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Question Number 78689 Answers: 0 Comments: 0

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 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 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 78256 Answers: 0 Comments: 0

Evaluate Σa_1 a_2 a_3 as a function of a_i

$$\mathrm{Evaluate}\:\Sigma\mathrm{a}_{\mathrm{1}} \mathrm{a}_{\mathrm{2}} \mathrm{a}_{\mathrm{3}} \: \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\:\mathrm{a}_{\mathrm{i}} \: \\ $$$$ \\ $$$$ \\ $$

Question Number 78161 Answers: 1 Comments: 0

Question Number 78160 Answers: 0 Comments: 1

find the term independent of x in [ ((x−1)/x)]^9

$$\mathrm{find}\:\mathrm{the}\:\mathrm{term}\:\mathrm{independent}\:\mathrm{of}\:\mathrm{x}\:\mathrm{in} \\ $$$$\:\:\left[\:\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right]^{\mathrm{9}} \\ $$

Question Number 78156 Answers: 0 Comments: 1

Given that u_(n + 1) = (a_n /2) + 5 evalatuate lim_(x→∞) a_n deduce if a_(n ) is convergent or divergent.

$$\mathrm{Given}\:\mathrm{that}\:{u}_{{n}\:+\:\mathrm{1}} =\:\frac{{a}_{{n}} }{\mathrm{2}}\:+\:\mathrm{5}\:\:\: \\ $$$${evalatuate}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \\ $$$${deduce}\:{if}\:{a}_{{n}\:} \:{is}\:{convergent}\:{or}\:{divergent}. \\ $$

Question Number 78075 Answers: 0 Comments: 5

expressing P(x) = ((x^2 + x)/((x−3)(x^2 −2))) in partial fractions gives A. (A/((x−3))) + ((Bx + C)/((x^2 −2))) B. (A/(x−3)) + (B/(x−2)) + (C/(x+2)) C. (A/(x−3)) + (B/(x−(√2))) + (C/(x + (√2))) D. ((Ax + B)/(x−3)) + (C/(x^2 −2))

$${expressing}\:\:{P}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} \:+\:{x}}{\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right)}\:{in}\:{partial}\:{fractions}\:{gives} \\ $$$${A}.\:\:\frac{{A}}{\left({x}−\mathrm{3}\right)}\:+\:\frac{{Bx}\:+\:{C}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)}\: \\ $$$${B}.\:\:\frac{{A}}{{x}−\mathrm{3}}\:+\:\frac{{B}}{{x}−\mathrm{2}}\:+\:\frac{{C}}{{x}+\mathrm{2}} \\ $$$${C}.\:\frac{{A}}{{x}−\mathrm{3}}\:+\:\frac{{B}}{{x}−\sqrt{\mathrm{2}}}\:+\:\frac{{C}}{{x}\:+\:\sqrt{\mathrm{2}}} \\ $$$${D}.\:\frac{{Ax}\:+\:{B}}{{x}−\mathrm{3}}\:+\:\frac{{C}}{{x}^{\mathrm{2}} −\mathrm{2}} \\ $$

Question Number 78074 Answers: 2 Comments: 0

evaluate ∫_1 ^4 sinh^(−1) x dx and ∫_1 ^(1/2) tanh^(−1) x dx

$${evaluate}\:\int_{\mathrm{1}} ^{\mathrm{4}} \mathrm{sinh}\:^{−\mathrm{1}} {x}\:{dx}\:\:{and}\:\underset{\mathrm{1}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\mathrm{tanh}\:^{−\mathrm{1}} {x}\:{dx} \\ $$

Question Number 77872 Answers: 2 Comments: 6

show that f(x)=2r^3 +5x−1 has a zero in the interval [0.1].

$${show}\:{that}\:{f}\left({x}\right)=\mathrm{2}{r}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{1}\:{has}\:{a}\:{zero}\:{in}\:{the}\:{interval}\:\left[\mathrm{0}.\mathrm{1}\right]. \\ $$

Question Number 77845 Answers: 0 Comments: 0

In the equation B=μ_0 H×μ_0 M why is the polarization of the vacuum accounted for by constant μ_0 if the vacuum is absolutely empty?

$$\boldsymbol{\mathrm{I}}\mathrm{n}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\boldsymbol{\mathrm{B}}=\mu_{\mathrm{0}} \boldsymbol{\mathrm{H}}×\mu_{\mathrm{0}} \boldsymbol{\mathrm{M}}\: \\ $$$$\mathrm{why}\:\mathrm{is}\:\mathrm{the}\:\mathrm{polarization}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vacuum}\: \\ $$$$\mathrm{accounted}\:\mathrm{for}\:\mathrm{by}\: \\ $$$$\mathrm{constant}\:\mu_{\mathrm{0}} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{vacuum}\:\mathrm{is}\:\mathrm{absolutely}\:\mathrm{empty}? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 77803 Answers: 1 Comments: 0

solve (4x^2 +4x+1)y′′−(12x+6)y′−8x^3 −1=12x^2 −16y+6x