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AlgebraQuestion and Answers: Page 226

Question Number 131343 Answers: 1 Comments: 0

A mapping is defined as G→S where (G,×) and (S,+), show that the mapping f(x) = ln x is an isomophism.

$$\mathrm{A}\:\mathrm{mapping}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:{G}\rightarrow{S}\:\mathrm{where}\:\left({G},×\right)\:\mathrm{and}\:\left({S},+\right), \\ $$$$\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{mapping}\:{f}\left({x}\right)\:=\:\mathrm{ln}\:{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{isomophism}. \\ $$

Question Number 131309 Answers: 2 Comments: 0

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Question Number 131214 Answers: 1 Comments: 0

Question Number 131201 Answers: 2 Comments: 0

Question Number 131158 Answers: 1 Comments: 0

If 4a_n +2a_(−n) =3n^2 +2n−3 find a_n =?

$${If}\:\mathrm{4}{a}_{{n}} +\mathrm{2}{a}_{−{n}} =\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{3} \\ $$$${find}\:{a}_{{n}} \:=? \\ $$

Question Number 131155 Answers: 2 Comments: 0

Given { ((a_(n+2) =a_(n+1) +(1/2)a_n )),((a_1 =3 ; a_2 =2)) :} find a_n .

$${Given}\:\begin{cases}{{a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}} }\\{{a}_{\mathrm{1}} =\mathrm{3}\:;\:{a}_{\mathrm{2}} =\mathrm{2}}\end{cases} \\ $$$$\:{find}\:{a}_{{n}} . \\ $$

Question Number 206939 Answers: 1 Comments: 7

In how many ways can the word KINECTIC be arranged so that no vowels can be together?

Question Number 131084 Answers: 2 Comments: 0

let solve this in R : x^4 −x^3 −4x^2 +x+1=0.

$$\mathrm{let}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{in}\:\mathbb{R}\::\: \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}. \\ $$

Question Number 131075 Answers: 1 Comments: 0

If a−3=−b−4=−c−5=d+6=e+7= a−b−c+d+e+8 then a−b−c+d+e =?

$$\mathrm{If}\:\mathrm{a}−\mathrm{3}=−\mathrm{b}−\mathrm{4}=−\mathrm{c}−\mathrm{5}=\mathrm{d}+\mathrm{6}=\mathrm{e}+\mathrm{7}= \\ $$$$\mathrm{a}−\mathrm{b}−\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{8}\:\mathrm{then}\:\mathrm{a}−\mathrm{b}−\mathrm{c}+\mathrm{d}+\mathrm{e}\:=? \\ $$

Question Number 131064 Answers: 3 Comments: 0

Question Number 131016 Answers: 0 Comments: 0

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Question Number 130846 Answers: 1 Comments: 1

prove that 3sec^(−1) ((√2))−4csc^(−1) ((√2))+5cot^(−1) (2)=1.533

$${prove}\:{that} \\ $$$$\mathrm{3}{sec}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)−\mathrm{4}{csc}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)+\mathrm{5}{cot}^{−\mathrm{1}} \left(\mathrm{2}\right)=\mathrm{1}.\mathrm{533} \\ $$

Question Number 130806 Answers: 1 Comments: 0

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Question Number 130733 Answers: 1 Comments: 1

Question Number 130719 Answers: 0 Comments: 2

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Question Number 130605 Answers: 2 Comments: 0

0^0 =??

$$\mathrm{0}^{\mathrm{0}} =?? \\ $$

Question Number 130593 Answers: 1 Comments: 3

if a_0 =1, a_1 =2 and a_(n+1) =(√(a_n a_(n−1) )) find a_n in terms of n.

$${if}\:{a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{2}\:{and}\:{a}_{{n}+\mathrm{1}} =\sqrt{{a}_{{n}} {a}_{{n}−\mathrm{1}} } \\ $$$${find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$

Question Number 130496 Answers: 3 Comments: 0

(d/dx)(x!)=?

$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}!\right)=? \\ $$

Question Number 130476 Answers: 1 Comments: 0

Question Number 130472 Answers: 2 Comments: 0

Question Number 130455 Answers: 0 Comments: 0

f(x)=x+ln∣e^x −2∣ { ((f(x)=x+ln(e^x −2) x∈]ln2;+∞[)),((f(x)=x+ln(−e^x +2) x∈]−∞;ln2[)) :} lim_(x→+∞) f(x)=lim_(x→+∞) (x+ln(e^x −2))=+∞ lim_(x→−∞) f(x)=lim_(x→−∞) (x+ln(−e^x +2))=−∞ lim_(x→−∞) e^x =0 lim_(x→−∞) ln(−e^x +2)=ln2 lim_(x→−∞) (x+ln(−e^x +2)=−∞ lim_(x→^> ln2) f(x)=lim_(x→^> ln2) (x+ln(e^x −2))=−∞ lim_(x→^< ln2) f(x)=lim_(x→^< ln2) (x+ln(−e^x +2))=−∞

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\mid\mathrm{e}^{\mathrm{x}} −\mathrm{2}\mid \\ $$$$\begin{cases}{\left.\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{2}\right)\:\:\:\:\:\:\mathrm{x}\in\right]\mathrm{ln2};+\infty\left[\right.}\\{\left.\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)\:\:\:\mathrm{x}\in\right]−\infty;\mathrm{ln2}\left[\right.}\end{cases} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{2}\right)\right)=+\infty \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)\right)=−\infty \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}e}^{\mathrm{x}} =\mathrm{0} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)=\mathrm{ln2} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)=−\infty\right. \\ $$$$\underset{{x}\overset{>} {\rightarrow}\mathrm{ln2}} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\overset{>} {\rightarrow}\mathrm{ln2}} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{2}\right)\right)=−\infty \\ $$$$\underset{{x}\overset{<} {\rightarrow}\mathrm{ln2}} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\overset{<} {\rightarrow}\mathrm{ln2}} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)\right)=−\infty \\ $$

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