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AlgebraQuestion and Answers: Page 254

Question Number 95048    Answers: 0   Comments: 1

a_1 =−(1/(30)) a_2 =−(1/(12)) a_3 =−(1/6) a_4 =−(7/(24)) a_5 =−(7/(15)) a_6 =−(7/(10)) a_7 =−1 find a_k

$${a}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{30}} \\ $$$${a}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${a}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${a}_{\mathrm{4}} =−\frac{\mathrm{7}}{\mathrm{24}} \\ $$$${a}_{\mathrm{5}} =−\frac{\mathrm{7}}{\mathrm{15}} \\ $$$${a}_{\mathrm{6}} =−\frac{\mathrm{7}}{\mathrm{10}} \\ $$$${a}_{\mathrm{7}} =−\mathrm{1} \\ $$$${find}\:{a}_{{k}} \\ $$

Question Number 95014    Answers: 0   Comments: 10

If y_(n + 1) − y_n = 6, and y_0 = 7 Find y_n

$$\boldsymbol{\mathrm{If}}\:\:\:\:\:\:\boldsymbol{\mathrm{y}}_{\boldsymbol{\mathrm{n}}\:\:+\:\:\mathrm{1}} \:\:−\:\:\boldsymbol{\mathrm{y}}_{\boldsymbol{\mathrm{n}}} \:\:\:=\:\:\:\mathrm{6},\:\:\:\:\:\:\:\boldsymbol{\mathrm{and}}\:\:\:\:\:\boldsymbol{\mathrm{y}}_{\mathrm{0}} \:\:=\:\:\:\mathrm{7} \\ $$$$\boldsymbol{\mathrm{Find}}\:\:\:\:\:\boldsymbol{\mathrm{y}}_{\boldsymbol{\mathrm{n}}} \\ $$

Question Number 94960    Answers: 2   Comments: 0

Question Number 94874    Answers: 1   Comments: 1

Question Number 94868    Answers: 0   Comments: 5

Question Number 94835    Answers: 0   Comments: 0

proof or disproof that if a quotient group (G/H) is abelian then G must be abelian.

$$\mathrm{proof}\:\mathrm{or}\:\mathrm{disproof}\:\mathrm{that}\:\mathrm{if}\:\mathrm{a}\:\mathrm{quotient} \\ $$$$\mathrm{group}\:\frac{\mathrm{G}}{\mathrm{H}}\:\mathrm{is}\:\mathrm{abelian}\:\mathrm{then}\:\mathrm{G}\:\mathrm{must}\:\mathrm{be}\:\mathrm{abelian}. \\ $$

Question Number 94818    Answers: 1   Comments: 7

find such a polynomial if is divided by (x−2) then the remainder is 5, if it isdivided by (x−3) the remainder is 9, if it is divided by (x−4) the remainder is 13, if divide by (x−10) and the remaider becomes 37 and if (x−(3/4)) divided by the remainder becomes zero?

$$\:\mathrm{find}\:\mathrm{such}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{if}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{2}\right)\:\mathrm{then}\:\mathrm{the}\: \\ $$$$\mathrm{remainder}\:\mathrm{is}\:\mathrm{5},\:\mathrm{if}\:\mathrm{it}\:\mathrm{isdivided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{3}\right)\:\mathrm{the}\:\mathrm{remainder} \\ $$$$\mathrm{is}\:\mathrm{9},\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{4}\right)\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{13},\:\mathrm{if}\: \\ $$$$\mathrm{divide}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{10}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{remaider}\:\mathrm{becomes}\:\mathrm{37}\:\mathrm{and} \\ $$$$\mathrm{if}\:\left(\mathrm{x}−\frac{\mathrm{3}}{\mathrm{4}}\right)\:\mathrm{divided}\:\mathrm{by}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{becomes}\:\mathrm{zero}? \\ $$

Question Number 94811    Answers: 3   Comments: 0

∫(dx/(x+(√x))) (2,3)=?

$$\int\frac{\mathrm{dx}}{\mathrm{x}+\sqrt{\mathrm{x}}}\:\:\:\left(\mathrm{2},\mathrm{3}\right)=? \\ $$

Question Number 94786    Answers: 0   Comments: 1

x, y ∈N\{0, 1} ∧ x≤y find z∈N with z!=x!y!

$${x},\:{y}\:\in\mathbb{N}\backslash\left\{\mathrm{0},\:\mathrm{1}\right\}\:\wedge\:{x}\leqslant{y} \\ $$$$\mathrm{find}\:{z}\in\mathbb{N}\:\mathrm{with}\:{z}!={x}!{y}! \\ $$

Question Number 94780    Answers: 0   Comments: 1

x!(x−4)!=12(2x−7)! x=?

$$\mathrm{x}!\left(\mathrm{x}−\mathrm{4}\right)!=\mathrm{12}\left(\mathrm{2x}−\mathrm{7}\right)! \\ $$$$\mathrm{x}=? \\ $$

Question Number 94742    Answers: 0   Comments: 1

Question Number 94705    Answers: 0   Comments: 1

a set X had one more subset than set Y. If X has 8 more subsets than Y. Find the number if element in the set X.

$$\mathrm{a}\:\mathrm{set}\:\mathrm{X}\:\mathrm{had}\:\mathrm{one}\:\mathrm{more}\:\mathrm{subset}\:\mathrm{than}\:\mathrm{set}\:\mathrm{Y}. \\ $$$$\mathrm{If}\:\mathrm{X}\:\mathrm{has}\:\mathrm{8}\:\mathrm{more}\:\mathrm{subsets}\:\mathrm{than}\:\mathrm{Y}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{if}\:\mathrm{element}\:\mathrm{in}\:\mathrm{the}\:\mathrm{set}\:\mathrm{X}. \\ $$

Question Number 94603    Answers: 1   Comments: 1

List the elements in C={x:x is an x^2 ≤4, integer}

$$\mathrm{List}\:\mathrm{the}\:\mathrm{elements}\:\mathrm{in}\: \\ $$$${C}=\left\{{x}:{x}\:\mathrm{is}\:\mathrm{an}\:{x}^{\mathrm{2}} \leqslant\mathrm{4},\:\mathrm{integer}\right\} \\ $$

Question Number 94601    Answers: 1   Comments: 0

Question Number 94522    Answers: 3   Comments: 0

if sinA+sinB=n and cosA+cosB=m then sin(A+B)=? a: ((3mn)/(m^2 +n^2 )) b:((2mn)/(m^2 +n^2 )) c:((mn)/(m^2 +n^2 )) d:((2mn)/(m+n)) with steps?

$$\mathrm{if}\:\:\mathrm{sinA}+\mathrm{sinB}=\mathrm{n}\:\mathrm{and}\:\mathrm{cosA}+\mathrm{cosB}=\mathrm{m} \\ $$$$\mathrm{then}\:\:\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)=? \\ $$$$\mathrm{a}:\:\frac{\mathrm{3mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\:\:\:\:\mathrm{b}:\frac{\mathrm{2mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\:\:\mathrm{c}:\frac{\mathrm{mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\:\:\mathrm{d}:\frac{\mathrm{2mn}}{\mathrm{m}+\mathrm{n}} \\ $$$$\mathrm{with}\:\mathrm{steps}? \\ $$

Question Number 94521    Answers: 3   Comments: 3

if a^(10) +a^5 +1=0 then a^(2005) +(1/a^(2005) )=? a: a^(10) +a^(11) b:a^(10) +a^5 c:3(a^(10) +a^5 ) d:0 with steps?

$$\mathrm{if}\:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{a}^{\mathrm{2005}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2005}} }=? \\ $$$$\mathrm{a}:\:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{11}} \:\:\:\:\mathrm{b}:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} \:\:\:\mathrm{c}:\mathrm{3}\left(\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} \right)\:\:\mathrm{d}:\mathrm{0} \\ $$$$\mathrm{with}\:\mathrm{steps}? \\ $$

Question Number 94479    Answers: 1   Comments: 0

find solution in C x^4 +x^3 +x^2 +x+1 = 0

$$\mathrm{find}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{C}\: \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 94463    Answers: 0   Comments: 2

lim_(x→∞) ((6x^(k−2) +3x^2 +10)/(3x^5 +x+20)) then k^2 +1=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{6x}^{\mathrm{k}−\mathrm{2}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{10}}{\mathrm{3x}^{\mathrm{5}} +\mathrm{x}+\mathrm{20}} \\ $$$$\mathrm{then}\:\mathrm{k}^{\mathrm{2}} +\mathrm{1}=? \\ $$

Question Number 94456    Answers: 1   Comments: 0

log_x 16+log_(16) x=log_(512) x+log_x 512 x=?

$$\mathrm{log}_{\mathrm{x}} \mathrm{16}+\mathrm{log}_{\mathrm{16}} \mathrm{x}=\mathrm{log}_{\mathrm{512}} \mathrm{x}+\mathrm{log}_{\mathrm{x}} \mathrm{512} \\ $$$$\mathrm{x}=? \\ $$

Question Number 94424    Answers: 1   Comments: 4

y=x^x y^′ =?

$$\mathrm{y}=\mathrm{x}^{\mathrm{x}} \\ $$$$\mathrm{y}^{'} =? \\ $$

Question Number 94406    Answers: 0   Comments: 5

Question Number 94309    Answers: 0   Comments: 0

Question Number 94257    Answers: 2   Comments: 0

If 9y^2 + (1/y^2 ) =3, then find the value of 27y^3 + (1/y^3 )

$$\mathrm{If}\:\mathrm{9y}^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\:=\mathrm{3},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{27y}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} } \\ $$

Question Number 94174    Answers: 1   Comments: 3

{ ((x+y = 10)),(((x)^(1/(3 )) + (y)^(1/(3 )) = (5/2) ((xy))^(1/(6 )) )) :} find x &y??

$$\begin{cases}{\mathrm{x}+\mathrm{y}\:=\:\mathrm{10}}\\{\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}}\:+\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{y}}\:=\:\frac{\mathrm{5}}{\mathrm{2}}\:\sqrt[{\mathrm{6}\:\:}]{\mathrm{xy}}}\end{cases}\: \\ $$$$\mathrm{find}\:\mathrm{x}\:\&\mathrm{y}?? \\ $$

Question Number 94124    Answers: 0   Comments: 3

20+a=a cosh(((75)/a)) a=?

$$\mathrm{20}+{a}={a}\:{cosh}\left(\frac{\mathrm{75}}{{a}}\right) \\ $$$${a}=? \\ $$

Question Number 94095    Answers: 0   Comments: 0

How many subgroups do Z_3 ⊕Z_(16 ) has? Justify.

$$\mathrm{How}\:\mathrm{many}\:\mathrm{subgroups}\:\mathrm{do}\:\mathrm{Z}_{\mathrm{3}} \oplus\mathrm{Z}_{\mathrm{16}\:} \:\mathrm{has}?\:\mathrm{Justify}. \\ $$

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