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AlgebraQuestion and Answers: Page 324

Question Number 50363 Answers: 0 Comments: 0

let p ∈K_n [x] snd A and H two rlements of K[x] 1) prove that p(A(x)+H(x))=Σ_(k=0) ^n ((p^((k)) (A(x)))/(k!)).(H(x))^k 2)find the condition that p(A(x)+H(x))is divided by H(x)≠0 3) if p(x)≠c prove that p(p(x))−x is divided by p(x)−x.

$${let}\:{p}\:\in{K}_{{n}} \left[{x}\right]\:{snd}\:{A}\:{and}\:{H}\:{two}\:{rlements}\:{of}\:{K}\left[{x}\right] \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{p}\left({A}\left({x}\right)+{H}\left({x}\right)\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \frac{{p}^{\left({k}\right)} \left({A}\left({x}\right)\right)}{{k}!}.\left({H}\left({x}\right)\right)^{{k}} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{condition}\:{that}\:{p}\left({A}\left({x}\right)+{H}\left({x}\right)\right){is} \\ $$$${divided}\:{by}\:{H}\left({x}\right)\neq\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:{if}\:{p}\left({x}\right)\neq{c}\:{prove}\:\:{that}\:{p}\left({p}\left({x}\right)\right)−{x}\:{is}\:{divided} \\ $$$${by}\:{p}\left({x}\right)−{x}. \\ $$

Question Number 50362 Answers: 0 Comments: 0

calculate S_1 =Σ_(k=0) ^n C_n ^k S_2 =Σ_(k=0) ^([(n/2)]) C_n ^(2k) S_3 = Σ_(k=0) ^([(n/3)]) C_n ^(3k)

$${calculate}\:{S}_{\mathrm{1}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \\ $$$${S}_{\mathrm{2}} =\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{k}} \\ $$$${S}_{\mathrm{3}} =\:\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{3}}\right]} \:{C}_{{n}} ^{\mathrm{3}{k}} \\ $$

Question Number 50359 Answers: 0 Comments: 0

1) calculate Σ_(n=0) ^∞ (−1)^k C_n ^k (1/(k+1)) 2)calculate S_n (p)=Σ_(k=0) ^n (−1)^k (C_n ^k /(p+k+1)) p integr natural.

$$\left.\mathrm{1}\right)\:{calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{S}_{{n}} \left({p}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:\:\:\frac{{C}_{{n}} ^{{k}} }{{p}+{k}+\mathrm{1}} \\ $$$${p}\:{integr}\:{natural}. \\ $$

Question Number 50356 Answers: 0 Comments: 0

devompose inside C[x] and R[x] the polynom 1)x^4 +1 2)x^6 −1 3)x^8 +x^4 +1

$${devompose}\:{inside}\:{C}\left[{x}\right]\:{and}\:{R}\left[{x}\right]\:{the}\:{polynom} \\ $$$$\left.\mathrm{1}\right){x}^{\mathrm{4}} +\mathrm{1}\:\:\:\: \\ $$$$\left.\mathrm{2}\right){x}^{\mathrm{6}} −\mathrm{1} \\ $$$$\left.\mathrm{3}\right){x}^{\mathrm{8}} \:+{x}^{\mathrm{4}} \:+\mathrm{1} \\ $$

Question Number 50355 Answers: 0 Comments: 0

let (α_k ) (k∈[[0,n−1]] the n^(eme) roots of 1 calculate p(x,y)=(x+α_0 y)(x+α_1 y)....(x+α_(n−1) y)

$${let}\:\left(\alpha_{{k}} \right)\:\:\left({k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{the}\:{n}^{{eme}} \:{roots}\:{of}\:\mathrm{1}\right. \\ $$$${calculate}\:{p}\left({x},{y}\right)=\left({x}+\alpha_{\mathrm{0}} {y}\right)\left({x}+\alpha_{\mathrm{1}} {y}\right)....\left({x}+\alpha_{{n}−\mathrm{1}} {y}\right) \\ $$

Question Number 50354 Answers: 0 Comments: 0

((x)=a_k ) _(1≤k≤n) is a sequence of reals let p(x) =Π_(k=1) ^n (cos(a_k )+xsin(a_k )) if p(x)=(x^2 +1)q +r find q and r

$$\left(\left({x}\right)={a}_{{k}} \right)\:_{\mathrm{1}\leqslant{k}\leqslant{n}} \:{is}\:{a}\:{sequence}\:{of}\:{reals}\:{let} \\ $$$${p}\left({x}\right)\:=\prod_{{k}=\mathrm{1}} ^{{n}} \left({cos}\left({a}_{{k}} \right)+{xsin}\left({a}_{{k}} \right)\right) \\ $$$${if}\:{p}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right){q}\:+{r}\:\:\:{find}\:{q}\:{and}\:{r} \\ $$$$ \\ $$

Question Number 50353 Answers: 0 Comments: 0

let p(x)=x^(4n) −x^(3n) +x^(2n) −x^n +1 and q(x)=x^4 −x^3 +x^2 −x+1 determine the integr n to have q divide p.

$${let}\:{p}\left({x}\right)={x}^{\mathrm{4}{n}} −{x}^{\mathrm{3}{n}} +{x}^{\mathrm{2}{n}} −{x}^{{n}} +\mathrm{1}\:{and} \\ $$$${q}\left({x}\right)={x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\:\:{determine}\:{the}\:{integr}\:{n} \\ $$$${to}\:{have}\:{q}\:{divide}\:{p}. \\ $$

Question Number 50328 Answers: 2 Comments: 0

Question Number 50279 Answers: 2 Comments: 0

{ ((x+y=6)),((y+z=10)) :} (x,y,z>0)

$$\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{6}}\\{\mathrm{y}+\mathrm{z}=\mathrm{10}}\end{cases}\:\:\left(\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\right) \\ $$

Question Number 50278 Answers: 1 Comments: 6

(√(a+(√(a−x)))) + (√(a−(√(a+x)))) = 2x please i beg u guys please solve this question

$$\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}\:+\:\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}\:=\:\mathrm{2x} \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{beg}\:\mathrm{u}\:\mathrm{guys}\: \\ $$$$\mathrm{please}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{question} \\ $$

Question Number 50226 Answers: 0 Comments: 1

Three prizes are awarded each for getting more than 80%marks, 98% attendance and good behaviour in the college.In how many ways the prozes can be awarded if 15 student of the college are eligible for the three prizes?

$$\mathrm{Three}\:\mathrm{prizes}\:\mathrm{are}\:\mathrm{awarded}\:\mathrm{each}\:\mathrm{for} \\ $$$$\mathrm{getting}\:\mathrm{more}\:\mathrm{than}\:\mathrm{80\%marks}, \\ $$$$\mathrm{98\%}\:\mathrm{attendance}\:\mathrm{and}\:\mathrm{good} \\ $$$$\mathrm{behaviour}\:\mathrm{in}\:\mathrm{the}\:\mathrm{college}.\mathrm{In}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{ways}\:\mathrm{the}\:\mathrm{prozes}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{awarded}\:\mathrm{if}\:\mathrm{15}\:\mathrm{student}\:\mathrm{of}\:\mathrm{the}\:\mathrm{college} \\ $$$$\mathrm{are}\:\mathrm{eligible}\:\mathrm{for}\:\mathrm{the}\:\mathrm{three}\:\mathrm{prizes}? \\ $$

Question Number 50349 Answers: 0 Comments: 0

Question Number 50163 Answers: 2 Comments: 0

Question Number 50089 Answers: 1 Comments: 2

lim_(x→8) ((x^2 −6x−16)/(∣x−8∣))

$$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{16}}{\mid\mathrm{x}−\mathrm{8}\mid} \\ $$

Question Number 50016 Answers: 2 Comments: 0

Can you please help me how can i solve this equation x=0.055(1.44x^2 −6.336x+6.9696)

$${Can}\:{you}\:{please}\:{help}\:{me}\: \\ $$$${how}\:{can}\:{i}\:{solve}\:{this}\:{equation}\: \\ $$$${x}=\mathrm{0}.\mathrm{055}\left(\mathrm{1}.\mathrm{44}{x}^{\mathrm{2}} −\mathrm{6}.\mathrm{336}{x}+\mathrm{6}.\mathrm{9696}\right) \\ $$

Question Number 50012 Answers: 0 Comments: 0

help me sir plz

$$\mathrm{help}\:\mathrm{me}\:\mathrm{sir}\:\mathrm{plz} \\ $$$$ \\ $$$$ \\ $$

Question Number 49857 Answers: 1 Comments: 4

Please guide me Sir. I was trying to solve this eq for searching possible values of x. eq is : ∣x − 2∣ < 3∣x + 7∣ the range of x whom i got : −((23)/2) < x < −((19)/4) but the result do not satisfy the eq, instead i put x > −4 , they satisfy the eq. please help me out of this pickle. Not because i didn′t try, yet i always stuck in this type of function.

$$\mathrm{Please}\:\mathrm{guide}\:\mathrm{me}\:\mathrm{Sir}.\:\mathrm{I}\:\mathrm{was}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{solve}\: \\ $$$$\mathrm{this}\:\mathrm{eq}\:\mathrm{for}\:\mathrm{searching}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:{x}. \\ $$$$\mathrm{eq}\:\mathrm{is}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{x}\:−\:\mathrm{2}\mid\:<\:\mathrm{3}\mid{x}\:+\:\mathrm{7}\mid \\ $$$$\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{x}\:\mathrm{whom}\:\mathrm{i}\:\mathrm{got}\::\:\:\:−\frac{\mathrm{23}}{\mathrm{2}}\:<\:{x}\:<\:−\frac{\mathrm{19}}{\mathrm{4}} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{result}\:\mathrm{do}\:\mathrm{not}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{eq},\:\mathrm{instead} \\ $$$$\mathrm{i}\:\mathrm{put}\:{x}\:>\:−\mathrm{4}\:,\:\mathrm{they}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{eq}.\: \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{out}\:\mathrm{of}\:\mathrm{this}\:\mathrm{pickle}. \\ $$$$\mathrm{Not}\:\mathrm{because}\:\mathrm{i}\:\mathrm{didn}'\mathrm{t}\:\mathrm{try},\:\mathrm{yet}\:\mathrm{i}\:\mathrm{always} \\ $$$$\mathrm{stuck}\:\mathrm{in}\:\mathrm{this}\:\mathrm{type}\:\mathrm{of}\:\mathrm{function}. \\ $$

Question Number 49851 Answers: 0 Comments: 0

Question Number 49823 Answers: 2 Comments: 0

Complete the square in the expression y^2 +8y+9k and hence find the value of k that makes it a perfect square.

$$\mathrm{Complete}\:\mathrm{the}\:\mathrm{square}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mathrm{y}^{\mathrm{2}} +\mathrm{8y}+\mathrm{9k}\:\mathrm{and}\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{k}\:\mathrm{that}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$

Question Number 49755 Answers: 1 Comments: 0

Solve simultaneously for s in terms of a and b. h^2 +(b−k)^2 = s^2 .....(i) (h^2 /a^2 )+(k^2 /b^2 ) = 1 .....(ii) (h−(s/2))^2 +(k+b(√(1−(s^2 /(4a^2 )))) )= s^2 ..(iii).

$${Solve}\:{simultaneously}\:{for}\:\boldsymbol{{s}}\:{in}\:{terms} \\ $$$${of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}. \\ $$$${h}^{\mathrm{2}} +\left({b}−{k}\right)^{\mathrm{2}} =\:{s}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:.....\left({i}\right) \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....\left({ii}\right) \\ $$$$\left({h}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({k}+{b}\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}\:\right)=\:{s}^{\mathrm{2}} \:\:\:..\left({iii}\right). \\ $$

Question Number 50924 Answers: 1 Comments: 0

factor the expression: E=x^5 +x^4 +1

$$\mathrm{factor}\:\mathrm{the}\:\mathrm{expression}: \\ $$$$\mathrm{E}={x}^{\mathrm{5}} +{x}^{\mathrm{4}} +\mathrm{1} \\ $$

Question Number 49647 Answers: 1 Comments: 3

let p(x) =x^(2n) −x^n +1 1) determine the roots of p(x) 2) factorize inside C[x] the polynom p(x) . 3)solve p(x)=0 and p(x) =2

$${let}\:{p}\left({x}\right)\:={x}^{\mathrm{2}{n}} \:−{x}^{{n}} \:+\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{inside}\:{C}\left[{x}\right]\:{the}\:{polynom}\:{p}\left({x}\right)\:. \\ $$$$\left.\mathrm{3}\right){solve}\:{p}\left({x}\right)=\mathrm{0}\:\:{and}\:{p}\left({x}\right)\:=\mathrm{2} \\ $$

Question Number 49642 Answers: 1 Comments: 0

if a+b =s and a^3 +b^3 =t find a^2 +b^2 and a^4 +b^4 interms of s and t .

$${if}\:\:{a}+{b}\:={s}\:{and}\:{a}^{\mathrm{3}} \:+{b}^{\mathrm{3}} \:={t}\:\:{find}\:{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:\:{and}\:{a}^{\mathrm{4}} \:+{b}^{\mathrm{4}} \:{interms}\:{of}\:{s}\:{and}\:{t}\:. \\ $$

Question Number 49604 Answers: 2 Comments: 0

Show that: (((a + b)^2 )/2) ≤ a^2 + b^2