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AlgebraQuestion and Answers: Page 341

Question Number 31490 Answers: 0 Comments: 0

simplify p(x)= (1+x^2 )(1+x^4 )....(1+x^(2n) ) with n fromN then find the roots of p(x).

$${simplify}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)....\left(\mathrm{1}+{x}^{\mathrm{2}{n}} \right)\:{with}\:{n}\:{fromN} \\ $$$${then}\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right). \\ $$

Question Number 31485 Answers: 2 Comments: 0

Question Number 31456 Answers: 0 Comments: 2

Find sum of S= (2/3) + (4/3^2 ) + (6/3^3 ) + (8/3^4 ) +......+∞ ?

$$\mathbb{F}{ind}\:{sum}\:{of} \\ $$$${S}=\:\frac{\mathrm{2}}{\mathrm{3}}\:+\:\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{6}}{\mathrm{3}^{\mathrm{3}} }\:+\:\frac{\mathrm{8}}{\mathrm{3}^{\mathrm{4}} }\:+......+\infty\:? \\ $$

Question Number 31409 Answers: 1 Comments: 0

The maximum area of the triangle whose sides a,b and c satisfy 0≤a≤1 , 1≤b≤2 , 2≤c≤3 is : A) 1 B) 2 C) 1.5 D) 0.5 ?

$${The}\:{maximum}\:{area}\:{of}\:{the}\:{triangle} \\ $$$${whose}\:{sides}\:{a},{b}\:{and}\:{c}\:{satisfy}\: \\ $$$$\mathrm{0}\leqslant{a}\leqslant\mathrm{1}\:,\:\mathrm{1}\leqslant{b}\leqslant\mathrm{2}\:,\:\mathrm{2}\leqslant{c}\leqslant\mathrm{3}\:{is}\:: \\ $$$$\left.{A}\right)\:\mathrm{1} \\ $$$$\left.{B}\right)\:\mathrm{2} \\ $$$$\left.{C}\right)\:\mathrm{1}.\mathrm{5} \\ $$$$\left.{D}\right)\:\mathrm{0}.\mathrm{5}\:\:\:\:\:\:\:? \\ $$

Question Number 31336 Answers: 0 Comments: 1

Find the principal value of z=(1−i)^(1+i) .Hence find the modulus of the result.

$${Find}\:{the}\:{principal}\:{value}\:{of} \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} .{Hence}\:{find}\:{the} \\ $$$${modulus}\:{of}\:{the}\:{result}. \\ $$

Question Number 31335 Answers: 0 Comments: 3

Find the pricipal value of z=(1−i)^i

$${Find}\:{the}\:{pricipal}\:{value}\:{of}\: \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{{i}} \\ $$

Question Number 31324 Answers: 1 Comments: 0

Question Number 31323 Answers: 1 Comments: 0

Question Number 31320 Answers: 1 Comments: 0

Let p and q are the roots of x^2 − 2mx − 5n = 0 and m and n are the roots of x^2 − 2px − 5q = 0 If p ≠ q ≠ m ≠ n, then the value of p + q + m + n is ...

$$\mathrm{Let}\:{p}\:\mathrm{and}\:{q}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{2}{mx}\:−\:\mathrm{5}{n}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{m}\:\mathrm{and}\:{n}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{2}{px}\:−\:\mathrm{5}{q}\:=\:\mathrm{0} \\ $$$$\mathrm{If}\:{p}\:\neq\:{q}\:\neq\:{m}\:\neq\:{n},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${p}\:+\:{q}\:+\:{m}\:+\:{n}\:\mathrm{is}\:... \\ $$

Question Number 31317 Answers: 0 Comments: 1

Question Number 31286 Answers: 1 Comments: 1

Find all set of ordered triple/s (x,y,z), x,y,z∈ℜ, such that x−y=1−z 3(x^2 −y^2 )=5(1−z^2 ) 7(x^3 −y^3 )=19(1−z^3 ). Please show your solution.

$${Find}\:{all}\:{set}\:{of}\:{ordered}\:{triple}/{s}\:\left({x},{y},{z}\right),\:\:{x},{y},{z}\in\Re,\:{such}\:{that} \\ $$$${x}−{y}=\mathrm{1}−{z} \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{5}\left(\mathrm{1}−{z}^{\mathrm{2}} \right) \\ $$$$\mathrm{7}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right)=\mathrm{19}\left(\mathrm{1}−{z}^{\mathrm{3}} \right). \\ $$$${Please}\:{show}\:{your}\:{solution}. \\ $$

Question Number 31290 Answers: 1 Comments: 1

Question Number 31259 Answers: 0 Comments: 4

Question Number 31214 Answers: 2 Comments: 2

Question Number 31194 Answers: 2 Comments: 1

Find the remainder when x^(203) −1 is divided by x^4 −1.

$${Find}\:{the}\:{remainder}\:{when}\:{x}^{\mathrm{203}} −\mathrm{1} \\ $$$${is}\:{divided}\:{by}\:{x}^{\mathrm{4}} −\mathrm{1}. \\ $$

Question Number 31047 Answers: 0 Comments: 0

let Δ={(x,y)∈N^2 /x+y=n , n∈N} find cardΔ 2) let A= {(x,y)∈N^2 / x+2y=n} find card A.

$${let}\:\Delta=\left\{\left({x},{y}\right)\in{N}^{\mathrm{2}} \:/{x}+{y}={n}\:,\:{n}\in{N}\right\}\:{find}\:{card}\Delta \\ $$$$\left.\mathrm{2}\right)\:{let}\:{A}=\:\left\{\left({x},{y}\right)\in{N}^{\mathrm{2}} /\:{x}+\mathrm{2}{y}={n}\right\}\:{find}\:{card}\:{A}. \\ $$

Question Number 31046 Answers: 0 Comments: 0

prove that C_n ^o C_n ^p +C_n ^1 C_(n−1) ^(p−1) +...C_n ^p C_(n−p) ^0 =2^p C_n ^p .

$${prove}\:{that}\:{C}_{{n}} ^{{o}} \:{C}_{{n}} ^{{p}} \:+{C}_{{n}} ^{\mathrm{1}} \:{C}_{{n}−\mathrm{1}} ^{{p}−\mathrm{1}} \:+...{C}_{{n}} ^{{p}} \:{C}_{{n}−{p}} ^{\mathrm{0}} =\mathrm{2}^{{p}} \:{C}_{{n}} ^{{p}} \:\:\:. \\ $$

Question Number 31042 Answers: 0 Comments: 0

solve in N^2 9y^2 −(x+1)^2 =32 .

$${solve}\:{in}\:{N}^{\mathrm{2}} \:\:\mathrm{9}{y}^{\mathrm{2}} \:−\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{32}\:. \\ $$

Question Number 31041 Answers: 0 Comments: 0

solve in Z^2 11x −5y =14

$${solve}\:{in}\:{Z}^{\mathrm{2}} \:\mathrm{11}{x}\:−\mathrm{5}{y}\:=\mathrm{14} \\ $$

Question Number 31040 Answers: 0 Comments: 0

solve in Z^2 x^2 −y^2 =1969

$${solve}\:{in}\:{Z}^{\mathrm{2}} \:{x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} =\mathrm{1969} \\ $$

Question Number 31039 Answers: 0 Comments: 0

find the sum s_0 = C_n ^o +C_n ^4 +C_n ^8 +... s_1 = C_n ^1 +C_n ^5 +C_n ^9 +.... s_3 = C_n ^2 +C_n ^6 + C_n ^(10) +....

$${find}\:{the}\:{sum} \\ $$$${s}_{\mathrm{0}} =\:{C}_{{n}} ^{{o}} \:+{C}_{{n}} ^{\mathrm{4}} \:+{C}_{{n}} ^{\mathrm{8}} \:+... \\ $$$${s}_{\mathrm{1}} =\:{C}_{{n}} ^{\mathrm{1}} \:+{C}_{{n}} ^{\mathrm{5}} \:+{C}_{{n}} ^{\mathrm{9}} \:+.... \\ $$$${s}_{\mathrm{3}} =\:{C}_{{n}} ^{\mathrm{2}} \:\:+{C}_{{n}} ^{\mathrm{6}} \:+\:{C}_{{n}} ^{\mathrm{10}} \:+.... \\ $$

Question Number 31038 Answers: 0 Comments: 0

let give α ∈]−π ,π[ 1)prove that sin^2 α −2(1+cosα) =−4cos^4 ((α/2)) 2)solve inside C z^2 −2z sinα +2(1+cosα)=0 find the module and arg of the roots.

$$\left.{let}\:{give}\:\alpha\:\in\right]−\pi\:,\pi\left[\right. \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:\:{sin}^{\mathrm{2}} \alpha\:−\mathrm{2}\left(\mathrm{1}+{cos}\alpha\right)\:=−\mathrm{4}{cos}^{\mathrm{4}} \left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right){solve}\:{inside}\:{C}\:\:{z}^{\mathrm{2}} \:−\mathrm{2}{z}\:{sin}\alpha\:+\mathrm{2}\left(\mathrm{1}+{cos}\alpha\right)=\mathrm{0}\:{find} \\ $$$${the}\:{module}\:{and}\:{arg}\:{of}\:{the}\:{roots}. \\ $$

Question Number 31037 Answers: 0 Comments: 0

let θ∈[0,π] and Z=1+cosθ +isinθ Z^′ =cosθ +(1+sinθ)i find ∣ZZ^′ ∣ arg(ZZ^′ ) ∣(Z/Z^′ )∣ , arg((Z/Z^, )) , ∣Z−Z^, ∣ and arg(Z−Z^, ).

$${let}\:\:\theta\in\left[\mathrm{0},\pi\right]\:\:{and}\:{Z}=\mathrm{1}+{cos}\theta\:+{isin}\theta \\ $$$${Z}^{'} ={cos}\theta\:+\left(\mathrm{1}+{sin}\theta\right){i}\:\:{find}\:\mid{ZZ}^{'} \mid\:\:{arg}\left({ZZ}^{'} \right) \\ $$$$\mid\frac{{Z}}{{Z}^{'} }\mid\:,\:{arg}\left(\frac{{Z}}{{Z}^{,} }\right)\:\:,\:\:\mid{Z}−{Z}^{,} \mid\:{and}\:{arg}\left({Z}−{Z}^{,} \right). \\ $$

Question Number 31036 Answers: 0 Comments: 0

solve inside C z^6 = (z^− )^2 .

$${solve}\:{inside}\:{C}\:\:{z}^{\mathrm{6}} =\:\left({z}^{−} \right)^{\mathrm{2}} \:\:. \\ $$

Question Number 31035 Answers: 0 Comments: 0

1) solve x^5 =1 2) if x_i are roots of this equation prove that Σ_i x_i =0 3)prove that cos(((2π)/5)) +cos(((4π)/5)) =−(1/2) 4)calculate cos(((4π)/5)) interms cos(((2π)/5)) then find its values.

$$\left.\mathrm{1}\right)\:{solve}\:{x}^{\mathrm{5}} =\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{if}\:{x}_{{i}} \:{are}\:{roots}\:{of}\:{this}\:{equation}\:{prove}\:{that}\:\sum_{{i}} \:{x}_{{i}} =\mathrm{0} \\ $$$$\left.\mathrm{3}\right){prove}\:{that}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\mathrm{4}\right){calculate}\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:{interms}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:{then}\:{find}\:{its} \\ $$$${values}. \\ $$

Question Number 31034 Answers: 0 Comments: 0

1) solve inside C z^4 =6 2)sove inside C (((z+i)/(z−i)))^3 +(((z+i)/(z−i)))^2 +(((z+i)/(z−i))) +1=0

$$\left.\mathrm{1}\right)\:{solve}\:{inside}\:{C}\:\:{z}^{\mathrm{4}} =\mathrm{6} \\ $$$$\left.\mathrm{2}\right){sove}\:{inside}\:{C}\:\left(\frac{{z}+{i}}{{z}−{i}}\right)^{\mathrm{3}} \:+\left(\frac{{z}+{i}}{{z}−{i}}\right)^{\mathrm{2}} \:+\left(\frac{{z}+{i}}{{z}−{i}}\right)\:+\mathrm{1}=\mathrm{0} \\ $$

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