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Question Number 26110    Answers: 0   Comments: 0

let s give n from N find the value of ∫_0 ^π ((sin(nx))/(sinx)) dx .

$${let}\:{s}\:{give}\:{n}\:{from}\:{N}\:\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sin}\left({nx}\right)}{{sinx}}\:{dx}\:. \\ $$

Question Number 26036    Answers: 1   Comments: 0

Find the area of a square if the sum of the diagonals is 100 cm.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{if}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{diagonals}\:\mathrm{is}\:\mathrm{100}\:\mathrm{cm}. \\ $$

Question Number 26031    Answers: 0   Comments: 0

calculate the number of protons which would have a charge of one coulomb charge(Proton Charge=1.6×10^(−19) C).

$${calculate}\:{the}\:{number}\:{of}\:{protons}\:{which}\:{would}\:{have}\:{a}\:{charge}\:{of}\:{one}\:{coulomb}\:{charge}\left({Proton}\:{Charge}=\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} {C}\right). \\ $$

Question Number 26024    Answers: 0   Comments: 0

e give a element from]0.∝[ find the value of ∫_0 ^∞ cos( ax^2 ) and ∫_0 ^∞ sin( ax^2 )dx.

$$\left.{e}\:{give}\:{a}\:{element}\:{from}\right]\mathrm{0}.\propto\left[\:\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left(\:{ax}^{\mathrm{2}} \right)\right. \\ $$$${and}\:\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left(\:{ax}^{\mathrm{2}} \right){dx}. \\ $$

Question Number 26023    Answers: 0   Comments: 0

answer to question25980 key of slution we develop the foction f(x) = sin(px) at fourier serie((f 2π periodic) f(x)= Σ_(n=1) ^(n=∝) a_n sin(nx) and a_n = 2/T ∫_([T]) sin(px)sin(nx)dx (T=2π) a_n =2π^(−1) ∫_0 ^π sin(px)sin(nx)dx−>a_n = (−1)^n sin(pπ).2n π^(−1) (n^2 − p^2 )^(−1) −−>sin(px)= 2 sin(pπ).π^(−1) Σ_(n=1) ^∝ n(−1)^(n−1) (n^2 −p^2 )^(−1) sin(nx) = 2π^(−1) sin(pπ)((1^2 −p^2 )^(−1) sin (x) −2(2^2 −p^2 )^(−1) sin(2x)+...)

$${answer}\:{to}\:{question}\mathrm{25980}\:{key}\:{of}\:{slution}\:{we}\:{develop}\:\:{the} \\ $$$${foction}\:{f}\left({x}\right)\:=\:{sin}\left({px}\right)\:{at}\:{fourier}\:{serie}\left(\left({f}\:\mathrm{2}\pi\:{periodic}\right)\right. \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:{a}_{{n}} {sin}\left({nx}\right)\:{and}\:\:{a}_{{n}} =\:\mathrm{2}/{T}\:\int_{\left[{T}\right]} {sin}\left({px}\right){sin}\left({nx}\right){dx}\:\:\:\left({T}=\mathrm{2}\pi\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${a}_{{n}} =\mathrm{2}\pi^{−\mathrm{1}} \:\int_{\mathrm{0}} ^{\pi} {sin}\left({px}\right){sin}\left({nx}\right){dx}−>{a}_{{n}} =\:\left(−\mathrm{1}\right)^{{n}} \:{sin}\left({p}\pi\right).\mathrm{2}{n}\:\pi^{−\mathrm{1}} \left({n}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$$−−>{sin}\left({px}\right)=\:\mathrm{2}\:{sin}\left({p}\pi\right).\pi^{−\mathrm{1}} \:\sum_{{n}=\mathrm{1}} ^{\propto} \:{n}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \:\:{sin}\left({nx}\right) \\ $$$$=\:\mathrm{2}\pi^{−\mathrm{1}} \:{sin}\left({p}\pi\right)\left(\left(\mathrm{1}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \:{sin}\:\left({x}\right)\:−\mathrm{2}\left(\mathrm{2}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} {sin}\left(\mathrm{2}{x}\right)+...\right) \\ $$

Question Number 26021    Answers: 1   Comments: 1

Question Number 26020    Answers: 0   Comments: 0

answer to 25962...we S=Σ_(n=1) ^∝ 1/_(n^2 (n+1)) and S_n = Σ_(k=1) ^(k=n) 1/_(k^2 (k+1)) we have S= lim_(n−>∝) S_n we decompose the the rational fraction F(X)= 1/_X 2_((X^2 +1)) ....F(X)= a/X +b/X^2 + c/X+1 we find a=−1..b=1..c=1 and F(X)= −1/X + 1/X+1+1/X^2 Σ_(k=1) ^(k=n) 1/_(k^2 (k+1)) = −Σ_(k=1) ^(k=n) 1/k + Σ_(k=1) ^(k=n) 1/_(k+1) + Σ_(k=1) ^(k=n) 1/_k^2 but Σ_(k=1) ^(k=n) 1/k = H_n Σ_(k=1) ^(k=n) 1/_(k+1) = Σ_(k=2) ^(k=n+1) 1/k = H_(n+1) −1 −> S_n = H_(n+1) −H_n −1 + Σ_(k=1) ^(k=n) 1/_k^2 but H_n =ln(n)+s+o_1 (1/n)...H_(n+1) = ln(n+1) +s +o_2 (1/n) −> H_(n+1) − H_n =ln((n+1)n^(−1) ) + o_3 (1/n) but lim _(n−>∝) o_3 (1/n)=0 so lim H_(n+1) − H_n =0 lim_ Σ_(k=1) ^(k=n) 1/_k^2 = Σ_(k=1) ^∝ 1/_k^2 =π^2 /6 finally... Σ_(n=1) ^(n=∝) 1/_(n^2 (n+1)) = π^2 /6 −1.

$${answer}\:{to}\:\mathrm{25962}...{we}\:{S}=\sum_{{n}=\mathrm{1}} ^{\propto} \:\mathrm{1}/_{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \:{and}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)} \\ $$$${we}\:{have}\:{S}=\:{lim}_{{n}−>\propto} \:{S}_{{n}} \:{we}\:{decompose}\:{the}\:{the}\:{rational}\:{fraction} \\ $$$${F}\left({X}\right)=\:\:\:\mathrm{1}/_{{X}} \mathrm{2}_{\left({X}^{\mathrm{2}} +\mathrm{1}\right)} ....{F}\left({X}\right)=\:\:{a}/{X}\:\:+{b}/{X}^{\mathrm{2}} \:\:+\:{c}/{X}+\mathrm{1} \\ $$$${we}\:{find}\:{a}=−\mathrm{1}..{b}=\mathrm{1}..{c}=\mathrm{1}\:\:{and}\:\:{F}\left({X}\right)=\:\:−\mathrm{1}/{X}\:+\:\mathrm{1}/{X}+\mathrm{1}+\mathrm{1}/{X}^{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)} =\:\:\:−\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/{k}\:\:+\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}+\mathrm{1}} \:+\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} } \\ $$$${but}\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\mathrm{1}/{k}\:\:=\:{H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}+\mathrm{1}} =\:\:\sum_{{k}=\mathrm{2}} ^{{k}={n}+\mathrm{1}} \:\mathrm{1}/{k}\:=\:{H}_{{n}+\mathrm{1}} −\mathrm{1} \\ $$$$−>\:{S}_{{n}} =\:{H}_{{n}+\mathrm{1}} −{H}_{{n}} −\mathrm{1}\:+\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} } \\ $$$${but}\:\:{H}_{{n}} \:={ln}\left({n}\right)+{s}+{o}_{\mathrm{1}} \left(\mathrm{1}/{n}\right)...{H}_{{n}+\mathrm{1}} =\:{ln}\left({n}+\mathrm{1}\right)\:+{s}\:+{o}_{\mathrm{2}} \left(\mathrm{1}/{n}\right) \\ $$$$−>\:{H}_{{n}+\mathrm{1}} \:−\:{H}_{{n}} ={ln}\left(\left({n}+\mathrm{1}\right){n}^{−\mathrm{1}} \right)\:\:+\:{o}_{\mathrm{3}} \left(\mathrm{1}/{n}\right) \\ $$$${but}\:{lim}\:_{{n}−>\propto} {o}_{\mathrm{3}} \left(\mathrm{1}/{n}\right)=\mathrm{0}\:\:{so}\:\:\:{lim}\:\:{H}_{{n}+\mathrm{1}} −\:{H}_{{n}} =\mathrm{0} \\ $$$${lim}_{} \:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\mathrm{1}/_{{k}^{\mathrm{2}} } \:\:=\:\sum_{{k}=\mathrm{1}} ^{\propto} \:\mathrm{1}/_{{k}^{\mathrm{2}} } \:\:=\pi^{\mathrm{2}} /\mathrm{6}\:\:{finally}... \\ $$$$\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:\:\mathrm{1}/_{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \:\:=\:\pi^{\mathrm{2}} /\mathrm{6}\:−\mathrm{1}. \\ $$$$ \\ $$

Question Number 26019    Answers: 0   Comments: 0

Question Number 26008    Answers: 2   Comments: 0

Question Number 26007    Answers: 1   Comments: 0

Question Number 25998    Answers: 1   Comments: 1

Question Number 26009    Answers: 1   Comments: 0

Question Number 25985    Answers: 1   Comments: 0

In a △ABC, a cot A+ b cot B+c cot C =

$$\mathrm{In}\:\mathrm{a}\:\bigtriangleup{ABC},\:{a}\:\mathrm{cot}\:{A}+\:{b}\:\mathrm{cot}\:{B}+{c}\:\mathrm{cot}\:{C}\:= \\ $$

Question Number 26097    Answers: 0   Comments: 0

Question Number 25975    Answers: 1   Comments: 0

Question Number 25974    Answers: 0   Comments: 0

A cowbell coffee at 85°C is placed in a freezer at 0°C.The temperature of the coffee decreases exponentially;so that after 5 minutes it is 30°C. i.)What is the temperature after 3minute? ii)how long will it take for the temperature to drop to 5°C?

$${A}\:{cowbell}\:{coffee}\:{at}\:\mathrm{85}°{C}\:{is}\:{placed}\:{in}\:{a} \\ $$$${freezer}\:{at}\:\mathrm{0}°{C}.{The}\:{temperature}\:{of}\:{the} \\ $$$${coffee}\:{decreases}\:{exponentially};{so}\:{that} \\ $$$${after}\:\mathrm{5}\:{minutes}\:{it}\:{is}\:\mathrm{30}°{C}. \\ $$$$\left.{i}.\right){What}\:{is}\:{the}\:{temperature}\:{after}\:\mathrm{3}{minute}? \\ $$$$\left.{ii}\right){how}\:{long}\:{will}\:{it}\:{take}\:{for}\:{the}\:{temperature} \\ $$$${to}\:{drop}\:{to}\:\mathrm{5}°{C}? \\ $$

Question Number 25973    Answers: 1   Comments: 0

((k.2^k +2.2^k +2k+4)/2)=(k+3)2^k

$$\frac{\mathrm{k}.\mathrm{2}^{\mathrm{k}} +\mathrm{2}.\mathrm{2}^{\mathrm{k}} +\mathrm{2k}+\mathrm{4}}{\mathrm{2}}=\left(\mathrm{k}+\mathrm{3}\right)\mathrm{2}^{\mathrm{k}} \\ $$

Question Number 25972    Answers: 2   Comments: 0

Question Number 25971    Answers: 1   Comments: 2

In finding the equations of the bisectors of the angles between two lines a_1 x+b_1 y+c_1 =0 and a_2 x+b_2 y+c_2 =0, why we observe a_1 a_2 +b_1 b_2 >0 or <0 for obtuse and acute angle bisectors?

$${In}\:{finding}\:{the}\:{equations}\:{of}\:{the} \\ $$$${bisectors}\:{of}\:{the}\:{angles}\:{between}\:{two} \\ $$$${lines}\:{a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} =\mathrm{0}\:{and}\:{a}_{\mathrm{2}} {x}+{b}_{\mathrm{2}} {y}+{c}_{\mathrm{2}} =\mathrm{0}, \\ $$$${why}\:{we}\:{observe}\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{b}_{\mathrm{1}} {b}_{\mathrm{2}} >\mathrm{0}\:{or}\:<\mathrm{0} \\ $$$${for}\:{obtuse}\:{and}\:{acute}\:{angle}\:{bisectors}? \\ $$

Question Number 25969    Answers: 0   Comments: 1

C_0 +2C_1 +3C_2 +..........+(n+1)C_n =(n+2)2^(n−1) using Bionomial teorm

$$\mathrm{C}_{\mathrm{0}} +\mathrm{2C}_{\mathrm{1}} +\mathrm{3C}_{\mathrm{2}} +..........+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{C}_{\mathrm{n}} =\left(\mathrm{n}+\mathrm{2}\right)\mathrm{2}^{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{using}\:\:\mathrm{Bionomial}\:\mathrm{teorm} \\ $$

Question Number 25981    Answers: 1   Comments: 0

Question Number 25965    Answers: 1   Comments: 0

solve the integral ∫sin^4 θdθ

$${solve}\:{the}\:{integral} \\ $$$$\int\mathrm{sin}^{\mathrm{4}} \theta{d}\theta \\ $$

Question Number 25963    Answers: 0   Comments: 0

A bus is traveling along a straight road at 100 km/hr and the bus conductor walks at 6 km/hr on the floor of the bus and in the same direction as the bus. Find the speed of the conductor relative to the road and relative to the bus.

$$\mathrm{A}\:\mathrm{bus}\:\mathrm{is}\:\mathrm{traveling}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}\:\mathrm{at}\:\mathrm{100}\:\mathrm{km}/\mathrm{hr}\:\mathrm{and}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{conductor} \\ $$$$\mathrm{walks}\:\mathrm{at}\:\mathrm{6}\:\mathrm{km}/\mathrm{hr}\:\mathrm{on}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\mathrm{as}\:\mathrm{the}\:\mathrm{bus}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{conductor}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{road}\:\mathrm{and}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bus}. \\ $$

Question Number 26016    Answers: 0   Comments: 0

Question Number 26013    Answers: 1   Comments: 0

Question Number 25961    Answers: 0   Comments: 0

8cos^4 x−8cos^2 x+1=0 solution:8cos^2 x(cos^2 x−1)+1=0⇒ −8cos^2 xsin^2 x=−1⇒ sin^2 2x=(1/2)⇒sinx=+_− ((√2)/2)⇒ { ((2x=2kπ+(π/4))),((2x=2kπ+((3π)/4))) :} { ((2x=2kπ−(π/4))),((2x=2kπ+((5π)/4))) :} and so we have { ((x=kπ+(π/8))),((x=kπ+((3π)/8))) :} { ((x=kπ−(π/8))),((x=kπ+((5π)/8))) :} why x=((kπ)/4)+(π/8)? can we solve by another way?

$$\mathrm{8cos}^{\mathrm{4}} \mathrm{x}−\mathrm{8cos}^{\mathrm{2}} \mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solution}:\mathrm{8cos}^{\mathrm{2}} \mathrm{x}\left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0}\Rightarrow \\ $$$$−\mathrm{8cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}=−\mathrm{1}\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{2x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sinx}=\underset{−} {+}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow \\ $$$$\begin{cases}{\mathrm{2x}=\mathrm{2k}\pi+\frac{\pi}{\mathrm{4}}}\\{\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2x}=\mathrm{2k}\pi−\frac{\pi}{\mathrm{4}}}\\{\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{5}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{k}\pi+\frac{\pi}{\mathrm{8}}}\\{\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{8}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{k}\pi−\frac{\pi}{\mathrm{8}}}\\{\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{5}\pi}{\mathrm{8}}}\end{cases} \\ $$$$\mathrm{why}\:\mathrm{x}=\frac{\mathrm{k}\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}? \\ $$$$ \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{another}\:\mathrm{way}? \\ $$

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