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Question Number 26055    Answers: 1   Comments: 0

calculate ∫_0 ^1 (1+t^2 )^(1/2) dt

$${calculate}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} {dt} \\ $$

Question Number 26054    Answers: 0   Comments: 0

if (1+cos(x))^(−1) = a_0 /_2 +Σ_(n=1) ^(n=∝) a_n cos(nx)) find a_0 and a_n ...you can use fourier series.

$$\left.{if}\:\:\left(\mathrm{1}+{cos}\left({x}\right)\right)^{−\mathrm{1}} \:=\:{a}_{\mathrm{0}} /_{\mathrm{2}} \:\:+\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:{a}_{{n}} \:{cos}\left({nx}\right)\right) \\ $$$${find}\:{a}_{\mathrm{0}} \:{and}\:\:{a}_{{n}} ...{you}\:{can}\:{use}\:{fourier} \\ $$$${series}. \\ $$$$ \\ $$

Question Number 26053    Answers: 1   Comments: 0

If x^2 + 9x + 2 = 0 and x^2 + kx + 5 = 0 have a common root, show that 2k^2 + 63k − 414 = 0 , hence find the value of k such that k > 9.3

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{9x}\:+\:\mathrm{2}\:=\:\mathrm{0}\:\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{kx}\:+\:\mathrm{5}\:=\:\mathrm{0}\:\:\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root},\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{2k}^{\mathrm{2}} \:+\:\mathrm{63k}\:−\:\mathrm{414}\:=\:\mathrm{0}\:,\:\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{k}\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{k}\:>\:\mathrm{9}.\mathrm{3} \\ $$

Question Number 26050    Answers: 1   Comments: 0

Question Number 26047    Answers: 2   Comments: 6

Question Number 26046    Answers: 1   Comments: 0

x^3 + (1/x^3 )=18 find the valu of x+(1/x)

$${x}^{\mathrm{3}} +\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{18}\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{valu}\:\mathrm{of}\:{x}+\frac{\mathrm{1}}{{x}} \\ $$

Question Number 26042    Answers: 0   Comments: 2

ax^2 −bx=0

$${ax}^{\mathrm{2}} −{bx}=\mathrm{0} \\ $$

Question Number 26040    Answers: 0   Comments: 0

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

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