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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.

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?

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

Question Number 25960 Answers: 0 Comments: 0

answer to 25955.we introduce the parametric function F(t) =∫_0 ^∞ ln(1+(1+x^2_ )t)(1+x^2 )^(−1) dx after verifying that F is derivable on[0.∝[ we find ∂F/∂t= ∫_0 ^∞ ( (1+(1+x^2 )t)^(−1) dx ∂F/∂t=1/2 ∫_R (tx^2 +t+1)^(−1) dx we put f(z) =(tz^2 +z+1)^(−1) let find the poles of f..tz^2 +z+1=0 <−> z=+−i((t+1)t^(−1) )^(1/2) and the poles are z_0 =i((t+1)t^(−1) )^(1/2) and z_1 =−i((t+1)t^(−1) )^(1/2) and f(t) =(t(t−z_0 )(t−z_1 ))^(−1) by residus theorem ∫_R f(z)dz =2iπ R(f.z_0 ) =2iπ (t(z_0 −z_1 ))^(−1) =π t^(−1/2) (t+1)^(−1/2 ) −>∂F/∂t =π 2^(−1) t^(−1/2) (1+t)^(−1/2) −>F(t) =π 2^(−1 ) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx +α α=F(0)=0 and F(t) =π2^(−1) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx and by the changement x^(1/2) =u we find F(t) = π ln( t^(1/2) +(1+t)^(1/2) ) so ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx=F(1)=πln(1+2^(1/2) )

Question Number 25962 Answers: 0 Comments: 0

find the value of Σ_(n=1) ^(n=∝) 1/_(n^2 (n+1)) we give Σ_(n=1) ^(n=∝) 1/_n 2= π^2 /6 and H_n =1+2^(−1) +3^(−1) +...+n^(−1) = ln(n) + s + θ(1/n) s is the constant number of Euler

$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:\:\:\mathrm{1}/_{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \:\:{we}\:{give}\:\:\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \mathrm{1}/_{{n}} \mathrm{2}=\:\pi^{\mathrm{2}} /\mathrm{6} \\ $$$${and}\:\:{H}_{{n}} =\mathrm{1}+\mathrm{2}^{−\mathrm{1}} +\mathrm{3}^{−\mathrm{1}} +...+{n}^{−\mathrm{1}} =\:{ln}\left({n}\right)\:+\:{s}\:+\:\theta\left(\mathrm{1}/{n}\right)\: \\ $$$${s}\:{is}\:{the}\:{constant}\:{number}\:{of}\:{Euler} \\ $$

Question Number 25955 Answers: 0 Comments: 0

find the value of ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\int_{\mathrm{0}} ^{\infty} \:\:\boldsymbol{\mathrm{ln}}\left(\mathrm{2}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)^{−\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$

Question Number 25954 Answers: 0 Comments: 0

nswer to 25929 by binome formula (1+x)^(n+m) =Σ_(k=0) ^(k=n+m) C_(n+m) ^k x^k the coefficent of x^n is C_(n+m) ^n and the coefficient of x^m is C_(n+m) ^m but we have for p<n C_n ^p = C_n ^(n−p) >>>>>C_(n+m) ^n = C_(n+m) ^m .

$${nswer}\:{to}\:\mathrm{25929}\:\:\:{by}\:{binome}\:{formula}\:\:\: \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}+{m}} \:\:=\sum_{{k}=\mathrm{0}} ^{{k}={n}+{m}} \:{C}_{{n}+{m}} ^{{k}} \:{x}^{{k}} \:\:\:\:{the}\:{coefficent}\:{of}\:{x}^{{n}} \:{is}\: \\ $$$${C}_{{n}+{m}} ^{{n}} \:{and}\:{the}\:{coefficient}\:{of}\:{x}^{{m}} \:\:{is}\:\:\:{C}_{{n}+{m}} ^{{m}} \:\:\:{but}\:{we}\:{have}\:\:{for} \\ $$$${p}<{n}\:\:\:\:{C}_{{n}} ^{{p}} \:\:\:=\:\:\:{C}_{{n}} ^{{n}−{p}} \:\:\:\:\:>>>>>{C}_{{n}+{m}} ^{{n}} \:\:=\:\:\:{C}_{{n}+{m}} ^{{m}} . \\ $$

Question Number 25949 Answers: 0 Comments: 0

Question Number 25948 Answers: 0 Comments: 1

Factorise: x^2 +y^2 −z^2 −2xy

$$\mathrm{Factorise}:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{z}^{\mathrm{2}} −\mathrm{2}{xy} \\ $$

Question Number 25943 Answers: 1 Comments: 0

Question Number 26096 Answers: 0 Comments: 0

Question Number 25940 Answers: 0 Comments: 0

every extreme point is a critical point but every critical is not a extreme point justify it

$${every}\:{extreme}\:{point}\:{is}\:{a}\:{critical}\: \\ $$$${point}\:{but}\:{every}\:{critical}\:{is}\:{not}\:{a}\: \\ $$$${extreme}\:{point}\:{justify}\:{it}\: \\ $$

Question Number 25937 Answers: 1 Comments: 0

For a certain amount of work,Ade takes 6hours less than Bode.if they work together it takes them 13hours 20 minutes.How long will it take Bode alone to complete the work?

$${For}\:{a}\:{certain}\:{amount}\:{of}\:{work},{Ade}\:{takes} \\ $$$$\mathrm{6}{hours}\:{less}\:{than}\:{Bode}.{if}\:{they}\:{work}\:{together} \\ $$$${it}\:{takes}\:{them}\:\mathrm{13}{hours}\:\mathrm{20}\:{minutes}.{How} \\ $$$${long}\:{will}\:{it}\:{take}\:{Bode}\:{alone}\:{to}\:{complete} \\ $$$${the}\:{work}? \\ $$

Question Number 25934 Answers: 1 Comments: 2

Question Number 25932 Answers: 0 Comments: 0

answer to 25824 we have a^(−x^2 ) = e^(−x^2_ ln(a)) so for a>1 ln(a)=( (ln(a))^(1/2) )^2 >>>>∫_R a^(−^ x^2 ) = ∫_R e^(−(x (ln(a)^(1/2) )^2 ) dx and with the changement t=x (ln(a)^(1/2) >>>>x=t ( ln(a))^(−1/2) we have ∫_R a^(−x^2 ) dx = π^(1/2) (ln(a))^(−1/2) ...if 0<a<1 ln(a)<0 and the integrale is divergente...

$${answer}\:{to}\:\mathrm{25824}\:\:{we}\:{have}\:{a}^{−{x}^{\mathrm{2}} } \:=\:{e}^{−{x}^{\mathrm{2}_{} } {ln}\left({a}\right)} \:\:{so}\:{for}\:{a}>\mathrm{1} \\ $$$${ln}\left({a}\right)=\left(\:\left({ln}\left({a}\right)\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} >>>>\int_{{R}} {a}^{−^{} {x}^{\mathrm{2}} } \:=\:\int_{{R}} {e}^{−\left({x}\:\left({ln}\left({a}\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} \right.} {dx} \\ $$$${and}\:{with}\:{the}\:{changement}\:\:{t}={x}\:\left({ln}\left({a}\right)^{\mathrm{1}/\mathrm{2}} \:\:>>>>{x}={t}\:\left(\:{ln}\left({a}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right. \\ $$$$\:{we}\:{have}\:\:\int_{{R}} {a}^{−{x}^{\mathrm{2}} } {dx}\:\:=\:\pi^{\mathrm{1}/\mathrm{2}} \left({ln}\left({a}\right)\right)^{−\mathrm{1}/\mathrm{2}} ...{if}\:\mathrm{0}<{a}<\mathrm{1}\:{ln}\left({a}\right)<\mathrm{0} \\ $$$$\:{and}\:{the}\:{integrale}\:{is}\:{divergente}... \\ $$

Question Number 25930 Answers: 2 Comments: 0

A line passes through A(−3, 0) and B(0, −4). A variable line perpendicular to AB is drawn to cut x and y-axes at M and N. Find the locus of the point of intersection of the lines AN and BM.

$$\mathrm{A}\:\mathrm{line}\:\mathrm{passes}\:\mathrm{through}\:{A}\left(−\mathrm{3},\:\mathrm{0}\right)\:\mathrm{and} \\ $$$${B}\left(\mathrm{0},\:−\mathrm{4}\right).\:\mathrm{A}\:\mathrm{variable}\:\mathrm{line}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:{AB}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{cut}\:{x}\:\mathrm{and}\:{y}-\mathrm{axes}\:\mathrm{at} \\ $$$${M}\:\mathrm{and}\:{N}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lines}\:{AN}\:\mathrm{and}\:{BM}. \\ $$

Question Number 25929 Answers: 0 Comments: 0

Show that the coefficients of x^m and x^n in (1+x)^(m+n) are equal.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\:\mathrm{coefficients}\:\mathrm{of}\:\mathrm{x}^{\mathrm{m}} \:\mathrm{and}\:\mathrm{x}^{\mathrm{n}} \:\mathrm{in}\:\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{m}+\mathrm{n}} \mathrm{are}\:\mathrm{equal}. \\ $$

Question Number 25928 Answers: 0 Comments: 0

Expand (1−x)^4 .Hence,find S if S=(1−x^3 )^4 −4(1−x^3 )^3 +6(1−x^3 )^2 −4(1−x^3 )^ +1.

$$\mathrm{Expand}\:\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} .\mathrm{H}\boldsymbol{\mathrm{ence}},\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{S}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{S}}=\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{3}} +\mathrm{6}\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{} +\mathrm{1}. \\ $$

Question Number 25924 Answers: 1 Comments: 0

x+3+4=5

$${x}+\mathrm{3}+\mathrm{4}=\mathrm{5} \\ $$

Question Number 25899 Answers: 1 Comments: 1

8a^6 +5a^3 +1=factorize