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Question Number 27073 Answers: 1 Comments: 0

∫ln x×cos 2ln xdx

$$\int\mathrm{ln}\:{x}×\mathrm{cos}\:\mathrm{2ln}\:{xdx} \\ $$

Question Number 27075 Answers: 0 Comments: 0

Laws of Motion question at ibb.co/cqq1NG I tried uploading here but it doesn′t get uploaded.

$$\mathrm{Laws}\:\mathrm{of}\:\mathrm{Motion}\:\mathrm{question}\:\mathrm{at} \\ $$$$\mathrm{ibb}.\mathrm{co}/\mathrm{cqq1NG} \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{uploading}\:\mathrm{here}\:\mathrm{but}\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\mathrm{get}\:\mathrm{uploaded}. \\ $$

Question Number 27074 Answers: 2 Comments: 0

Question Number 27065 Answers: 0 Comments: 0

Question Number 27057 Answers: 1 Comments: 3

Try to write new year number (2018)as: (i) Sum of two primes (ii)Sum of three primes (iii)Sum of primes (iv)Sum of as many distinct primes as possible.

$$\mathrm{Try}\:\mathrm{to}\:\mathrm{write}\:\mathrm{new}\:\mathrm{year}\:\mathrm{number} \\ $$$$\left(\mathrm{2018}\right)\mathrm{as}: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{primes} \\ $$$$\left(\mathrm{ii}\right)\mathrm{Sum}\:\mathrm{of}\:\mathrm{three}\:\mathrm{primes} \\ $$$$\left(\mathrm{iii}\right)\mathrm{Sum}\:\mathrm{of}\:\mathrm{primes} \\ $$$$\left(\mathrm{iv}\right)\mathrm{Sum}\:\mathrm{of}\:\mathrm{as}\:\mathrm{many}\:\mathrm{distinct}\:\mathrm{primes}\:\mathrm{as} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{possible}. \\ $$

Question Number 27055 Answers: 1 Comments: 1

Question Number 27050 Answers: 0 Comments: 2

Question Number 27059 Answers: 0 Comments: 2

Question Number 27046 Answers: 0 Comments: 0

Considering y=x^3 +px+q If (dy/dx)∣_(x=α) =0 ⇒ α^2 =−(p/3) if ((d(y/x))/dx)∣_(x=β) =0 ⇒ β^( 3) =(q/2) roots of the cubic eq^n are: x=[−β^( 3) ±(√(β^( 6) −α^6 )) ]^(1/3) −[β^( 3) ±(√(β^( 6) −α^6 )) ]^(1/3) . Why such a connection? If equation is quadratic even_ y=ax^2 +bx+c (dy/dx)∣_(x=α) =0 ⇒ α=−(b/(2a)) ((d(y/x))/dx)∣_(x=β) =0 ⇒ β^( 2) =(c/a) roots of quadratic eq. are: x=𝛂±(√(𝛂^2 −𝛃^( 2) )) why such a connection ?

$${Considering}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{px}}+\boldsymbol{{q}} \\ $$$${If}\:\:\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\Rightarrow\:\:\alpha^{\mathrm{2}} =−\frac{{p}}{\mathrm{3}} \\ $$$${if}\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\:\Rightarrow\:\beta^{\:\mathrm{3}} =\frac{{q}}{\mathrm{2}} \\ $$$${roots}\:{of}\:{the}\:{cubic}\:\:{eq}^{{n}} \:{are}: \\ $$$$\:\:\:\:{x}=\left[−\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}} −\alpha^{\mathrm{6}} }\:\right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left[\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}} −\alpha^{\mathrm{6}} }\:\right]^{\mathrm{1}/\mathrm{3}} \:. \\ $$$$\:{Why}\:{such}\:{a}\:{connection}? \\ $$$${If}\:{equation}\:{is}\:{quadratic}\:{even\_} \\ $$$$\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}} \\ $$$$\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\:\Rightarrow\:\:\alpha=−\frac{{b}}{\mathrm{2}{a}} \\ $$$$\:\:\:\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\Rightarrow\:\beta^{\:\mathrm{2}} =\frac{{c}}{{a}} \\ $$$${roots}\:{of}\:{quadratic}\:{eq}.\:{are}: \\ $$$$\:\:\:\:{x}=\boldsymbol{\alpha}\pm\sqrt{\boldsymbol{\alpha}^{\mathrm{2}} −\boldsymbol{\beta}^{\:\mathrm{2}} }\: \\ $$$${why}\:{such}\:{a}\:{connection}\:?\: \\ $$

Question Number 27044 Answers: 1 Comments: 0