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AllQuestion and Answers: Page 71

Question Number 206421    Answers: 1   Comments: 0

If tanpθ = ptanθ then prove that ((sin^2 pθ)/(sin^2 θ)) = (p^2 /(1 + (p^2 − 1)sin^2 θ)) .

$$\mathrm{If}\:\mathrm{tan}{p}\theta\:=\:{p}\mathrm{tan}\theta\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{sin}^{\mathrm{2}} {p}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\frac{{p}^{\mathrm{2}} }{\mathrm{1}\:+\:\left({p}^{\mathrm{2}} \:−\:\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \theta}\:.\: \\ $$

Question Number 206399    Answers: 2   Comments: 1

Question Number 206396    Answers: 3   Comments: 0

Question Number 206394    Answers: 0   Comments: 1

Question Number 206393    Answers: 1   Comments: 0

find S=1+Σ_ℓ (((−)^ℓ )/ℓ)((1/ℓ)−(1/(ℓ+1))) , ℓ∈[1,∞) 1+Σ_ℓ (((−)^ℓ )/ℓ)((1/ℓ)−(1/(ℓ+1))) 1−(1−(1/2))+(1/2)((1/2)−(1/3))−(1/3)((1/3)−(1/4))+(1/4)((1/4)−(1/5))−......

$$\mathrm{find}\:\mathrm{S}=\mathrm{1}+\underset{\ell} {\sum}\:\frac{\left(−\right)^{\ell} }{\ell}\left(\frac{\mathrm{1}}{\ell}−\frac{\mathrm{1}}{\ell+\mathrm{1}}\right)\:,\:\ell\in\left[\mathrm{1},\infty\right) \\ $$$$\mathrm{1}+\underset{\ell} {\sum}\:\frac{\left(−\right)^{\ell} }{\ell}\left(\frac{\mathrm{1}}{\ell}−\frac{\mathrm{1}}{\ell+\mathrm{1}}\right) \\ $$$$\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}\right)−...... \\ $$

Question Number 206391    Answers: 2   Comments: 0

Find: ∫_(−3) ^( −2) (∣x∣ + ∣x − 4∣) dx = ?

$$\mathrm{Find}: \\ $$$$\int_{−\mathrm{3}} ^{\:−\mathrm{2}} \:\left(\mid\mathrm{x}\mid\:+\:\mid\mathrm{x}\:−\:\mathrm{4}\mid\right)\:\mathrm{dx}\:=\:? \\ $$

Question Number 206365    Answers: 2   Comments: 4

Number series: a_3 = 2a + b − 6 a_9 = a + b + 5 a_(15) = 3a + b − 7 Find: a = ?

$$\mathrm{Number}\:\mathrm{series}: \\ $$$$\mathrm{a}_{\mathrm{3}} \:=\:\mathrm{2a}\:+\:\mathrm{b}\:−\:\mathrm{6} \\ $$$$\mathrm{a}_{\mathrm{9}} \:=\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{5} \\ $$$$\mathrm{a}_{\mathrm{15}} \:=\:\mathrm{3a}\:+\:\mathrm{b}\:−\:\mathrm{7} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}\:=\:? \\ $$

Question Number 206364    Answers: 2   Comments: 0

Question Number 206363    Answers: 0   Comments: 2

If 0<a<1 Compare: (1/(a−1)) , (a/(a−1)) , (1/(1−a)) , (a/(1−a)) , (a/(2a))

$$\mathrm{If}\:\:\:\mathrm{0}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{Compare}: \\ $$$$\frac{\mathrm{1}}{\mathrm{a}−\mathrm{1}}\:\:,\:\:\frac{\mathrm{a}}{\mathrm{a}−\mathrm{1}}\:\:,\:\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{a}}\:\:,\:\:\frac{\mathrm{a}}{\mathrm{1}−\mathrm{a}}\:\:,\:\:\frac{\mathrm{a}}{\mathrm{2a}} \\ $$

Question Number 206357    Answers: 2   Comments: 0

Find: 1 + cos444° − cos84° + cot45° = ?

$$\mathrm{Find}: \\ $$$$\mathrm{1}\:+\:\mathrm{cos444}°\:−\:\mathrm{cos84}°\:+\:\mathrm{cot45}°\:=\:? \\ $$

Question Number 206355    Answers: 2   Comments: 0

if the sum of three positive real numbers is equal to their product, prove that at least one of the numbers is larger than 1.7.

$${if}\:{the}\:{sum}\:{of}\:{three}\:{positive}\:{real}\: \\ $$$${numbers}\:{is}\:{equal}\:{to}\:{their}\:{product}, \\ $$$${prove}\:{that}\:{at}\:{least}\:{one}\:{of}\:{the}\: \\ $$$${numbers}\:{is}\:{larger}\:{than}\:\mathrm{1}.\mathrm{7}. \\ $$

Question Number 206353    Answers: 0   Comments: 6

Question Number 206351    Answers: 0   Comments: 1

expression of the sequence (a_n ) defined by { ((a_0 >0 , a_1 >0)),((a_(n+2) =((2(−1)^n )/(n+2))−((2(−1)^n (2n+3))/(n+2))a_(n+1) +((n+1)/(n+2))a_n )) :}

$${expression}\:{of}\:{the}\:{sequence}\:\left({a}_{{n}} \right)\:{defined} \\ $$$${by}\: \\ $$$$\begin{cases}{{a}_{\mathrm{0}} >\mathrm{0}\:,\:{a}_{\mathrm{1}} >\mathrm{0}}\\{{a}_{{n}+\mathrm{2}} =\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}}−\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}+\mathrm{3}\right)}{{n}+\mathrm{2}}{a}_{{n}+\mathrm{1}} +\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}{a}_{{n}} }\end{cases} \\ $$

Question Number 206340    Answers: 2   Comments: 0

∫_0 ^( 1) (( ln(1−x )ln(1+x ))/x)dx = Σ_(n=1) ^∞ Ω_n find : Σ_(n=1) ^∞ n Ω_n = ?

$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\:{ln}\left(\mathrm{1}−{x}\:\right){ln}\left(\mathrm{1}+{x}\:\right)}{{x}}{dx}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\Omega_{{n}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:{find}\::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:{n}\:\Omega_{{n}} \:=\:? \\ $$

Question Number 206339    Answers: 1   Comments: 0

E ⊆ Y ⊆ ( X , d )∣_(metric space) prove E is open in Y if and only if ∃ G (open set ) in X such that E = G ∩ Y .... (mathematical analysis (I))

$$ \\ $$$$\:\:\:\:\:{E}\:\subseteq\:{Y}\:\subseteq\:\left(\:{X}\:,\:{d}\:\right)\mid_{{metric}\:{space}} \\ $$$$\:\:\:\:{prove}\:\:{E}\:{is}\:{open}\:{in}\:{Y}\:{if}\:{and}\:\:{only}\:{if} \\ $$$$\:\:\:\:\:\:\:\exists\:{G}\:\left({open}\:{set}\:\right)\:{in}\:{X}\:\:{such}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:{E}\:=\:{G}\:\cap\:{Y}\:\:\:....\:\left({mathematical}\:{analysis}\:\left({I}\right)\right) \\ $$

Question Number 206338    Answers: 1   Comments: 0

Question Number 206367    Answers: 4   Comments: 2

Find: (1/6) + (1/(24)) + (1/(60)) + ... + ... (1/(720)) = ?

$$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{24}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{60}}\:\:+\:...\:+\:...\:\:\frac{\mathrm{1}}{\mathrm{720}}\:=\:? \\ $$

Question Number 206332    Answers: 1   Comments: 0

If log(a + b + c) = loga + logb + logc then prove that log(((2a)/(1 − a^2 )) + ((2b)/(1 − b^2 )) + ((2c)/(1 − c^2 ))) = log(((2a)/(1 − a^2 ))) + log(((2b)/(1 − b^2 ))) + log(((2c)/(1 − c^2 ))).

$$\mathrm{If}\:\mathrm{log}\left({a}\:+\:{b}\:+\:{c}\right)\:=\:\mathrm{log}{a}\:+\:\mathrm{log}{b}\:+\:\mathrm{log}{c}\: \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\right)\:=\: \\ $$$$\mathrm{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\right)\:+\:\mathrm{log}\left(\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\right)\:+\:\mathrm{log}\left(\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\right). \\ $$

Question Number 206328    Answers: 2   Comments: 0

f(x)= ⌊ (( 1+ x +x^2 )/(x+ x^( 2) )) ⌋ is given. ⇒ { (( D_f = ? (domain ))),(( R_( f) = ?( range))) :}

$$ \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\:\lfloor\:\frac{\:\mathrm{1}+\:{x}\:+{x}^{\mathrm{2}} }{{x}+\:{x}^{\:\mathrm{2}} }\:\rfloor\:{is}\:{given}. \\ $$$$\:\:\:\:\:\:\:\Rightarrow\begin{cases}{\:\:{D}_{{f}} \:=\:?\:\left({domain}\:\right)}\\{\:\:\:{R}_{\:{f}} \:=\:?\left(\:{range}\right)}\end{cases} \\ $$$$ \\ $$

Question Number 206323    Answers: 0   Comments: 0

Question Number 206322    Answers: 1   Comments: 1

If abc = 1 then prove that (1/(1 + a + b^(−1) )) + (1/(1 + b + c^(−1) )) + (1/(1 + c + a^(−1) )) = 1

$$\mathrm{If}\:{abc}\:=\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{a}\:+\:{b}^{−\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{b}\:+\:{c}^{−\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{c}\:+\:{a}^{−\mathrm{1}} }\:=\:\mathrm{1} \\ $$

Question Number 206321    Answers: 2   Comments: 0

Question Number 206319    Answers: 0   Comments: 0

Question Number 206318    Answers: 0   Comments: 0

∫_0 ^1 ∫_0 ^1 ((xln(1+y)ln(1−x))/(1+x^2 y))dxdy

$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\mathrm{1}+{y}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} {y}}{dxdy} \\ $$

Question Number 206362    Answers: 2   Comments: 0

sin(π/7) × sin((2π)/7) × sin((3π)/7) = ?

$$\mathrm{sin}\frac{\pi}{\mathrm{7}}\:×\:\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\:×\:\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{7}}\:=\:? \\ $$

Question Number 206311    Answers: 0   Comments: 0

Hi everyone... This is not a question per say (sorry ′bout that). It′s just that for past 4 years, to type math messages to my friends (text messages) I′ve been painfully using the two most famous android ones (matheditor and dxmath). And I was thinking, I realised how typing on this app on android (matheditor) is so much more practical. So if any one here knows how to code, I would love to see an android−keyboard adaptation (only with the unicode charters on this keyboard that is... and with the extra ^(1234567890−+×()inx) _(1234567890inx) unicode ones that might come in handy once we can′t use proper indexation) I would personnaly be ready to pay for it. And I would be intersted to know if anyone of you would be too. I don′t think it would be very hard to do... I am ready to help if I can. I don′t know how to make an android app, but being a mathematician and a physicist, I′m sure I can make myself usefull...

$$\mathrm{Hi}\:\mathrm{everyone}... \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{question}\:\mathrm{per}\:\mathrm{say}\:\left(\mathrm{sorry}\:'\mathrm{bout}\right. \\ $$$$\left.\mathrm{that}\right).\:\mathrm{It}'\mathrm{s}\:\mathrm{just}\:\mathrm{that}\:\mathrm{for}\:\mathrm{past}\:\mathrm{4}\:\mathrm{years},\:\mathrm{to}\:\mathrm{type} \\ $$$$\mathrm{math}\:\mathrm{messages}\:\mathrm{to}\:\mathrm{my}\:\mathrm{friends}\:\left(\mathrm{text}\:\mathrm{messages}\right) \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{painfully}\:\mathrm{using}\:\mathrm{the}\:\mathrm{two}\:\mathrm{most}\:\mathrm{famous} \\ $$$$\mathrm{android}\:\mathrm{ones}\:\left({matheditor}\:\mathrm{and}\:{dxmath}\right). \\ $$$$\mathrm{And}\:\mathrm{I}\:\mathrm{was}\:\mathrm{thinking},\:\mathrm{I}\:\mathrm{realised}\:\mathrm{how}\:\mathrm{typing} \\ $$$$\mathrm{on}\:\mathrm{this}\:\mathrm{app}\:\mathrm{on}\:\mathrm{android}\:\left({matheditor}\right)\:\mathrm{is}\:\mathrm{so} \\ $$$$\mathrm{much}\:\mathrm{more}\:\mathrm{practical}. \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{if}\:\mathrm{any}\:\mathrm{one}\:\mathrm{here}\:\mathrm{knows}\:\mathrm{how}\:\mathrm{to}\:\mathrm{code},\:\mathrm{I}\:\mathrm{would} \\ $$$$\mathrm{love}\:\mathrm{to}\:\mathrm{see}\:\mathrm{an}\:\mathrm{android}−\mathrm{keyboard}\:\mathrm{adaptation} \\ $$$$\left(\mathrm{only}\:\mathrm{with}\:\mathrm{the}\:\mathrm{unicode}\:\mathrm{charters}\:\mathrm{on}\:\mathrm{this}\right. \\ $$$$\mathrm{keyboard}\:\mathrm{that}\:\mathrm{is}...\:\mathrm{and}\:\mathrm{with}\:\mathrm{the}\:\mathrm{extra} \\ $$$$\:^{\mathrm{1234567890}−+×\left(\right)\mathrm{inx}} \:_{\mathrm{1234567890inx}} \:\mathrm{unicode}\:\mathrm{ones} \\ $$$$\mathrm{that}\:\mathrm{might}\:\mathrm{come}\:\mathrm{in}\:\mathrm{handy}\:\mathrm{once}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{use} \\ $$$$\left.\mathrm{proper}\:\mathrm{indexation}\right) \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{would}\:\mathrm{personnaly}\:\mathrm{be}\:\mathrm{ready}\:\mathrm{to}\:\mathrm{pay}\:\mathrm{for}\:\mathrm{it}. \\ $$$$\mathrm{And}\:\mathrm{I}\:\mathrm{would}\:\mathrm{be}\:\mathrm{intersted}\:\mathrm{to}\:\mathrm{know}\:\mathrm{if}\:\mathrm{anyone} \\ $$$$\mathrm{of}\:\mathrm{you}\:\mathrm{would}\:\mathrm{be}\:\mathrm{too}. \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{it}\:\mathrm{would}\:\mathrm{be}\:\mathrm{very}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{do}... \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{ready}\:\mathrm{to}\:\mathrm{help}\:\mathrm{if}\:\mathrm{I}\:\mathrm{can}.\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{how} \\ $$$$\mathrm{to}\:\mathrm{make}\:\mathrm{an}\:\mathrm{android}\:\mathrm{app},\:\mathrm{but}\:\mathrm{being}\:\mathrm{a} \\ $$$$\mathrm{mathematician}\:\mathrm{and}\:\mathrm{a}\:\mathrm{physicist},\:\mathrm{I}'\mathrm{m}\:\mathrm{sure} \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{make}\:\mathrm{myself}\:\mathrm{usefull}... \\ $$

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