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Question Number 79079 Answers: 1 Comments: 0

Question Number 79015 Answers: 0 Comments: 4

Rigorously over one month′s time, I developed a formula for general cubic. x^3 +ax^2 +bx+c=0 let x=((pt+q)/(t+1)) pq=m, p+q=s ________________________ m^2 {(a^2 +b)^2 −6a(ab−c)} +m{2(b^2 +ac)(a^2 +b)− 3(ab−c)(ab+3c)} +(b^2 +ac)^2 −6bc(ab−c)=0 ________________________ s=−(2/3){((m(a^2 +b)+b^2 +ac)/(ab−c))} +{(8/(27))[((m(a^2 +b)+b^2 +ac)/(ab−c))]^3 −8[((m^3 +bm^2 +acm+c^2 )/(ab−c))]}^(1/3) p,q = (s/2)±(√((s^2 /4)−m)) t=−(((3pq^2 +2apq+ap^2 +2bp+bq+3c))/((p^3 +ap^2 +bp+c))) x=((pt+q)/(t+1)) . (Please help checking..) (edited a digit 1 in place of 4)

$${Rigorously}\:{over}\:{one}\:{month}'{s} \\ $$$${time},\:{I}\:{developed}\:{a}\:{formula}\:{for} \\ $$$${general}\:{cubic}. \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$$\boldsymbol{{pq}}=\boldsymbol{{m}},\:\boldsymbol{{p}}+\boldsymbol{{q}}=\boldsymbol{{s}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\boldsymbol{{m}}^{\mathrm{2}} \left\{\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)^{\mathrm{2}} −\mathrm{6}\boldsymbol{{a}}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)\right\} \\ $$$$+\boldsymbol{{m}}\left\{\mathrm{2}\left(\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}\right)\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)−\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)\left(\boldsymbol{{ab}}+\mathrm{3}\boldsymbol{{c}}\right)\right\} \\ $$$$\:\:\:+\left(\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}\right)^{\mathrm{2}} −\mathrm{6}\boldsymbol{{bc}}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)=\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\boldsymbol{{s}}=−\frac{\mathrm{2}}{\mathrm{3}}\left\{\frac{\boldsymbol{{m}}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)+\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}}{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right\} \\ $$$$\:\:+\left\{\frac{\mathrm{8}}{\mathrm{27}}\left[\frac{\boldsymbol{{m}}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)+\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}}{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right]^{\mathrm{3}} \right. \\ $$$$\left.\:\:\:−\mathrm{8}\left[\frac{\boldsymbol{{m}}^{\mathrm{3}} +\boldsymbol{{bm}}^{\mathrm{2}} +\boldsymbol{{acm}}+\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right]\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\boldsymbol{{p}},\boldsymbol{{q}}\:=\:\frac{\boldsymbol{{s}}}{\mathrm{2}}\pm\sqrt{\frac{\boldsymbol{{s}}^{\mathrm{2}} }{\mathrm{4}}−\boldsymbol{{m}}} \\ $$$$\boldsymbol{{t}}=−\frac{\left(\mathrm{3}\boldsymbol{{pq}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{apq}}+\boldsymbol{{ap}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{bp}}+\boldsymbol{{bq}}+\mathrm{3}\boldsymbol{{c}}\right)}{\left(\boldsymbol{{p}}^{\mathrm{3}} +\boldsymbol{{ap}}^{\mathrm{2}} +\boldsymbol{{bp}}+\boldsymbol{{c}}\right)} \\ $$$$\boldsymbol{{x}}=\frac{\boldsymbol{{pt}}+\boldsymbol{{q}}}{\boldsymbol{{t}}+\mathrm{1}}\:. \\ $$$$\left({Please}\:{help}\:{checking}..\right) \\ $$$$\left({edited}\:{a}\:{digit}\:\mathrm{1}\:{in}\:{place}\:{of}\:\mathrm{4}\right) \\ $$

Question Number 78880 Answers: 1 Comments: 0

x + (1/x) = 3 , x ∈ R (x^(2020) + (1/x^(2020) )) mod (10) = y y^2 − 1 = ?

$${x}\:+\:\frac{\mathrm{1}}{{x}}\:\:=\:\:\mathrm{3}\:\:\:,\:\:\:\:{x}\:\in\:\mathbb{R} \\ $$$$\left({x}^{\mathrm{2020}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2020}} }\right)\:\:{mod}\:\left(\mathrm{10}\right)\:\:=\:\:{y} \\ $$$${y}^{\mathrm{2}} \:−\:\mathrm{1}\:\:=\:\:? \\ $$

Question Number 78851 Answers: 0 Comments: 1

Question Number 78683 Answers: 0 Comments: 4

solve x^4 −18x−35=0 by using substitution x= u+v

$${solve}\:{x}^{\mathrm{4}} −\mathrm{18}{x}−\mathrm{35}=\mathrm{0}\:{by}\:{using}\:{substitution}\:{x}=\:{u}+{v} \\ $$

Question Number 78569 Answers: 1 Comments: 0

which of the following is increasing or decreasing a. u_n = ((n!)/n^n ) b. u_n = (4^n /(3^n +1)) c. u_n = (2^n /n^2 )

$${which}\:{of}\:{the}\:{following}\:{is}\:{increasing}\:{or}\:{decreasing} \\ $$$${a}.\:\:{u}_{{n}} \:=\:\frac{{n}!}{{n}^{{n}} } \\ $$$${b}.\:\:{u}_{{n}} =\:\frac{\mathrm{4}^{{n}} }{\mathrm{3}^{{n}} +\mathrm{1}} \\ $$$${c}.\:{u}_{{n}} =\:\frac{\mathrm{2}^{{n}} }{{n}^{\mathrm{2}} } \\ $$

Question Number 78496 Answers: 1 Comments: 4

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