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Question Number 11979    Answers: 1   Comments: 0

If the sum of first p terms, first q terms and first r terms of an AP be x, y and z respectively. Then (x/p)(q−r) + (y/q)(r−p) + (z/r)(p−q) is

$$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:{p}\:\mathrm{terms},\:\mathrm{first}\:\:{q}\:\mathrm{terms}\:\mathrm{and} \\ $$$$\mathrm{first}\:{r}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{be}\:\:{x},\:{y}\:\:\mathrm{and}\:{z}\: \\ $$$$\mathrm{respectively}.\:\mathrm{Then} \\ $$$$\frac{{x}}{{p}}\left({q}−{r}\right)\:+\:\frac{{y}}{{q}}\left({r}−{p}\right)\:+\:\frac{{z}}{{r}}\left({p}−{q}\right)\:\:\mathrm{is} \\ $$

Question Number 11921    Answers: 1   Comments: 0

given that y=Acos5x + Bsin5x, show that (d^2 y/dx^2 )+25y=0

$${given}\:{that}\:{y}={Acos}\mathrm{5}{x}\:+\:{Bsin}\mathrm{5}{x}, \\ $$$${show}\:{that}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{25}{y}=\mathrm{0} \\ $$

Question Number 11920    Answers: 4   Comments: 0

differentiate each function from first principle 1)f (x) = 1+(1/x) 2)f(x) = (1/(2x+3)) 3) f(x)=sin2x 4)f(x)=co2x

$${differentiate}\:{each}\:{function}\:{from}\:{first}\:{principle} \\ $$$$\left.\mathrm{1}\right){f}\:\left({x}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{{x}} \\ $$$$\left.\mathrm{2}\right){f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{3}}\: \\ $$$$\left.\mathrm{3}\right)\:{f}\left({x}\right)={sin}\mathrm{2}{x} \\ $$$$\left.\mathrm{4}\right){f}\left({x}\right)={co}\mathrm{2}{x} \\ $$$$ \\ $$

Question Number 11915    Answers: 2   Comments: 2

Question Number 11914    Answers: 0   Comments: 0

Turevlenebilir bir f fonksiyonu icin f(x+y)=f(x)+f(y)+2xy ve f′(0)=−3 old.gore f′(2)=? czm∵ f(x+y)=f(x)+f(y)+2xy y yi sabit kabul edersek f′(x+y)=f′(x)+2y x=0,y=2 icn f′(2)=f′(0)+2.2 f′(2)=−3+4=1

$${Turevlenebilir}\:{bir}\:{f}\:{fonksiyonu}\:{icin} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+\mathrm{2}{xy}\:{ve}\:{f}'\left(\mathrm{0}\right)=−\mathrm{3} \\ $$$${old}.{gore}\:{f}'\left(\mathrm{2}\right)=? \\ $$$${czm}\because\:\:{f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+\mathrm{2}{xy} \\ $$$${y}\:{yi}\:{sabit}\:{kabul}\:{edersek} \\ $$$${f}'\left({x}+{y}\right)={f}'\left({x}\right)+\mathrm{2}{y} \\ $$$${x}=\mathrm{0},{y}=\mathrm{2}\:{icn} \\ $$$${f}'\left(\mathrm{2}\right)={f}'\left(\mathrm{0}\right)+\mathrm{2}.\mathrm{2} \\ $$$${f}'\left(\mathrm{2}\right)=−\mathrm{3}+\mathrm{4}=\mathrm{1} \\ $$

Question Number 11913    Answers: 0   Comments: 0

∫x^(x^x ) dx

$$\int\mathrm{x}^{\mathrm{x}^{\mathrm{x}} \:\:} \:\mathrm{dx} \\ $$

Question Number 11902    Answers: 1   Comments: 1

Assuming it rained at a constant rate, and the rain fell at angle θ to the ground (see diagram), determine if walking or running causes you to get more/less wet, or of it makes no difference for: 1. θ=90° (downwards) 2. θ<90° (the rain is moving on the same direction as you) 3. θ>90° (the rain is moving/blowing into you)

$$\mathrm{Assuming}\:\mathrm{it}\:\mathrm{rained}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{rate}, \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{rain}\:\mathrm{fell}\:\mathrm{at}\:\mathrm{angle}\:\theta\:\mathrm{to}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\left(\mathrm{see}\:\mathrm{diagram}\right),\:\mathrm{determine}\:\mathrm{if}\:\mathrm{walking}\:\mathrm{or} \\ $$$$\mathrm{running}\:\mathrm{causes}\:\mathrm{you}\:\mathrm{to}\:\mathrm{get}\:\mathrm{more}/\mathrm{less}\:\mathrm{wet}, \\ $$$$\mathrm{or}\:\mathrm{of}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{difference}\:\mathrm{for}: \\ $$$$\: \\ $$$$\mathrm{1}.\:\:\:\theta=\mathrm{90}°\:\:\left(\mathrm{downwards}\right) \\ $$$$\mathrm{2}.\:\theta<\mathrm{90}°\:\:\left(\mathrm{the}\:\mathrm{rain}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{on}\:\mathrm{the}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\mathrm{same}\:\mathrm{direction}\:\mathrm{as}\:\mathrm{you}\right) \\ $$$$\mathrm{3}.\:\theta>\mathrm{90}°\:\:\left(\mathrm{the}\:\mathrm{rain}\:\mathrm{is}\:\mathrm{moving}/\mathrm{blowing}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\mathrm{into}\:\mathrm{you}\right) \\ $$

Question Number 11901    Answers: 0   Comments: 0

((7cos^2 x+sin^2 x−3)/(2cos^2 x−sin^2 x))=? czm∵ ((7cos^2 x+1−cos^2 x−3)/(2cos^2 x−sin^2 x)) ((6cos^2 x−2)/(2cos^2 x−(1−cos^2 x)))=((6cos^2 x−2)/(3cos^2 x−1)) ((2(3cos^2 x−1))/(3cos^2 x−1))=2

$$\frac{\mathrm{7}{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}−\mathrm{3}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}=? \\ $$$${czm}\because\:\:\frac{\mathrm{7}{cos}^{\mathrm{2}} {x}+\mathrm{1}−{cos}^{\mathrm{2}} {x}−\mathrm{3}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}} \\ $$$$\frac{\mathrm{6}{cos}^{\mathrm{2}} {x}−\mathrm{2}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)}=\frac{\mathrm{6}{cos}^{\mathrm{2}} {x}−\mathrm{2}}{\mathrm{3}{cos}^{\mathrm{2}} {x}−\mathrm{1}} \\ $$$$\frac{\mathrm{2}\left(\mathrm{3}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)}{\mathrm{3}{cos}^{\mathrm{2}} {x}−\mathrm{1}}=\mathrm{2} \\ $$

Question Number 11900    Answers: 1   Comments: 0

Calculate. cos(𝛑/7)×cos((4𝛑)/7)×cos((5𝛑)/7).

$$\boldsymbol{\mathrm{Calculate}}. \\ $$$$\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\pi}}{\mathrm{7}}×\boldsymbol{\mathrm{cos}}\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{7}}×\boldsymbol{\mathrm{cos}}\frac{\mathrm{5}\boldsymbol{\pi}}{\mathrm{7}}. \\ $$

Question Number 11899    Answers: 0   Comments: 0

f′(x)={_(3 ;x>2) ^(2x ; x≤2) f(2)=1 ise f(1)+f(3)=? czm∵ f(x)={_(3x+c_2 ; x>2) ^(x^2 +c_1 ;x≤2) f(2)=x^2 +c_1 dir 1=4+c_1 => c_1 =−3 1=2.3+c_2 => c_2 =−5 f(x)={_(3x−5 x>2) ^(x^2 −3 ;x≤2) f(1)=1^2 −3=−2, f(3)=3.3−5=4 f(1)+f(3)=−2+4=2

$${f}'\left({x}\right)=\left\{_{\mathrm{3}\:\:\:\:\:;{x}>\mathrm{2}} ^{\mathrm{2}{x}\:\:;\:{x}\leqslant\mathrm{2}} \right. \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{1}\:{ise}\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{3}\right)=? \\ $$$${czm}\because\:{f}\left({x}\right)=\left\{_{\mathrm{3}{x}+{c}_{\mathrm{2}} \:;\:{x}>\mathrm{2}} ^{{x}^{\mathrm{2}} +{c}_{\mathrm{1}} \:;{x}\leqslant\mathrm{2}} \right. \\ $$$${f}\left(\mathrm{2}\right)={x}^{\mathrm{2}} +{c}_{\mathrm{1}} \:{dir} \\ $$$$\mathrm{1}=\mathrm{4}+{c}_{\mathrm{1}} \:=>\:{c}_{\mathrm{1}} =−\mathrm{3} \\ $$$$\mathrm{1}=\mathrm{2}.\mathrm{3}+{c}_{\mathrm{2}} \:=>\:{c}_{\mathrm{2}} =−\mathrm{5} \\ $$$${f}\left({x}\right)=\left\{_{\mathrm{3}{x}−\mathrm{5}\:\:\:\:{x}>\mathrm{2}} ^{{x}^{\mathrm{2}} −\mathrm{3}\:\:\:;{x}\leqslant\mathrm{2}} \right. \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}^{\mathrm{2}} −\mathrm{3}=−\mathrm{2},\:{f}\left(\mathrm{3}\right)=\mathrm{3}.\mathrm{3}−\mathrm{5}=\mathrm{4} \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{3}\right)=−\mathrm{2}+\mathrm{4}=\mathrm{2} \\ $$$$ \\ $$

Question Number 11889    Answers: 1   Comments: 0

The number of ways in which 8 different flowers can be strung to form a garland so that 4 particular flowers are never separated is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{in}\:\mathrm{which}\:\mathrm{8}\: \\ $$$$\mathrm{different}\:\mathrm{flowers}\:\mathrm{can}\:\mathrm{be}\:\mathrm{strung}\:\mathrm{to} \\ $$$$\mathrm{form}\:\mathrm{a}\:\mathrm{garland}\:\mathrm{so}\:\mathrm{that}\:\mathrm{4}\:\mathrm{particular} \\ $$$$\mathrm{flowers}\:\mathrm{are}\:\mathrm{never}\:\mathrm{separated}\:\mathrm{is} \\ $$

Question Number 11886    Answers: 0   Comments: 0

∫x^x^x dx

$$\int\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:\mathrm{dx} \\ $$

Question Number 11887    Answers: 1   Comments: 0

(dy/dt) +3t^2 y = t^(2 ) , y(0) = 1 y(t) = ?

$$\frac{{dy}}{{dt}}\:+\mathrm{3}{t}^{\mathrm{2}} {y}\:=\:{t}^{\mathrm{2}\:} \:\:\:\:,\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$$${y}\left({t}\right)\:=\:? \\ $$

Question Number 11898    Answers: 1   Comments: 0

(a) You are listening to your favourite song on a CD. You note that the sound wave has a pleasant frequency of 12 Hertz. (i) What is the velovity of the sound wave (ii) What wavelenght are the wave moving at (iii) What is the period (b) 60 complete waves pass a particular point in 4 secs, if the distance between 3 successive troughs of the water is 15m. Calculate the speed of the wave.

$$\left(\mathrm{a}\right) \\ $$$$\mathrm{You}\:\mathrm{are}\:\mathrm{listening}\:\mathrm{to}\:\mathrm{your}\:\mathrm{favourite}\:\mathrm{song}\:\mathrm{on}\:\mathrm{a}\:\mathrm{CD}.\:\mathrm{You}\:\mathrm{note}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sound} \\ $$$$\mathrm{wave}\:\mathrm{has}\:\mathrm{a}\:\mathrm{pleasant}\:\mathrm{frequency}\:\mathrm{of}\:\mathrm{12}\:\mathrm{Hertz}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{velovity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sound}\:\mathrm{wave} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{What}\:\mathrm{wavelenght}\:\mathrm{are}\:\mathrm{the}\:\mathrm{wave}\:\mathrm{moving}\:\mathrm{at} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{period} \\ $$$$\left(\mathrm{b}\right) \\ $$$$\mathrm{60}\:\mathrm{complete}\:\mathrm{waves}\:\mathrm{pass}\:\mathrm{a}\:\mathrm{particular}\:\mathrm{point}\:\mathrm{in}\:\mathrm{4}\:\mathrm{secs},\:\mathrm{if}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{3}\:\mathrm{successive}\:\mathrm{troughs}\:\mathrm{of}\:\mathrm{the}\:\mathrm{water}\:\mathrm{is}\:\mathrm{15m}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wave}. \\ $$

Question Number 11883    Answers: 1   Comments: 0

Draw the structural formula of the compound 2,2,7 - trimethyl - 4 - (1 - methylpropyl) nonane

$$\mathrm{Draw}\:\mathrm{the}\:\mathrm{structural}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{the}\:\mathrm{compound} \\ $$$$\mathrm{2},\mathrm{2},\mathrm{7}\:-\:\mathrm{trimethyl}\:-\:\mathrm{4}\:-\:\left(\mathrm{1}\:-\:\mathrm{methylpropyl}\right)\:\mathrm{nonane} \\ $$

Question Number 11880    Answers: 0   Comments: 0

S_(ABCD) =3+2(√2) ∠BAO=∠MAO=22,5° ∠BCM=∠DCM S_(AOB) =?

$$\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{ABCD}}} =\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\angle\boldsymbol{\mathrm{BAO}}=\angle\boldsymbol{\mathrm{MAO}}=\mathrm{22},\mathrm{5}° \\ $$$$\angle\boldsymbol{\mathrm{BCM}}=\angle\boldsymbol{\mathrm{DCM}} \\ $$$$\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{AOB}}} =? \\ $$

Question Number 11869    Answers: 0   Comments: 0

Why the expansion of (a+b)^(−n) follows newton′s expansion rule?

$${Why}\:{the}\:{expansion}\:{of}\:\:\left({a}+{b}\right)^{−{n}} \:{follows} \\ $$$${newton}'{s}\:{expansion}\:{rule}? \\ $$

Question Number 11868    Answers: 1   Comments: 0

Question Number 11867    Answers: 0   Comments: 1

Question Number 11865    Answers: 1   Comments: 0

∫_2 ^4 (√(16−x^2 ))dx/x^4 =

$$\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }{dx}/{x}^{\mathrm{4}} = \\ $$

Question Number 11864    Answers: 2   Comments: 0

∫_7 ^(10) x^2 dx/x^2 −3x+2=

$$\underset{\mathrm{7}} {\overset{\mathrm{10}} {\int}}{x}^{\mathrm{2}} {dx}/{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}= \\ $$

Question Number 11862    Answers: 2   Comments: 0

Lesson1. AM−GM ′ s inequality (Cauchy) form : ((a_1 +a_2 +...+a_n )/n) ≥ ((a_1 a_2 ...a_n ))^(1/n) where a_1 ,a_2 ,....,a_n >0 Equal at a_1 =a_2 =.....=a_n e.g. 1. Given a,b,c>0, prove that (a+b)(b+c)(c+a)≥8abc Solu. by AM−GM a+b ≥ 2(√(ab)) (1) b+c ≥ 2(√(bc)) (2) c+a ≥ 2(√(ca)) (3) (1)×(2)×(3) ⇒ (a+b)(b+c)(c+a)≥8(√(a^2 b^2 c^2 ))=8abc Now practice. . Given a,b,c>0 prove that 1. a^2 +b^2 +c^2 ≥ab+bc+ca 2. (a+(1/b))(b+(1/c))(c+(1/a))≥8 3. 4(a^3 +b^3 )≥(a+b)^3 4. 9(a^3 +b^3 +c^3 )≥(a+b+c)^3 let′s try, I will post my solution for which one that you can′t do ;)

$$\boldsymbol{{Lesson}}\mathrm{1}.\:\boldsymbol{\mathrm{AM}}−\boldsymbol{\mathrm{GM}}\:'\:\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{inequality}}\:\left(\boldsymbol{\mathrm{Cauchy}}\right) \\ $$$$\boldsymbol{\mathrm{form}}\::\:\frac{\boldsymbol{{a}}_{\mathrm{1}} +\boldsymbol{{a}}_{\mathrm{2}} +...+\boldsymbol{{a}}_{\boldsymbol{{n}}} }{\boldsymbol{{n}}}\:\geqslant\:\sqrt[{\boldsymbol{{n}}}]{\boldsymbol{{a}}_{\mathrm{1}} \boldsymbol{{a}}_{\mathrm{2}} ...\boldsymbol{{a}}_{\boldsymbol{{n}}} } \\ $$$$\boldsymbol{{where}}\:\boldsymbol{{a}}_{\mathrm{1}} ,\boldsymbol{{a}}_{\mathrm{2}} ,....,\boldsymbol{{a}}_{\boldsymbol{{n}}} >\mathrm{0} \\ $$$$\boldsymbol{{E}\mathrm{qual}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{{a}}_{\mathrm{1}} =\boldsymbol{{a}}_{\mathrm{2}} =.....=\boldsymbol{{a}}_{\boldsymbol{{n}}} \\ $$$$\boldsymbol{{e}}.\boldsymbol{{g}}.\:\mathrm{1}.\:{Given}\:{a},{b},{c}>\mathrm{0},\:{prove}\:{that}\: \\ $$$$\:\:\:\:\:\:\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)\geqslant\mathrm{8}{abc} \\ $$$$\boldsymbol{{Solu}}.\:{by}\:{AM}−{GM} \\ $$$$\:\:\:\:\:\:\:\:{a}+{b}\:\geqslant\:\mathrm{2}\sqrt{{ab}}\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:{b}+{c}\:\geqslant\:\mathrm{2}\sqrt{{bc}}\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:{c}+{a}\:\geqslant\:\mathrm{2}\sqrt{{ca}}\:\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\:\left(\mathrm{1}\right)×\left(\mathrm{2}\right)×\left(\mathrm{3}\right)\:\Rightarrow\:\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)\geqslant\mathrm{8}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }=\mathrm{8}{abc} \\ $$$$\boldsymbol{{Now}}\:\boldsymbol{{practice}}. \\ $$$$\:.\:{Given}\:{a},{b},{c}>\mathrm{0}\:{prove}\:{that} \\ $$$$\:\:\:\:\:\mathrm{1}.\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ab}+{bc}+{ca} \\ $$$$\:\:\:\:\:\mathrm{2}.\:\left({a}+\frac{\mathrm{1}}{{b}}\right)\left({b}+\frac{\mathrm{1}}{{c}}\right)\left({c}+\frac{\mathrm{1}}{{a}}\right)\geqslant\mathrm{8} \\ $$$$\:\:\:\:\:\mathrm{3}.\:\mathrm{4}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)\geqslant\left({a}+{b}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\mathrm{4}.\:\:\mathrm{9}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)\geqslant\left({a}+{b}+{c}\right)^{\mathrm{3}} \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{try},\:\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{which}\:\mathrm{one} \\ $$$$\left.\mathrm{that}\:\mathrm{you}\:\mathrm{can}'\mathrm{t}\:\mathrm{do}\:;\right) \\ $$

Question Number 11855    Answers: 1   Comments: 0

Ten men are present at a club. In how many ways can four be chosen to play bridge if two men refuse to sit at the same table.

$$\mathrm{Ten}\:\mathrm{men}\:\mathrm{are}\:\mathrm{present}\:\mathrm{at}\:\mathrm{a}\:\mathrm{club}.\:\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{four}\:\mathrm{be}\:\mathrm{chosen}\:\mathrm{to} \\ $$$$\mathrm{play}\:\mathrm{bridge}\:\mathrm{if}\:\mathrm{two}\:\mathrm{men}\:\mathrm{refuse}\:\mathrm{to}\:\mathrm{sit}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{table}. \\ $$

Question Number 11854    Answers: 1   Comments: 0

Solve by mathematical induction that 1 + (1/(1 + 2)) + (1/(1 + 2 + 3)) + ... + (1/(1 + 2 + 3 + ... + n)) = ((2n)/(n + 1))

$$\mathrm{Solve}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that} \\ $$$$\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}}\:+\:...\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:...\:+\:\mathrm{n}}\:=\:\frac{\mathrm{2n}}{\mathrm{n}\:+\:\mathrm{1}} \\ $$

Question Number 11844    Answers: 2   Comments: 0

cotα+cosecα=k then find cosecα−cotα and also find cotα

$$\mathrm{cot}\alpha+\mathrm{cosec}\alpha={k}\:{then}\:{find}\:\mathrm{cosec}\alpha−{cot}\alpha\:{and}\:{also}\:{find}\:{cot}\alpha \\ $$

Question Number 11843    Answers: 1   Comments: 0

Differentiate, ln(cosx) from the first principle.

$$\mathrm{Differentiate},\:\:\mathrm{ln}\left(\mathrm{cosx}\right)\:\:\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{principle}. \\ $$

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