Crossword Solver > found answer > PLUGGING

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CROSSWORD
ANSWER

plugging

Searching in Crosswords ...

The answer PLUGGING has 2 possible clue(s) in existing crosswords.

Searching in Word Games ...

The word PLUGGING is VALID in some board games. Check PLUGGING in word games in Scrabble, Words With Friends, see scores, anagrams etc.

Searching in Dictionaries ...

Definitions of plugging in various dictionaries:

verb - fill or close tightly with or as if with a plug

verb - persist in working hard

verb - deliver a quick blow to

more

Word Research / Anagrams and more ...

Keep reading for additional results and analysis below.

Possible Crossword Clues
Blocking promotion
Stopping for advertising

Last Seen in these Crosswords & Puzzles
Jan 1 2004 The Guardian - Cryptic crossword
Jul 7 2002 The Telegraph - Cryptic

Possible Dictionary Clues
A tired or old horse.
Block or fill in (a hole or cavity)
Mention (a product, event, or establishment) publicly in order to promote it.
Shoot or hit (someone or something)
Proceed steadily and laboriously with a journey or task.

Plugging might refer to
In mathematics, a Change of variables is a basic technique used to simplify problems in which the original variables are replaced with functions of other variables. The intent is that when expressed in new variables, the problem may become simpler, or equivalent to a better understood problem. * Change of variables is an operation that is related to substitution. However these are different operations, as can be seen when considering differentiation (chain rule) or integration (integration by substitution). * A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth degree polynomial:* * * * * x * * 6 * * * − * 9 * * x * * 3 * * * + * 8 * = * 0. * * * * {\displaystyle x^{6}-9x^{3}+8=0.\,} * Sixth degree polynomial equations are generally impossible to solve in terms of radicals (see Abel–Ruffini theorem). This particular equation, however, may be written * * * * * ( * * x * * 3 * * * * ) * * 2 * * * − * 9 * ( * * x * * 3 * * * ) * + * 8 * = * 0 * * * {\displaystyle (x^{3})^{2}-9(x^{3})+8=0} * (this is a simple case of a polynomial decomposition). Thus the equation may be * simplified by defining a new variable u =x3. Substituting x by * * * * * * u * * 3 * * * * * * {\displaystyle {\sqrt[{3}]{u}}} * into the polynomial gives * * * * * * u * * 2 * * * − * 9 * u * + * 8 * = * 0 * , * * * {\displaystyle u^{2}-9u+8=0,} * which is just a quadratic equation with the two solutions: * * * * * u * = * 1 * * * and * * * u * = * 8. * * * {\displaystyle u=1\quad {\text{and}}\quad u=8.} * The solutions in terms of the original variable are obtained by substituting x3 back in for u, which gives * * * * * * x * * 3 * * * = * 1 * * * and * * * * x * * 3 * * * = * 8. * * * {\displaystyle x^{3}=1\quad {\text{and}}\quad x^{3}=8.} * Then, assuming that one is interested only in real solutions, the solutions of the original equation are * * * * * x * = * ( * 1 * * ) * * 1 * * / * * 3 * * * = * 1 * * * ...

Plugging might refer to

In mathematics, a Change of variables is a basic technique used to simplify problems in which the original variables are replaced with functions of other variables. The intent is that when expressed in new variables, the problem may become simpler, or equivalent to a better understood problem.
* Change of variables is an operation that is related to substitution. However these are different operations, as can be seen when considering differentiation (chain rule) or integration (integration by substitution).
* A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth degree polynomial:*
*
*
*
* x
*
* 6
*
*
* −
* 9
*
* x
*
* 3
*
*
* +
* 8
* =
* 0.
*
*
*
* {\displaystyle x^{6}-9x^{3}+8=0.\,}
* Sixth degree polynomial equations are generally impossible to solve in terms of radicals (see Abel–Ruffini theorem). This particular equation, however, may be written
*
*
*
*
* (
*
* x
*
* 3
*
*
*
* )
*
* 2
*
*
* −
* 9
* (
*
* x
*
* 3
*
*
* )
* +
* 8
* =
* 0
*
*
* {\displaystyle (x^{3})^{2}-9(x^{3})+8=0}
* (this is a simple case of a polynomial decomposition). Thus the equation may be
* simplified by defining a new variable u =x3. Substituting x by
*
*
*
*
*
* u
*
* 3
*
*
*
*
*
* {\displaystyle {\sqrt[{3}]{u}}}
* into the polynomial gives
*
*
*
*
*
* u
*
* 2
*
*
* −
* 9
* u
* +
* 8
* =
* 0
* ,
*
*
* {\displaystyle u^{2}-9u+8=0,}
* which is just a quadratic equation with the two solutions:
*
*
*
*
* u
* =
* 1
*
*
* and
*
*
* u
* =
* 8.
*
*
* {\displaystyle u=1\quad {\text{and}}\quad u=8.}
* The solutions in terms of the original variable are obtained by substituting x3 back in for u, which gives
*
*
*
*
*
* x
*
* 3
*
*
* =
* 1
*
*
* and
*
*
*
* x
*
* 3
*
*
* =
* 8.
*
*
* {\displaystyle x^{3}=1\quad {\text{and}}\quad x^{3}=8.}
* Then, assuming that one is interested only in real solutions, the solutions of the original equation are
*
*
*
*
* x
* =
* (
* 1
*
* )
*
* 1
*
* /
*
* 3
*
*
* =
* 1
*
*
* ...

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