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Question Number 74615 Answers: 2 Comments: 0

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Question Number 74609 Answers: 0 Comments: 1

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Question Number 74554 Answers: 1 Comments: 0

Find the superimum of the set {(n^2 /2^n )}

$$\boldsymbol{{Find}}\:\boldsymbol{{the}}\:\boldsymbol{{superimum}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{set}}\:\left\{\frac{\boldsymbol{{n}}^{\mathrm{2}} }{\mathrm{2}^{\boldsymbol{{n}}} }\right\} \\ $$

Question Number 74412 Answers: 0 Comments: 0

a,b,c are given real constants. p,q,r,t are unknowns from which we can choose values of two of them (non zero) and have to determine the other two (non zero), obeying two equations given below p^2 +q(1+bq)t^2 +q(ap+br)t +r(1+bq)t+r(ap+br) = 0 pt+q(a+cq)t^2 +r(a+cq)t +cqrt+cr^2 = 0 Can this be done solving a quadratic eq. and none higher..?

$${a},{b},{c}\:{are}\:{given}\:{real}\:{constants}. \\ $$$${p},{q},{r},{t}\:{are}\:{unknowns}\:{from}\:{which} \\ $$$${we}\:{can}\:{choose}\:{values}\:{of}\:{two}\:{of} \\ $$$${them}\:\left({non}\:{zero}\right)\:{and}\:{have}\:{to}\: \\ $$$${determine}\:{the}\:{other}\:{two}\:\left({non}\:{zero}\right), \\ $$$${obeying}\:\:{two}\:{equations}\:{given}\:{below} \\ $$$$\:{p}^{\mathrm{2}} +{q}\left(\mathrm{1}+{bq}\right){t}^{\mathrm{2}} +{q}\left({ap}+{br}\right){t} \\ $$$$\:\:\:\:\:+{r}\left(\mathrm{1}+{bq}\right){t}+{r}\left({ap}+{br}\right)\:=\:\mathrm{0} \\ $$$${pt}+{q}\left({a}+{cq}\right){t}^{\mathrm{2}} +{r}\left({a}+{cq}\right){t} \\ $$$$\:\:\:\:+{cqrt}+{cr}^{\mathrm{2}} =\:\mathrm{0} \\ $$$${Can}\:{this}\:{be}\:{done}\:{solving}\:{a} \\ $$$${quadratic}\:{eq}.\:{and}\:{none}\:{higher}..? \\ $$

Question Number 74329 Answers: 1 Comments: 1

Solve : ax+by=r bx−ay=s

$${Solve}\:: \\ $$$${ax}+{by}={r} \\ $$$${bx}−{ay}={s} \\ $$

Question Number 74266 Answers: 0 Comments: 0

please kindly help me with the solutions to these question?very urgent please (1) if dy=x^3 dx. find the equation of y in terms of x if the curve passes through (1,1) 2) Given that the volume v(t) of cell at a time t changes according to ((dV(t))/dt)= sin t, with v(t)=4. find v(t) 3) Given (dP/dt) + 3P = 0. determine P (t) if p(0)= 4 4) Radium decomposes at a rate proportion to the amount present. if the half−life of the radium is 1000 years. what is the percentage lost in 100 years? 5) Calculate the pressure of a gas after phase transition at 171°K from 101.3kPa pressure at 472°K taking R = 0.1886kJ\kgK and L = 35.73kg

$${please}\:{kindly}\:{help}\:{me}\:{with}\:{the}\:{solutions}\:{to}\:{these}\:{question}?{very}\:{urgent}\:{please}\:\left(\mathrm{1}\right)\:{if}\:{dy}={x}^{\mathrm{3}} {dx}.\:{find}\:{the}\:{equation}\:{of}\:{y}\:{in}\:{terms}\:{of}\:{x} \\ $$$$\:{if}\:{the}\:{curve}\:{passes}\:{through}\:\left(\mathrm{1},\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:{Given}\:{that}\:{the}\:{volume}\:{v}\left({t}\right)\:{of}\:{cell}\:{at}\:{a}\:{time}\:{t}\:{changes}\:{according}\:{to} \\ $$$$\frac{{dV}\left({t}\right)}{{dt}}=\:{sin}\:{t},\:{with}\:{v}\left({t}\right)=\mathrm{4}.\:{find}\:{v}\left({t}\right) \\ $$$$\left.\mathrm{3}\right)\:{Given}\:\frac{{dP}}{{dt}}\:+\:\mathrm{3}{P}\:=\:\mathrm{0}.\:{determine}\:{P}\:\left({t}\right)\: \\ $$$${if}\:{p}\left(\mathrm{0}\right)=\:\mathrm{4} \\ $$$$\left.\mathrm{4}\right)\:{Radium}\:{decomposes}\:{at}\:{a}\:{rate}\:{proportion}\:{to}\:{the}\:{amount}\: \\ $$$${present}.\:{if}\:{the}\:{half}−{life}\:{of}\:{the}\:{radium}\:{is} \\ $$$$\mathrm{1000}\:{years}.\:{what}\:{is}\:{the}\:{percentage}\:{lost}\:{in}\:\mathrm{100}\:{years}? \\ $$$$\left.\mathrm{5}\right)\:{Calculate}\:{the}\:{pressure}\:{of}\:{a}\:{gas}\:{after}\:{phase} \\ $$$${transition}\:{at}\:\mathrm{171}°{K}\:{from}\:\mathrm{101}.\mathrm{3}{kPa} \\ $$$${pressure}\:{at}\:\mathrm{472}°{K}\:{taking}\:{R}\:=\:\mathrm{0}.\mathrm{1886}{kJ}\backslash{kgK}\:{and}\:{L}\:=\:\mathrm{35}.\mathrm{73}{kg} \\ $$

Question Number 74213 Answers: 2 Comments: 1

Question Number 74209 Answers: 1 Comments: 0

Question Number 74196 Answers: 0 Comments: 1

solve for x: 4^x +6^x =9^x

$${solve}\:{for}\:{x}: \\ $$$$\mathrm{4}^{{x}} +\mathrm{6}^{{x}} =\mathrm{9}^{{x}} \\ $$

Question Number 74164 Answers: 2 Comments: 0

factorize 6x^2 −11xy −10y^2

$${factorize} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} \:−\mathrm{11}{xy}\:−\mathrm{10}{y}^{\mathrm{2}} \\ $$

Question Number 74168 Answers: 3 Comments: 1

Question Number 74148 Answers: 1 Comments: 0

{ ((−x(√3)+2my(√2)=((√3)/3))),((2mx−3y(√6)=1)) :} help me solve it.

$$\begin{cases}{−{x}\sqrt{\mathrm{3}}+\mathrm{2}{my}\sqrt{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}\\{\mathrm{2}{mx}−\mathrm{3}{y}\sqrt{\mathrm{6}}=\mathrm{1}}\end{cases}\:\:\:\:\:\: \\ $$$$ \\ $$$${help}\:{me}\:{solve}\:{it}. \\ $$

Question Number 74129 Answers: 1 Comments: 0

2C_4 ^n = 35C_3 ^(n/2) ⇒ n = ?

$$\mathrm{2}\boldsymbol{{C}}_{\mathrm{4}} ^{\boldsymbol{{n}}} \:=\:\mathrm{35}\boldsymbol{{C}}_{\mathrm{3}} ^{\frac{\boldsymbol{{n}}}{\mathrm{2}}} \: \\ $$$$\Rightarrow\:\boldsymbol{{n}}\:=\:? \\ $$

Question Number 74121 Answers: 0 Comments: 1

Factor the polynomial ((c/2))x^2 +(b−((3c)/2))x+(c−b+a)

$$\mathrm{Factor}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\left(\frac{{c}}{\mathrm{2}}\right){x}^{\mathrm{2}} +\left({b}−\frac{\mathrm{3}{c}}{\mathrm{2}}\right){x}+\left({c}−{b}+{a}\right) \\ $$

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Question Number 74024 Answers: 1 Comments: 1

{ ((h^2 +y^2 +(k−z)^2 =s^2 )),((a^2 +(b−y)^2 +z^2 =s^2 )),((ah+y(y−b)+z(z−k)=0)),((((h+a)/2)+yz−(b−y)(k−z)=1)),((b+a(k−z)+hz=1)),((k+h(b−y)+ay=1)) :} Find s_(min) or at least express s=f(y) or g(z).

$$\begin{cases}{{h}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({k}−{z}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} +\left({b}−{y}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} ={s}^{\mathrm{2}} }\\{{ah}+{y}\left({y}−{b}\right)+{z}\left({z}−{k}\right)=\mathrm{0}}\\{\frac{{h}+{a}}{\mathrm{2}}+{yz}−\left({b}−{y}\right)\left({k}−{z}\right)=\mathrm{1}}\\{{b}+{a}\left({k}−{z}\right)+{hz}=\mathrm{1}}\\{{k}+{h}\left({b}−{y}\right)+{ay}=\mathrm{1}}\end{cases} \\ $$$${Find}\:\:{s}_{{min}} \:{or}\:{at}\:{least}\:{express} \\ $$$$\:{s}={f}\left({y}\right)\:{or}\:{g}\left({z}\right). \\ $$

Question Number 74006 Answers: 1 Comments: 1

If 3x^2 e^(log _x 27) =27000 then find x

$${If}\:\mathrm{3}{x}^{\mathrm{2}} {e}^{\mathrm{log}\:_{{x}} \mathrm{27}} =\mathrm{27000}\:{then}\:{find}\:{x} \\ $$

Question Number 74339 Answers: 0 Comments: 2

x^3 +ax^2 +bx+c=0 Let x=((pt+q)/(t+1)) p^3 t^3 +3p^2 qt^2 +3pq^2 t+q^3 +a(t+1)(p^2 t^2 +2pqt+q^2 ) +b(pt+q)(t^2 +2t+1) +c(t^3 +3t^2 +3t+1) = 0 ⇒ (p^3 +ap^2 +bp+c)t^3 +(3p^2 q+ap^2 +2apq+bq+2bp+3c)t^2 +(3q^2 p+aq^2 +2apq+bp+2bq+3c)t +(q^3 +aq^2 +bq+c) = 0 Let coeffs. of t^2 and t be zero. Subtracting and adding them 3pq+a(p+q)+b=0 & 3pq(p+q)+a{(p+q)^2 −2pq} +4apq+3b(p+q)+6c = 0 lets call pq=m , p+q=s ⇒ 3m+as+b=0 ....(i) 3ms+a(s^2 −2m)+4am+3bs+6c=0 ⇒ am+bs+3c=0 ....(ii) ⇒ s=((9c−ab)/(a^2 −3b)) ; m=((b^2 −3ac)/(a^2 −3b)) Now p,q are roots of eq. z^2 −sz+m=0 p,q = (s/2)±(√((s^2 /4)−m)) t^3 =−(((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bq+c))) (t≠−1) x=((pt+q)/(t+1)) .

$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${Let}\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$${p}^{\mathrm{3}} {t}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} {qt}^{\mathrm{2}} +\mathrm{3}{pq}^{\mathrm{2}} {t}+{q}^{\mathrm{3}} \\ $$$$+{a}\left({t}+\mathrm{1}\right)\left({p}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}{pqt}+{q}^{\mathrm{2}} \right) \\ $$$$+{b}\left({pt}+{q}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}\right) \\ $$$$+{c}\left({t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}\right)\:\:\:\:=\:\:\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\:\:\left({p}^{\mathrm{3}} +{ap}^{\mathrm{2}} +{bp}+{c}\right){t}^{\mathrm{3}} \\ $$$$+\left(\mathrm{3}{p}^{\mathrm{2}} {q}+{ap}^{\mathrm{2}} +\mathrm{2}{apq}+{bq}+\mathrm{2}{bp}+\mathrm{3}{c}\right){t}^{\mathrm{2}} \\ $$$$+\left(\mathrm{3}{q}^{\mathrm{2}} {p}+{aq}^{\mathrm{2}} +\mathrm{2}{apq}+{bp}+\mathrm{2}{bq}+\mathrm{3}{c}\right){t} \\ $$$$+\left({q}^{\mathrm{3}} +{aq}^{\mathrm{2}} +{bq}+{c}\right)\:=\:\mathrm{0} \\ $$$${Let}\:{coeffs}.\:{of}\:{t}^{\mathrm{2}} \:{and}\:{t}\:{be}\:{zero}. \\ $$$${Subtracting}\:{and}\:{adding}\:{them} \\ $$$$\:\mathrm{3}{pq}+{a}\left({p}+{q}\right)+{b}=\mathrm{0}\:\:\:\& \\ $$$$\mathrm{3}{pq}\left({p}+{q}\right)+{a}\left\{\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq}\right\} \\ $$$$\:+\mathrm{4}{apq}+\mathrm{3}{b}\left({p}+{q}\right)+\mathrm{6}{c}\:=\:\mathrm{0} \\ $$$${lets}\:{call}\:\:{pq}={m}\:,\:\:{p}+{q}={s}\:\:\Rightarrow \\ $$$$\:\:\mathrm{3}{m}+{as}+{b}=\mathrm{0}\:\:\:\:....\left({i}\right) \\ $$$$\mathrm{3}{ms}+{a}\left({s}^{\mathrm{2}} −\mathrm{2}{m}\right)+\mathrm{4}{am}+\mathrm{3}{bs}+\mathrm{6}{c}=\mathrm{0} \\ $$$$\Rightarrow\:{am}+{bs}+\mathrm{3}{c}=\mathrm{0}\:\:\:\:....\left({ii}\right) \\ $$$$\Rightarrow\:\:{s}=\frac{\mathrm{9}{c}−{ab}}{{a}^{\mathrm{2}} −\mathrm{3}{b}}\:\:\:;\:\:{m}=\frac{{b}^{\mathrm{2}} −\mathrm{3}{ac}}{{a}^{\mathrm{2}} −\mathrm{3}{b}} \\ $$$$\:{Now}\:\:{p},{q}\:\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\:\:\:{z}^{\mathrm{2}} −{sz}+{m}=\mathrm{0} \\ $$$$\:\:\:{p},{q}\:=\:\frac{{s}}{\mathrm{2}}\pm\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{4}}−{m}} \\ $$$$\:\:{t}^{\mathrm{3}} =−\left(\frac{{q}^{\mathrm{3}} +{aq}^{\mathrm{2}} +{bq}+{c}}{{p}^{\mathrm{3}} +{ap}^{\mathrm{2}} +{bq}+{c}}\right)\:\:\:\:\:\left({t}\neq−\mathrm{1}\right) \\ $$$$\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}}\:. \\ $$

Question Number 73775 Answers: 0 Comments: 0

Question Number 73766 Answers: 3 Comments: 0

{ (((1/(x−1))=(2/(y−2))=(3/(z−3)))),((x+2y+3z=56)) :} please help me to solve it in R^3

$$\begin{cases}{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{y}−\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{z}−\mathrm{3}}}\\{\mathrm{x}+\mathrm{2y}+\mathrm{3z}=\mathrm{56}}\end{cases} \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \\ $$

Question Number 73679 Answers: 0 Comments: 1

We have updated backend code to disallow delete of question which are already answered or commented.

$$\mathrm{We}\:\mathrm{have}\:\mathrm{updated}\:\mathrm{backend}\:\mathrm{code}\:\mathrm{to}\: \\ $$$$\mathrm{disallow}\:\mathrm{delete}\:\mathrm{of}\:\mathrm{question}\:\mathrm{which}\:\mathrm{are} \\ $$$$\mathrm{already}\:\mathrm{answered}\:\mathrm{or}\:\mathrm{commented}. \\ $$$$ \\ $$

Question Number 73665 Answers: 1 Comments: 0

if cos^2 (θ)=((m^2 −1)/3) , tan^3 ((θ/2))=tan(a) prove that ((cos^2 (a)))^(1/3) + ((sin^2 (a)))^(1/3) = ((((2/m))^2 ))^(1/3)

$${if} \\ $$$$ \\ $$$${cos}^{\mathrm{2}} \left(\theta\right)=\frac{{m}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}}\:\:,\:\:{tan}^{\mathrm{3}} \left(\frac{\theta}{\mathrm{2}}\right)={tan}\left({a}\right) \\ $$$$ \\ $$$${prove}\:{that} \\ $$$$ \\ $$$$\sqrt[{\mathrm{3}}]{{cos}^{\mathrm{2}} \left({a}\right)}\:+\:\sqrt[{\mathrm{3}}]{{sin}^{\mathrm{2}} \left({a}\right)}\:=\:\sqrt[{\mathrm{3}}]{\left(\frac{\mathrm{2}}{{m}}\right)^{\mathrm{2}} } \\ $$

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