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AlgebraQuestion and Answers: Page 62

Question Number 192542    Answers: 2   Comments: 0

Question Number 192508    Answers: 1   Comments: 1

Question Number 192481    Answers: 1   Comments: 0

Question Number 192477    Answers: 2   Comments: 0

Find: ((((25)/(42)) − (5/(16)) + ((10)/9) − (2/3))/((3/8) + (4/5) − (5/7) − (4/3))) = ?

Find:2542516+1092338+455743=?

Question Number 192463    Answers: 3   Comments: 0

Question Number 192440    Answers: 1   Comments: 0

Given { ((A=(((p^2 +q^2 +r^2 )^2 )/((pq)^2 +(pr)^2 +(qr)^2 )))),((B=((q^2 −pr)/(p^2 +q^2 +r^2 )) )) :} If p+q+r=0 then A^2 −4B=?

Given{A=(p2+q2+r2)2(pq)2+(pr)2+(qr)2B=q2prp2+q2+r2Ifp+q+r=0thenA24B=?

Question Number 192437    Answers: 1   Comments: 0

(1/a) + (1/b) + (1/c) = (1/(a + b + c)) . Prove that (1/a^5 ) + (1/b^5 ) + (1/c^5 ) = (1/(a^5 + b^5 + c^5 )) = (1/((a + b + c)^5 ))

1a+1b+1c=1a+b+c.Provethat1a5+1b5+1c5=1a5+b5+c5=1(a+b+c)5

Question Number 192425    Answers: 1   Comments: 0

Question if “k” is odd & A=1^k +2^k +...+n^(k ) & B=1+2+...+n prove that : B ∣ A

Questionifkisodd&A=1k+2k+...+nk&B=1+2+...+nprovethat:BA

Question Number 192409    Answers: 1   Comments: 0

Question Number 192399    Answers: 1   Comments: 1

Algebra (1 ) G, is a group and o(G ) = p^( 2) . prove that G is an abelian group. hint: ( p is prime number ) −−−−−−−−−−−−−

Algebra(1)G,isagroupando(G)=p2.provethatGisanabeliangroup.hint:(pisprimenumber)

Question Number 192387    Answers: 1   Comments: 0

Simplify (√(2(1+(√(4+(((2017^4 −1)/(2017^2 )))^2 ))))) is ....

Simplify2(1+4+(20174120172)2)is....

Question Number 192374    Answers: 3   Comments: 1

why “ 200!<100^(200) ” ?

why200!<100200?

Question Number 192370    Answers: 2   Comments: 0

Solve the equation x^4 − 2x^3 + 4x^2 + 6x − 21 = 0, Given that the sum of two of its roots is zero

Solvetheequationx42x3+4x2+6x21=0,Giventhatthesumoftwoofitsrootsiszero

Question Number 192363    Answers: 0   Comments: 0

Solve for x x_i −x+(2cx−cb)(y_i +cx^2 −cbx)=0 the following is true for this equaition i)c>0 ii)b>0 iii)there is only one real solution

Solveforxxix+(2cxcb)(yi+cx2cbx)=0thefollowingistrueforthisequaitioni)c>0ii)b>0iii)thereisonlyonerealsolution

Question Number 192330    Answers: 0   Comments: 0

Question Number 192350    Answers: 2   Comments: 0

Question Number 192186    Answers: 3   Comments: 0

If α, β and γ are the roots of x^3 + px + q = 0, find Σα^4 .

Ifα,βandγaretherootsofx3+px+q=0,findΣα4.

Question Number 192171    Answers: 1   Comments: 2

2^x^x^x =2^(√2) x=?

2xxx=22x=?

Question Number 192173    Answers: 1   Comments: 0

prove that (x^2 +a^2 )^4 = (x^4 −6x^2 a^2 +a^4 )^2 +(4x^3 a−4xa^3 )^2

provethat(x2+a2)4=(x46x2a2+a4)2+(4x3a4xa3)2

Question Number 192149    Answers: 5   Comments: 0

Find: (1/2) + (3/2^3 ) + (5/2^5 ) + (7/2^7 ) + ...

Find:12+323+525+727+...

Question Number 192143    Answers: 1   Comments: 2

Find: (7/2) + ((77)/(22)) + ((777)/(222)) + ((7777)/(2222)) +...+ ((77777777)/(22222222))

Find:72+7722+777222+77772222+...+7777777722222222

Question Number 192142    Answers: 1   Comments: 0

Question let x=<a_n a_(n−1) ...a_1 a_0 > ∈N ; a_0 ≠0 & y=<a_n a_(n−1) ...a_1 > ∈N be two natural numbers such that (x/y)∈N find the number “ x ” ?

Questionletx=<anan1...a1a0>N;a00&y=<anan1...a1>NbetwonaturalnumberssuchthatxyNfindthenumberx?

Question Number 192134    Answers: 1   Comments: 0

when ((√2)+1)^7 =(√(57125))+(√(57124)) then ((√2)−1)^7 =?

when(2+1)7=57125+57124then(21)7=?

Question Number 192172    Answers: 1   Comments: 0

Q1 ∴ x=<1a_1 a_2 ...a_n >∈N & y=<a_1 a_2 ...a_n 1>∈N if y=3x then , find the smallest value of x Q2 ∴ with the above conditions ,what other values can be placed besides the number “ 1 ”

Q1x=<1a1a2...an>∈N&y=<a1a2...an1>∈Nify=3xthen,findthesmallestvalueofxQ2withtheaboveconditions,whatothervaluescanbeplacedbesidesthenumber1

Question Number 192087    Answers: 0   Comments: 3

Let {H_α } ∈ Ω, be a family of subgroup of a group G, then prove that ∩_(α ∈ Ω) H_α .

Let{Hα}Ω,beafamilyofsubgroupofagroupG,thenprovethatαΩHα.

Question Number 192062    Answers: 2   Comments: 3

prove it : times_n ; (√(4+(√(4+(√(4+...+(√4))))) )) < 3

proveit:times_n;4+4+4+...+4<3

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