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Question Number 170800    Answers: 1   Comments: 0

Prove that, for any real number x and odd positive integer n, cos^n x=(1/2^(n+1) )Σ_(k=0) ^((n−1)/2) C_k ^n cos (n−2k)x

$$\mathrm{Prove}\:\mathrm{that},\:\mathrm{for}\:\mathrm{any}\:\mathrm{real}\:\mathrm{number}\:{x}\:\mathrm{and}\:\mathrm{odd}\:\mathrm{positive}\:\mathrm{integer}\:{n}, \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}^{{n}} {x}=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{cos}\:\left({n}−\mathrm{2}{k}\right){x} \\ $$

Question Number 170796    Answers: 2   Comments: 0

Question Number 170795    Answers: 2   Comments: 0

solve (x^2 +y^2 +1)dx+x(x−2y)dy=0

$${solve} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right){dx}+{x}\left({x}−\mathrm{2}{y}\right){dy}=\mathrm{0} \\ $$

Question Number 170794    Answers: 2   Comments: 0

2n objects of each of three kinds are given to two persons, so that each person gets 3n objects. Prove that this can be done in 3n^2 + 3n + 1 ways.

$$\mathrm{2}{n}\:\mathrm{objects}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{three}\:\mathrm{kinds}\:\mathrm{are} \\ $$$$\mathrm{given}\:\mathrm{to}\:\mathrm{two}\:\mathrm{persons},\:\mathrm{so}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{person}\:\mathrm{gets}\:\mathrm{3}{n}\:\mathrm{objects}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{in}\:\mathrm{3}{n}^{\mathrm{2}} \:+\:\mathrm{3}{n}\:+\:\mathrm{1}\:\mathrm{ways}. \\ $$

Question Number 170781    Answers: 2   Comments: 0

(√((2−x))) = (2−x)^2 solve for x

$$\sqrt{\left(\mathrm{2}−\mathrm{x}\right)}\:\:\:=\:\:\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} \:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}} \\ $$

Question Number 170780    Answers: 1   Comments: 0

Question Number 170776    Answers: 0   Comments: 2

Question Number 170772    Answers: 0   Comments: 0

Question Number 170767    Answers: 2   Comments: 0

Question Number 170766    Answers: 2   Comments: 0

lim_(x→−∞) ((x^7 +x^6 −1))^(1/7) +((x^5 +1−x^9 ))^(1/9) =?

$$\:\:\:\:\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{7}}]{{x}^{\mathrm{7}} +{x}^{\mathrm{6}} −\mathrm{1}}\:+\sqrt[{\mathrm{9}}]{{x}^{\mathrm{5}} +\mathrm{1}−{x}^{\mathrm{9}} }\:=? \\ $$

Question Number 170754    Answers: 1   Comments: 5

Solve: x + (√y) = 3 ...... (i) (√x) + y = 5 ...... (ii)

$$\mathrm{Solve}: \\ $$$$\:\:\:\:\:\mathrm{x}\:\:\:\:+\:\:\:\sqrt{\mathrm{y}}\:\:\:\:=\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:......\:\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\sqrt{\mathrm{x}}\:\:\:\:+\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:......\:\:\left(\mathrm{ii}\right) \\ $$

Question Number 170753    Answers: 0   Comments: 0

Question Number 170752    Answers: 2   Comments: 0

Solve:− ((sinx + cosx)/(sinx − cosx)) = (((√(3 )) − 1)/( (√(3 )) + 1))

$${Solve}:−\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}}{\:\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}}\: \\ $$

Question Number 170748    Answers: 1   Comments: 0

A number of four different digits is to be formed from the digits 1,2,3,4,5,6,7,8,9.Find how many of them are: i)greater than 4000 ii)less than 6000 iii) divisible by 2 iv) divisible by 5

$$ \\ $$A number of four different digits is to be formed from the digits 1,2,3,4,5,6,7,8,9.Find how many of them are: i)greater than 4000 ii)less than 6000 iii) divisible by 2 iv) divisible by 5

Question Number 170743    Answers: 0   Comments: 0

2. Uma soluca^ o tampao foi preparada misturarando 200ml NH_(3 ) 0,6 moles e 300ml de uma solucao de NH_3 Cl 0,2 moles P^(kh) =9,24 e log 2=0,3 a) Qual e o P^h desta solucao tampa^ o supondo-se um volume de 500ml? b)Qual sera o P^h depois de ser adicionado 0,2 molar de ion [H^− ] Dados n(CH_3 -CH_2 COOH)=0,02mok/l=0,02M Ka=1,3∙10^(−5) n(CH_3 −CH_2 -COONa)=0,015mol/l=0,015M V=1l log 2=0,3 log 0,11 P^h =? 1°Passo 2°Passo P^(ka) =−log Ka P^h =P^(ka) +log(([Base])/([Acido])) P^(ka) =−log 1,3∙10^(−5) ? P^h =4,89+log(([0,02])/([0,015])) P^(ka) =(−5+0,11) P^h =4,89+log1,3 P^(ka) =5−11 P^h =4,89+0,11 P^(ka) =4,49 P^h =5 3°Passo P^h =P^h +log(([Basica])/([Acida])) P^h =4,89+log(([0,02])/([0,025])) P^h =4,89+log0,8 P^h =4,89+0,096 P^h =4,986≈5

$$ \\ $$$$\mathrm{2}.\:\mathrm{Uma}\:\:\mathrm{soluc}\overset{ } {\mathrm{a}o}\:\:\mathrm{tampao}\:\mathrm{foi}\:\mathrm{preparada}\:\mathrm{misturarando}\:\mathrm{200}\boldsymbol{\mathrm{m}{l}}\:\mathrm{NH}_{\mathrm{3}\:} \:\:\mathrm{0},\mathrm{6}\:{moles} \\ $$$${e}\:\mathrm{300}{m}\boldsymbol{{l}}\:{de}\:{uma}\:{solucao}\:{de}\:\mathrm{NH}_{\mathrm{3}} \mathrm{C}{l}\:\:\:\mathrm{0},\mathrm{2}\:{moles}\:{P}^{\mathrm{kh}} =\mathrm{9},\mathrm{24}\:{e}\:\mathrm{log}\:\mathrm{2}=\mathrm{0},\mathrm{3} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Q}{ual}\:{e}\:{o}\:{P}^{\mathrm{h}} \:\mathrm{desta}\:\mathrm{solucao}\:\mathrm{tamp}\overset{ } {\mathrm{a}o}\:\mathrm{supondo}-{se}\:{um}\:{volume}\:{de}\:\mathrm{500}\boldsymbol{{ml}}? \\ $$$$\left.\mathrm{b}\right)\mathrm{Q}{ual}\:{sera}\:{o}\:{P}^{\mathrm{h}} \:{depois}\:{de}\:{ser}\:{adicionado}\:\mathrm{0},\mathrm{2}\:{molar}\:{de}\:{ion}\:\left[\mathrm{H}^{−} \right] \\ $$$$\mathrm{Dados} \\ $$$$\mathrm{n}\left(\mathrm{CH}_{\mathrm{3}} -\mathrm{CH}_{\mathrm{2}} \mathrm{COOH}\right)=\mathrm{0},\mathrm{02}\boldsymbol{{mok}}/\boldsymbol{{l}}=\mathrm{0},\mathrm{02M} \\ $$$${K}\mathrm{a}=\mathrm{1},\mathrm{3}\centerdot\mathrm{10}^{−\mathrm{5}} \\ $$$${n}\left(\mathrm{CH}_{\mathrm{3}} −\mathrm{CH}_{\mathrm{2}} -\mathrm{COONa}\right)=\mathrm{0},\mathrm{015}{mol}/\boldsymbol{{l}}=\mathrm{0},\mathrm{015M} \\ $$$$\mathrm{V}=\mathrm{1}\boldsymbol{{l}} \\ $$$$\mathrm{log}\:\mathrm{2}=\mathrm{0},\mathrm{3} \\ $$$$\mathrm{log}\:\mathrm{0},\mathrm{11} \\ $$$$\mathrm{P}^{\mathrm{h}} =? \\ $$$$\:\:\mathrm{1}°\boldsymbol{\mathrm{Passo}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}°\mathrm{Passo} \\ $$$$\mathrm{P}^{\mathrm{ka}} =−\mathrm{log}\:\mathrm{Ka}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{P}^{\mathrm{ka}} +\mathrm{log}\frac{\left[\mathrm{Base}\right]}{\left[\mathrm{Acido}\right]} \\ $$$$\mathrm{P}^{\mathrm{ka}} =−\mathrm{log}\:\mathrm{1},\mathrm{3}\centerdot\mathrm{10}^{−\mathrm{5}} \:\:\:\:\:?\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log}\frac{\left[\mathrm{0},\mathrm{02}\right]}{\left[\mathrm{0},\mathrm{015}\right]} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\left(−\mathrm{5}+\mathrm{0},\mathrm{11}\right)\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log1},\mathrm{3} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\mathrm{5}−\mathrm{11}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{0},\mathrm{11} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\mathrm{4},\mathrm{49}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{5} \\ $$$$\mathrm{3}°\boldsymbol{\mathrm{Passo}} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{P}^{\mathrm{h}} +\mathrm{log}\frac{\left[\mathrm{Basica}\right]}{\left[\mathrm{Acida}\right]} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log}\frac{\left[\mathrm{0},\mathrm{02}\right]}{\left[\mathrm{0},\mathrm{025}\right]} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log0},\mathrm{8} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{0},\mathrm{096} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{986}\approx\mathrm{5} \\ $$

Question Number 170738    Answers: 0   Comments: 9

Question Number 170737    Answers: 1   Comments: 0

(3/(sin^2 40°)) − (1/(cos^2 40°)) + 64 sin^2 40° = ?

$$\frac{\mathrm{3}}{\mathrm{sin}^{\mathrm{2}} \mathrm{40}°}\:−\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{40}°}\:+\:\mathrm{64}\:\mathrm{sin}^{\mathrm{2}} \mathrm{40}°\:\:=\:\:? \\ $$

Question Number 170729    Answers: 3   Comments: 2

Question Number 170726    Answers: 0   Comments: 0

Question Number 170725    Answers: 1   Comments: 0

Question Number 170722    Answers: 2   Comments: 0

Question Number 170721    Answers: 0   Comments: 1

Which is the true magnefication′s formula for the concave mirror? m=(f/(p−f))? m=−(f/(p−f))?

$${Which}\:{is}\:{the}\:{true}\:{magnefication}'{s}\: \\ $$$${formula}\:{for}\:{the}\:{concave}\:{mirror}? \\ $$$${m}=\frac{{f}}{{p}−{f}}? \\ $$$${m}=−\frac{{f}}{{p}−{f}}? \\ $$

Question Number 170720    Answers: 0   Comments: 0

Prove the magnefication for concave mirror. m=(f/(∣f−p∣))

$${Prove}\:{the}\:{magnefication}\:{for}\:{concave} \\ $$$${mirror}. \\ $$$${m}=\frac{{f}}{\mid{f}−{p}\mid} \\ $$

Question Number 170706    Answers: 1   Comments: 2

Between 10:PM and 7:45AM the water level in a swimming pool decreased by 13÷16 (13/16)inch. Assuming that the water level decreased at a constant rate, how much did it drop each hour? The water level decreased by inch each hour.

$$ \\ $$Between 10:PM and 7:45AM the water level in a swimming pool decreased by 13÷16 (13/16)inch. Assuming that the water level decreased at a constant rate, how much did it drop each hour? The water level decreased by inch each hour.

Question Number 170701    Answers: 0   Comments: 2

In the first week of July, a record 1060 people went to the local swimming pool. In the second week, 105 fewer people went to the pool. In the third week, 135 more people went to the pool than in the second week. In the fourth week, 136 fewer people went to the pool than in the third week.What is the percent change in the number of people who went to the pool between the first and last weeks?

$$ \\ $$In the first week of July, a record 1060 people went to the local swimming pool. In the second week, 105 fewer people went to the pool. In the third week, 135 more people went to the pool than in the second week. In the fourth week, 136 fewer people went to the pool than in the third week.What is the percent change in the number of people who went to the pool between the first and last weeks?

Question Number 170677    Answers: 0   Comments: 0

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