and Subtraction Only like terms can be added or sub-trac ted in two algebraic expressions. [Pg.430]

Like terms in powers of a are collected to form the individual problems. [Pg.455]

Upon expansion of the differentials and collec tion of like terms, this becomes... [Pg.515]

Combining these expressions with Eq. (4-48) and collecting like terms gives... [Pg.517]

The differences between like terms represent residuals between derived and experimental values. Defining these residuals as... [Pg.536]

Upon cancellation of like terms and division by dxdy. [Pg.500]

By collecting like terms this can also be writtel ... [Pg.97]

Correct (like terms) Incorrect (unlike terms)... [Pg.29]

A fraction must be stated in like terms therefore, proportion may be used to convert grams to milligrams. [Pg.40]

As with all fractions, the numerator and die denominator must be of like terms, for example, milligrams over milligrams or grams over grams. Errors in using diis and odier drug formulas, as well as proportions, will be reduced if die entire dose is written ratiier than just the numbers. [Pg.40]

Axial Dispersion Effects In adsorption bed calculations, axial dispersion effects are typically accounted for by the axial diffusion-like term in the bed conservation equations [Eqs. (16-51) and (16-52)]. For nearly linear isotherms (0.5 < B < 1.5), the combined effects of axial dispersion and mass-transfer resistances on the adsorption behavior of packed beds can be expressed approximately in terms of an apparent rate coefficient k0 for use with a fluid-phase driving force (column 1, Table 16-12) ... [Pg.25]

It is important to notice that the Coulomb-like term (the third term in the equation) is written in terms of the four-current, and the exchange and correlation energy is given by... [Pg.139]

The second term drops out because of the initial condition. Grouping like terms in A(.) gives... [Pg.317]

The damping factors take into account 1) the mean free path k(k) of the photoelectron the exponential factor selects the contributions due to those photoelectron waves which make the round trip from the central atom to the scatterer and back without energy losses 2) the mean square value of the relative displacements of the central atom and of the scatterer. This is called Debye-Waller like term since it is not referred to the laboratory frame, but it is a relative value, and it is temperature dependent, of course It is important to remember the peculiar way of probing the matter that EXAFS does the source of the probe is the excited atom which sends off a photoelectron spherical wave, the detector of the distribution of the scattering centres in the environment is again the same central atom that receives the back-diffused photoelectron amplitude. This is a unique feature since all other crystallographic probes are totally (source and detector) or partially (source or detector) external probes , i.e. the measured quantities are referred to the laboratory reference system. [Pg.105]

You first write the two numbers in terms of the same unknown or variable. If the first number is x, then the other number is 7 - x. How did I pull the 7 - x out of my hat Think about two numbers having a sum of 7. If one of them is 5, then the other is 7 - 5, or 2. If one of them is 3, then the other number is 7 - 3, or 4. Sometimes, when you do easy problems in your head, it s hard to figure out how to write what you re doing in math speak. So, if the two numbers are x and 7 - x, then you have to square each of them, add them together, and set the sum equal to 29. The equation to use is x2 + (7 - x)2 = 29. To solve this equation, you square the binomial, combine like terms, subtract 29 from each side, factor the quadratic equation, and then set each of the factors equal to 0. [Pg.155]

First write the two consecutive integers as n and n + 1. The squares of those two numbers are n2 and (n + l)2. To write the equation, add the two squares together and set them equal to 145. Then do the squaring on the left, combine like terms, subtract 145 from each side to set the equation equal to 0, and solve the quadratic equation. [Pg.165]

A polynomial is the sum or difference of two or more unlike terms. Like terms have exactly the same variable(s). [Pg.167]


See other pages where Like terms is mentioned: [Pg.455]    [Pg.63]    [Pg.151]    [Pg.50]    [Pg.558]    [Pg.345]    [Pg.8]    [Pg.31]    [Pg.205]    [Pg.206]    [Pg.208]    [Pg.210]    [Pg.210]    [Pg.216]    [Pg.216]    [Pg.5]    [Pg.117]    [Pg.390]    [Pg.591]    [Pg.602]    [Pg.555]    [Pg.81]    [Pg.52]    [Pg.52]    [Pg.56]    [Pg.249]    [Pg.101]    [Pg.145]    [Pg.231]    [Pg.85]   
See also in sourсe #XX -- [ Pg.14 ]

See also in sourсe #XX -- [ Pg.11 ]




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