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AlgebraQuestion and Answers: Page 234

Question Number 106378    Answers: 2   Comments: 0

Question Number 106364    Answers: 1   Comments: 2

0^i =?????

$$\mathrm{0}^{{i}} =????? \\ $$

Question Number 106314    Answers: 3   Comments: 0

Question Number 106281    Answers: 2   Comments: 0

(√(x+50))+(√(y+100))+(√(z+150))=((x+y+z)/4)+78 find x+y−z

$$\sqrt{\mathrm{x}+\mathrm{50}}+\sqrt{\mathrm{y}+\mathrm{100}}+\sqrt{\mathrm{z}+\mathrm{150}}=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{4}}+\mathrm{78} \\ $$$$\mathrm{find}\:\mathrm{x}+\mathrm{y}−\mathrm{z}\: \\ $$

Question Number 106272    Answers: 1   Comments: 1

(√(1 + (√(5 +(√(11 + (√(19 + (√(29 + (√…))))))))))) = ?

$$\sqrt{\mathrm{1}\:+\:\sqrt{\mathrm{5}\:+\sqrt{\mathrm{11}\:+\:\sqrt{\mathrm{19}\:+\:\sqrt{\mathrm{29}\:+\:\sqrt{\ldots}}}}}}\:=\:? \\ $$

Question Number 106259    Answers: 2   Comments: 1

Question Number 106233    Answers: 2   Comments: 0

(x^2 + 1)(x − 1)^2 = 2017yz (y^2 + 1)(y − 1)^2 = 2017xz (z^2 + 1)(z − 1)^2 = 2017xy x ≥ 1 ; y ≥ 1 ; z ≥ 1 prove that x = y = z = ((1+ (√(2018)) + (√(2015+ 2(√(2018)))))/2)

$$\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{2017}{yz} \\ $$$$\left({y}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({y}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{2017}{xz} \\ $$$$\left({z}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({z}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{2017}{xy} \\ $$$${x}\:\geqslant\:\mathrm{1}\:;\:{y}\:\geqslant\:\mathrm{1}\:;\:{z}\:\geqslant\:\mathrm{1} \\ $$$${prove}\:{that}\:\:\:{x}\:=\:{y}\:=\:{z}\:= \\ $$$$\frac{\mathrm{1}+\:\sqrt{\mathrm{2018}}\:+\:\sqrt{\mathrm{2015}+\:\mathrm{2}\sqrt{\mathrm{2018}}}}{\mathrm{2}} \\ $$

Question Number 106188    Answers: 2   Comments: 0

Solve for x x^x^(...x^a ) =a with a∈R^+

$${Solve}\:{for}\:{x} \\ $$$${x}^{{x}^{...{x}^{{a}} } } ={a}\:{with}\:{a}\in\mathbb{R}^{+} \\ $$

Question Number 106239    Answers: 1   Comments: 0

A rope of length 10cm is used to form as ector of circle of radius 35cm What ish te size of the angle of the sector

$$ \\ $$$$\mathrm{A}\:\mathrm{rope}\:\mathrm{of}\:\mathrm{length}\:\mathrm{10cm}\:\mathrm{is}\:\mathrm{used}\:\mathrm{to}\:\mathrm{form}\:\mathrm{as} \\ $$$$\mathrm{ector}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{35cm}\:\mathrm{What}\:\mathrm{ish} \\ $$$$\mathrm{te}\:\mathrm{size}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sector} \\ $$

Question Number 106176    Answers: 3   Comments: 1

If 20 men can lay 36m of a pipe in 8 hours. How long would 25 men take to lay the next 54m of the pipe?

$$\mathrm{If}\:\mathrm{20}\:\mathrm{men}\:\mathrm{can}\:\mathrm{lay}\:\mathrm{36m}\:\mathrm{of}\:\mathrm{a}\:\mathrm{pipe} \\ $$$$\mathrm{in}\:\mathrm{8}\:\mathrm{hours}.\:\mathrm{How}\:\mathrm{long}\:\mathrm{would}\:\mathrm{25} \\ $$$$\mathrm{men}\:\mathrm{take}\:\mathrm{to}\:\mathrm{lay}\:\mathrm{the}\:\mathrm{next}\:\mathrm{54m}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{pipe}? \\ $$

Question Number 106139    Answers: 0   Comments: 0

a,b∈Z ^3 (√(9^3 (√2) − 9)) = 1 −^3 (√a) +^3 (√b), a = ? b = ?

$${a},{b}\in\mathbb{Z}\:\:\:\:\:^{\mathrm{3}} \sqrt{\mathrm{9}\:^{\mathrm{3}} \sqrt{\mathrm{2}}\:−\:\mathrm{9}}\:=\:\mathrm{1}\:−\:^{\mathrm{3}} \sqrt{{a}}\:+\:^{\mathrm{3}} \sqrt{{b}},\:\:\:\:\:{a}\:=\:?\:\:\:\:\:\:{b}\:=\:? \\ $$

Question Number 106156    Answers: 2   Comments: 1

Given { (((√(xy)) +(1/(√x)) +(1/(√y)) =9)),(((√x)+(√y) = 20)) :} where x > y. find the value of x(√y) −y(√x) .

$$\mathrm{Given}\:\begin{cases}{\sqrt{\mathrm{xy}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{x}}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{y}}}\:=\mathrm{9}}\\{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\:=\:\mathrm{20}}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{x}\:>\:\mathrm{y}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}\sqrt{\mathrm{y}}\:−\mathrm{y}\sqrt{\mathrm{x}}\:. \\ $$

Question Number 106076    Answers: 2   Comments: 0

solve this pls x^x^(1/2) =(1/2)

$${solve}\:{this}\:{pls} \\ $$$${x}^{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} } =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Question Number 106022    Answers: 0   Comments: 0

Question Number 105929    Answers: 1   Comments: 0

Question Number 105927    Answers: 3   Comments: 2

((x/(x−2)))^2 +((x/(x+2)))^2 = 2

$$\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}\: \\ $$

Question Number 105851    Answers: 1   Comments: 0

Each of the students in the class was in the theater exactly two times during the winter holidays,while performances A,B and C were seen by 25,12 and 23 students,respectively .How many students are there in the class?How many of them saw performances A and B,B and C,C and A?

$$\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{students}\:\mathrm{in}\:\mathrm{the}\:\mathrm{class}\:\mathrm{was} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{theater}\:\mathrm{exactly}\:\mathrm{two}\:\mathrm{times}\:\mathrm{during}\: \\ $$$$\:\mathrm{the}\:\mathrm{winter}\:\mathrm{holidays},\mathrm{while} \\ $$$$\mathrm{performances}\:\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{were}\:\mathrm{seen}\:\mathrm{by} \\ $$$$\mathrm{25},\mathrm{12}\:\mathrm{and}\:\mathrm{23}\:\mathrm{students},\mathrm{respectively} \\ $$$$.\mathrm{How}\:\mathrm{many}\:\mathrm{students}\:\mathrm{are}\:\mathrm{there}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{class}?\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{them}\:\mathrm{saw}\:\mathrm{performances} \\ $$$$\:\mathrm{A}\:\mathrm{and}\:\mathrm{B},\mathrm{B}\:\mathrm{and}\:\mathrm{C},\mathrm{C}\:\mathrm{and}\:\mathrm{A}? \\ $$

Question Number 105822    Answers: 2   Comments: 0

Given f(x+(1/(2x))) = x^2 +(1/(4x^2 ))+3x+(3/(2x)) find f(2)

$$\mathcal{G}{iven}\:{f}\left({x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)\:=\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }+\mathrm{3}{x}+\frac{\mathrm{3}}{\mathrm{2}{x}} \\ $$$${find}\:{f}\left(\mathrm{2}\right)\: \\ $$

Question Number 105704    Answers: 1   Comments: 2

Solve the following system of equations: { ((x^3 y+x^3 y^2 +2x^2 y^2 +x^2 y^3 +xy^3 =30)),((x^2 y+xy+x+y+xy^2 =11)) :}

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{3}} \mathrm{y}+\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} +\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} +\mathrm{xy}^{\mathrm{3}} =\mathrm{30}}\\{\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{xy}+\mathrm{x}+\mathrm{y}+\mathrm{xy}^{\mathrm{2}} =\mathrm{11}}\end{cases} \\ $$

Question Number 105687    Answers: 3   Comments: 0

{ ((x^2 +y^2 = 13)),((x^3 +y^3 = 35 )) :}

$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{13}}\\{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} \:=\:\mathrm{35}\:}\end{cases} \\ $$

Question Number 105686    Answers: 1   Comments: 0

solve 7^x + 24^x = 25^x

$${solve}\:\mathrm{7}^{{x}} \:+\:\mathrm{24}^{{x}} \:=\:\mathrm{25}^{{x}} \: \\ $$

Question Number 105634    Answers: 0   Comments: 0

Question Number 105625    Answers: 1   Comments: 0

prove that ((m),(m) ) ((( n)),((n−k)) )+ ((( m)),((m−1)) ) ((( n)),((n−k+1)) ) + ((( m)),((m−2)) ) ((( n)),((n−k+2)) )+......+ ((( m)),((m−k)) ) ((n),(n) ) = (((m+n)),(( k)) )

$${prove}\:{that} \\ $$$$\begin{pmatrix}{{m}}\\{{m}}\end{pmatrix}\begin{pmatrix}{\:\:\:{n}}\\{{n}−{k}}\end{pmatrix}+\begin{pmatrix}{\:\:\:\:{m}}\\{{m}−\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\:\:\:\:\:\:\:{n}}\\{{n}−{k}+\mathrm{1}}\end{pmatrix} \\ $$$$+\begin{pmatrix}{\:\:\:{m}}\\{{m}−\mathrm{2}}\end{pmatrix}\begin{pmatrix}{\:\:\:\:\:\:\:\:{n}}\\{{n}−{k}+\mathrm{2}}\end{pmatrix}+......+\begin{pmatrix}{\:\:\:\:{m}}\\{{m}−{k}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{{m}+{n}}\\{\:\:\:\:\:{k}}\end{pmatrix} \\ $$

Question Number 105605    Answers: 2   Comments: 0

prove by mathematical induction 2^3 +4^3 +6^3 +8^3 +...+(2n)^3 = 2n^2 (n+1)^2

$${prove}\:{by}\:{mathematical}\:{induction}\: \\ $$$$\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +\mathrm{8}^{\mathrm{3}} +...+\left(\mathrm{2}{n}\right)^{\mathrm{3}} =\:\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Question Number 105536    Answers: 1   Comments: 0

Question Number 105534    Answers: 1   Comments: 2

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