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Question Number 78342 Answers: 0 Comments: 0

sppose that (R,+,.)be aring and we have the ring (R×Z,+^(′ ) ,.^′ ) prove that (R×0,+^′ ,.′) it was ideal in (R×Z,+^′ ,.^′ ) and prove (0×Z,+,.)be isomorphic in (Z,+,.) and if a∈R identity element (a^2 =a)prove that (−a,1)be identity element in the ring (R×Z,+^′ ,.^′ ) pleas sir help me am neding this pleas?

$${sppose}\:{that}\:\left({R},+,.\right){be}\:{aring}\:{and}\:{we}\:{have}\:{the}\:{ring}\:\left({R}×{Z},+^{'\:} ,.^{'} \right)\:{prove}\:{that}\:\left({R}×\mathrm{0},+^{'} ,.'\right)\:{it}\:{was}\:{ideal}\:{in}\:\left({R}×{Z},+^{'} ,.^{'} \right) \\ $$$${and}\:{prove}\:\left(\mathrm{0}×{Z},+,.\right){be}\:{isomorphic}\:{in}\:\left({Z},+,.\right) \\ $$$${and}\:{if}\:{a}\in{R}\:{identity}\:{element}\:\left({a}^{\mathrm{2}} ={a}\right){prove}\:{that}\:\left(−{a},\mathrm{1}\right){be}\:{identity}\:{element}\:{in}\:{the}\:{ring}\:\left({R}×{Z},+^{'} ,.^{'} \right) \\ $$$${pleas}\:{sir}\:{help}\:{me}\:{am}\:{neding}\:{this}\:{pleas}? \\ $$

Question Number 78340 Answers: 2 Comments: 0

Let a,b,c > 0 and c^2 = ((ab+bc+ca)/3) . Prove that ((a^3 +b^3 −2c^3 )/(a^3 +b^3 +c^3 )) ≤ 3(((a^2 +b^2 −2c^2 )/(a^2 +b^2 +c^2 )))

$$\mathrm{Let}\:\:{a},{b},{c}\:>\:\mathrm{0}\:\:\mathrm{and}\:\:{c}^{\mathrm{2}} \:=\:\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\mathrm{2}{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\:\leqslant\:\mathrm{3}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right) \\ $$

Question Number 78335 Answers: 1 Comments: 0

find the solution of (x/((x−2)^3 +(x−3)^3 −1)) ≥ 0

$${find}\:{the}\:{solution}\:{of} \\ $$$$\frac{{x}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}}\:\geqslant\:\mathrm{0} \\ $$

Question Number 78334 Answers: 0 Comments: 1

∫_0 ^(π/2) [cos^2 (cos x)+sin^2 (sin x) ] dx

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\left[\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{cos}\:{x}\right)+\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)\:\right]\:{dx} \\ $$

Question Number 78333 Answers: 1 Comments: 0

prove by contradiction that (√(2 )) is irrational.

$$\mathrm{prove}\:\mathrm{by}\:\mathrm{contradiction}\:\mathrm{that}\:\sqrt{\mathrm{2}\:}\:\mathrm{is}\:\mathrm{irrational}. \\ $$

Question Number 78332 Answers: 0 Comments: 2

Find the values of k and n for which x^3 and higher powers of x are negligeble given that (1+kx)^n =1+2x+6x^2 .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{k}\:\mathrm{and}\:\mathrm{n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{x}^{\mathrm{3}} \mathrm{and}\:\mathrm{higher}\:\mathrm{powers}\:\mathrm{of}\:\mathrm{x}\:\mathrm{are}\:\mathrm{negligeble} \\ $$$$\mathrm{given}\:\mathrm{that}\:\left(\mathrm{1}+\mathrm{kx}\right)^{\mathrm{n}} =\mathrm{1}+\mathrm{2x}+\mathrm{6x}^{\mathrm{2}} . \\ $$

Question Number 78330 Answers: 0 Comments: 0

given the regular pyramid T.ABCD with a square base . length AB = 8 , TC = 6. point P is mid BC. if x is the angle between TP and BD. determine the value of cos x.

$${given}\:{the}\:{regular}\:{pyramid} \\ $$$${T}.{ABCD}\:{with}\:{a}\:{square}\:{base} \\ $$$$.\:{length}\:{AB}\:=\:\mathrm{8}\:,\:{TC}\:=\:\mathrm{6}.\:{point} \\ $$$${P}\:{is}\:{mid}\:{BC}.\:{if}\:{x}\:{is}\:{the}\:{angle}\: \\ $$$${between}\:{TP}\:{and}\:{BD}.\:{determine} \\ $$$${the}\:{value}\:{of}\:\mathrm{cos}\:{x}. \\ $$

Question Number 78325 Answers: 0 Comments: 3

lim_(x→∞) (((5x^2 +3x)/(5x^2 −2x)))^(3x−1)

$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)^{\mathrm{3}{x}−\mathrm{1}} \\ $$

Question Number 78324 Answers: 0 Comments: 2

lim_(x→0) ((1+sin 4x−cos 2x)/((√(x+1))−(√(1−x))))=

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{4}{x}−\mathrm{cos}\:\mathrm{2}{x}}{\sqrt{{x}+\mathrm{1}}−\sqrt{\mathrm{1}−{x}}}= \\ $$

Question Number 78319 Answers: 0 Comments: 1

57+6h=16h−33

$$\mathrm{57}+\mathrm{6}{h}=\mathrm{16}{h}−\mathrm{33} \\ $$

Question Number 78316 Answers: 0 Comments: 2

the circle represents a farm where (LK) is symetric axe of circle such as ∀ M of this circle verifying ML^2 −4MK^2 =0 with LK=150m. calculate the radius of circle. please help me...

$$\mathrm{the}\:\mathrm{circle}\:\mathrm{represents}\:\mathrm{a}\:\mathrm{farm}\:\mathrm{where} \\ $$$$\left(\mathrm{LK}\right)\:\mathrm{is}\:\mathrm{symetric}\:\mathrm{axe}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{such} \\ $$$$\mathrm{as}\:\forall\:\mathrm{M}\:\mathrm{of}\:\mathrm{this}\:\mathrm{circle}\:\mathrm{verifying} \\ $$$$\mathrm{ML}^{\mathrm{2}} −\mathrm{4MK}^{\mathrm{2}} =\mathrm{0}\:\:\mathrm{with}\:\mathrm{LK}=\mathrm{150m}. \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}. \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}... \\ $$

Question Number 78314 Answers: 1 Comments: 1

resolve {_(logx_y =logy_x ) ^(x^y =y^x )

$${resolve} \\ $$$$\left\{_{{logx}_{{y}} ={logy}_{{x}} } ^{{x}^{{y}} ={y}^{{x}} } \right. \\ $$

Question Number 78311 Answers: 1 Comments: 2

Show that ((3+sin2x−2cos2x)/(1+3sin^2 x−cos2x ))=(2/5)(2+(1/(tanx)))

$$\mathrm{Show}\:\mathrm{that}\:\frac{\mathrm{3}+\mathrm{sin2}{x}−\mathrm{2cos2}{x}}{\mathrm{1}+\mathrm{3}{sin}^{\mathrm{2}} {x}−{cos}\mathrm{2}{x}\:}=\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{2}+\frac{\mathrm{1}}{{tanx}}\right) \\ $$

Question Number 78307 Answers: 0 Comments: 0

Question Number 78306 Answers: 1 Comments: 0

The sum of age of Hamadou his wife and theirs son is 100. n years ago the wife had the quadruple of his son′s age and Hamadou was 6 time older than his son. Determine theirs ages. i want that you help me to found equations. i found the first : x+y+z=100 please help me for the rest.

$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{age}\:\mathrm{of}\:\mathrm{Hamadou}\:\:\: \\ $$$$\mathrm{his}\:\mathrm{wife}\:\mathrm{and}\:\mathrm{theirs}\:\mathrm{son}\:\mathrm{is}\:\mathrm{100}. \\ $$$$\mathrm{n}\:\mathrm{years}\:\mathrm{ago}\:\mathrm{the}\:\mathrm{wife}\:\mathrm{had}\:\mathrm{the}\: \\ $$$$\mathrm{quadruple}\:\mathrm{of}\:\mathrm{his}\:\mathrm{son}'\mathrm{s}\:\mathrm{age}\:\mathrm{and}\: \\ $$$$\mathrm{Hamadou}\:\mathrm{was}\:\mathrm{6}\:\mathrm{time}\:\mathrm{older}\:\mathrm{than} \\ $$$$\mathrm{his}\:\mathrm{son}. \\ $$$$\mathrm{Determine}\:\mathrm{theirs}\:\mathrm{ages}. \\ $$$$ \\ $$$$\mathrm{i}\:\mathrm{want}\:\mathrm{that}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{found} \\ $$$$\mathrm{equations}. \\ $$$$\mathrm{i}\:\mathrm{found}\:\mathrm{the}\:\mathrm{first}\::\:{x}+{y}+{z}=\mathrm{100} \\ $$$${please}\:{help}\:{me}\:{for}\:{the}\:{rest}. \\ $$

Question Number 78305 Answers: 0 Comments: 0

∫^∞^ 5646778727711=778888877−{116567×622−[66262−712]66666}

$$\overset{\hat {\infty}} {\int}\mathrm{5646778727711}=\mathrm{778888877}−\left\{\mathrm{116567}×\mathrm{622}−\left[\mathrm{66262}−\mathrm{712}\right]\mathrm{66666}\right\} \\ $$$$ \\ $$$$ \\ $$

Question Number 78309 Answers: 1 Comments: 0