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AlgebraQuestion and Answers: Page 62

Question Number 192594 Answers: 0 Comments: 0

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))) = ?

$$\mathrm{Find}:\:\:\:\:\:\frac{\frac{\mathrm{25}}{\mathrm{42}}\:−\:\frac{\mathrm{5}}{\mathrm{16}}\:+\:\frac{\mathrm{10}}{\mathrm{9}}\:−\:\frac{\mathrm{2}}{\mathrm{3}}}{\frac{\mathrm{3}}{\mathrm{8}}\:+\:\frac{\mathrm{4}}{\mathrm{5}}\:−\:\frac{\mathrm{5}}{\mathrm{7}}\:−\:\frac{\mathrm{4}}{\mathrm{3}}}\:=\:? \\ $$

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=?

$$\:\:\:\:\mathrm{Given}\:\begin{cases}{\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{pr}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} }}\\{\mathrm{B}=\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:}\end{cases}\:\:\:\:\:\: \\ $$$$\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{0}\:\mathrm{then}\:\mathrm{A}^{\mathrm{2}} −\mathrm{4B}=? \\ $$$$ \\ $$

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 ))

$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} \:+\:{b}^{\mathrm{5}} \:+\:{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{5}} } \\ $$

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

$${Question} \\ $$$${if}\:\:``{k}''\:{is}\:{odd}\:\:\&\:{A}=\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +...+{n}^{{k}\:\:} \:\&\:\:{B}=\mathrm{1}+\mathrm{2}+...+{n} \\ $$$${prove}\:{that}\:\::\:\:{B}\:\mid\:{A}\: \\ $$

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 ) −−−−−−−−−−−−−

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Algebra}\:\left(\mathrm{1}\:\right) \\ $$$$\:\:{G},\:{is}\:{a}\:{group}\:\:{and}\:\:\:{o}\left({G}\:\right)\:=\:{p}^{\:\mathrm{2}} \:. \\ $$$$\:\:\:{prove}\:{that}\:{G}\:{is}\:{an}\:{abelian}\:{group}. \\ $$$$\:\:\:{hint}:\:\:\left(\:{p}\:{is}\:{prime}\:{number}\:\:\right) \\ $$$$\:\:\:\:\:−−−−−−−−−−−−− \\ $$

Question Number 192387 Answers: 1 Comments: 0

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

$$\:\mathrm{Simplify}\: \\ $$$$\:\sqrt{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{4}+\left(\frac{\mathrm{2017}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }\right)^{\mathrm{2}} }\right)}\: \\ $$$$\:\mathrm{is}\:....\: \\ $$

Question Number 192374 Answers: 3 Comments: 1

why “ 200!<100^(200) ” ?

$${why}\:\:\:``\:\mathrm{200}!<\mathrm{100}^{\mathrm{200}} \:''\:? \\ $$

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

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

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

$$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}_{{i}} −{x}+\left(\mathrm{2}{cx}−{cb}\right)\left({y}_{{i}} +{cx}^{\mathrm{2}} −{cbx}\right)=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{this}\:\mathrm{equaition} \\ $$$$\left.{i}\right)\mathrm{c}>\mathrm{0} \\ $$$$\left.{ii}\right)\mathrm{b}>\mathrm{0} \\ $$$$\left.{iii}\right)\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution} \\ $$

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 .

$$\mathrm{If}\:\:\alpha,\:\beta\:\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\:\:\mathrm{x}^{\mathrm{3}} \:\:+\:\:\mathrm{px}\:\:+\:\:\mathrm{q}\:\:=\:\:\mathrm{0},\:\:\:\:\mathrm{find}\:\:\:\Sigma\alpha^{\mathrm{4}} . \\ $$

Question Number 192171 Answers: 1 Comments: 2

2^x^x^x =2^(√2) x=?

$$\mathrm{2}^{{x}^{{x}^{{x}} } } =\mathrm{2}^{\sqrt{\mathrm{2}}} \\ $$$${x}=? \\ $$

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

$${prove}\:{that} \\ $$$$\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} \:=\:\left({x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{2}} {a}^{\mathrm{2}} +{a}^{\mathrm{4}} \right)^{\mathrm{2}} +\left(\mathrm{4}{x}^{\mathrm{3}} {a}−\mathrm{4}{xa}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$

Question Number 192149 Answers: 5 Comments: 0

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

$$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }\:+\:\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }\:+\:\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}} }\:+\:... \\ $$

Question Number 192143 Answers: 1 Comments: 2

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

$$\mathrm{Find}: \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\:+\:\frac{\mathrm{77}}{\mathrm{22}}\:+\:\frac{\mathrm{777}}{\mathrm{222}}\:+\:\frac{\mathrm{7777}}{\mathrm{2222}}\:+...+\:\frac{\mathrm{77777777}}{\mathrm{22222222}} \\ $$

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 ” ?

$${Question} \\ $$$${let}\:\:\:{x}=<{a}_{{n}} {a}_{{n}−\mathrm{1}} ...{a}_{\mathrm{1}} {a}_{\mathrm{0}} >\:\in\mathbb{N}\:;\:{a}_{\mathrm{0}} \neq\mathrm{0}\:\:\&\: \\ $$$$\:{y}=<{a}_{{n}} {a}_{{n}−\mathrm{1}} ...{a}_{\mathrm{1}} >\:\in\mathbb{N}\:\:{be}\: \\ $$$${two}\:{natural}\:{numbers}\: \\ $$$${such}\:{that}\:\:\frac{{x}}{{y}}\in\mathbb{N}\: \\ $$$${find}\:{the}\:{number}\:``\:{x}\:''\:? \\ $$$$ \\ $$

Question Number 192134 Answers: 1 Comments: 0

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

$$\mathrm{when}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{7}} =\sqrt{\mathrm{57125}}+\sqrt{\mathrm{57124}} \\ $$$$\mathrm{then}\:\:\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{7}} =? \\ $$

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 ”

$${Q}\mathrm{1}\:\therefore\:\:{x}=<\mathrm{1}{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{{n}} >\in\mathbb{N}\:\:\&\:\:{y}=<{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{{n}} \mathrm{1}>\in\mathbb{N} \\ $$$${if}\:\:{y}=\mathrm{3}{x}\:\:{then}\:\:,\:{find}\:{the}\:{smallest}\: \\ $$$${value}\:{of}\:\:{x} \\ $$$${Q}\mathrm{2}\:\therefore\:{with}\:{the}\:{above}\:{conditions}\:,{what}\:{other}\:{values}\: \\ $$$${can}\:{be}\:{placed}\:\:{besides}\:{the}\:{number}\:``\:\mathrm{1}\:''\: \\ $$

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 .

$$\mathrm{Let}\:\left\{\mathrm{H}_{\alpha} \right\}\:\in\:\Omega,\:\mathrm{be}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\: \\ $$$$\mathrm{subgroup}\:\mathrm{of}\:\mathrm{a}\:\mathrm{group}\:\mathrm{G},\:\mathrm{then}\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\cap_{\alpha\:\in\:\Omega} \mathrm{H}_{\alpha} . \\ $$$$ \\ $$$$ \\ $$

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