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What is the negative form -"Everybody will admit that i am right"

What is the negative form -"Everybody will admit that i am right" Correct Answer Nobody will not deny i am right

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If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?
The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with
$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$     $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$     $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$     is simplified to :
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaaigdacaaIXaGaaiilaiaabckacaqG0bGaaeiAaiaa % bwgacaqGUbGaaeiOaiaabAgacaqGPbGaaeOBaiaabsgacaqGGcGaae % iDaiaabIgacaqGLbGaaeiOaiaabAhacaqGHbGaaeiBaiaabwhacaqG % LbGaaeiOaiaab+gacaqGMbGaaeiOamaabmaapaqaa8qacaqG4bGaey % OeI0IaaGymaiaaiodaaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGa % aGynaaaakiaabckacqGHRaWkcaqGGcWaaSaaa8aabaWdbiaaigdaa8 % aabaWdbmaabmaapaqaa8qacaWG4bGaeyOeI0IaaGymaiaaigdaaiaa % wIcacaGLPaaapaWaaWbaaSqabeaapeGaaGynaaaaaaGccaGGGcGaai % Olaaaa!6E72! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = 11,{\rm{\;then\;find\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)If x + 1/(x - 13) = 11, then what will be the value of (x – 13)5 + 1/(x – 11)5?
The quark content of $$\sum {^ + } ,\,{K^ - },\,{\pi ^ - }$$   and p is indicated: $$\left| {\sum {^ + } } \right\rangle = \left| {uus} \right\rangle ;\,\left| {{K^ + }} \right\rangle = \left| {s\overline u } \right\rangle ;\,\left| \pi \right\rangle = \left| d \right\rangle ;\,\left| p \right\rangle = \left| {uud} \right\rangle $$
In the process, $${\pi ^ - } + p \to {K^ - } + \sum {^ + } ,$$     considering strong interactions only, which of the following statements is true?
Consider the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}\left( {\text{t}} \right)}}{{{\text{dt}}}} + {\text{y}}\left( {\text{t}} \right) = \delta \left( {\text{t}} \right)$$      with $${\left. {{\text{y}}\left( {\text{t}} \right)} \right|_{{\text{t}} = 0}} = - 2$$   and $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 0.$$
The numerical value of $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}}$$   is
Let the function
\[{\text{f}}\left( \theta \right) = \left| {\begin{array}{*{20}{c}} {\sin \theta }&{\cos \theta }&{\tan \theta } \\ {\sin \left( {\frac{\pi }{6}} \right)}&{\cos \left( {\frac{\pi }{6}} \right)}&{\tan \left( {\frac{\pi }{6}} \right)} \\ {\sin \left( {\frac{\pi }{3}} \right)}&{\cos \left( {\frac{\pi }{3}} \right)}&{\tan \left( {\frac{\pi }{3}} \right)} \end{array}} \right|\
In the following passage, some of the words have been left out. Read the passage carefully and select the correct answer for the given blank out of the four alternatives. This is the arresting and dominant fact about modern social discussion; that the quarrel is __________ merely about the difficulties, but about the aim. We agree about the evil; it is about the good that we should tear each ____________ eyes out. We all admit that a lazy aristocracy is a bad thing. We should not by ________ means all admit that an active aristocracy would be a good thing. We all feel angry with an irreligious priesthood; but some of us would _______ mad with disgust at a really religious one. Everyone is indignant if our army is weak, _____________ the people who would be even more indignant if it were strong. that the quarrel is __________ merely about the difficulties,