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Question Number 63424    Answers: 0   Comments: 0

A colony of bacteria if left undisturbed will grow at a rate proportional to the number of bacteria, P present at time,t. However,a toxic substance is being added slowly such that at time t, the bacteria also die at the rate μPt where μ is a positive constant. (a) Show that at time t the rate of growth of the bacteria in the colony is governed by the differential equation (dP/dt)= (k−μt)p where k is apositive constant. when t=0, (dP/dt)=2P and when t=1, (dP/dt)=((19)/(10))P (b) show that (dP/dt)= (1/(10))(20−t)P. Sir Forkum Michael.

$${A}\:{colony}\:{of}\:{bacteria}\:{if}\:{left}\:{undisturbed}\:{will}\:{grow}\:{at}\:{a}\:{rate} \\ $$$${proportional}\:{to}\:{the}\:{number}\:{of}\:{bacteria},\:{P}\:{present}\:{at}\:{time},{t}. \\ $$$${However},{a}\:{toxic}\:{substance}\:{is}\:{being}\:{added}\:{slowly}\:{such}\:{that} \\ $$$${at}\:{time}\:{t},\:{the}\:{bacteria}\:{also}\:{die}\:{at}\:{the}\:{rate}\:\mu{Pt}\:{where}\:\mu\:{is} \\ $$$${a}\:{positive}\:{constant}. \\ $$$$\left({a}\right)\:\:{Show}\:{that}\:{at}\:{time}\:{t}\:{the}\:{rate}\:{of}\:{growth}\:{of}\:{the}\:{bacteria}\:{in} \\ $$$${the}\:{colony}\:{is}\:{governed}\:{by}\:{the}\:{differential}\:{equation} \\ $$$$\:\frac{{dP}}{{dt}}=\:\left({k}−\mu{t}\right){p}\:{where}\:{k}\:{is}\:{apositive}\:{constant}. \\ $$$${when}\:{t}=\mathrm{0},\:\frac{{dP}}{{dt}}=\mathrm{2}{P}\:{and}\:{when}\:{t}=\mathrm{1},\:\frac{{dP}}{{dt}}=\frac{\mathrm{19}}{\mathrm{10}}{P} \\ $$$$\left({b}\right)\:{show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dP}}{{dt}}=\:\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{20}−{t}\right){P}. \\ $$$$\:{Sir}\:{Forkum}\:{Michael}. \\ $$

Question Number 63423    Answers: 0   Comments: 1

Prove that ((sin^2 (36°))/(cos^2 (72°)))+sin^2 (72°)cos(36°)=((45+11(√5))/(16))

$${Prove}\:{that}\: \\ $$$$\frac{{sin}^{\mathrm{2}} \left(\mathrm{36}°\right)}{{cos}^{\mathrm{2}} \left(\mathrm{72}°\right)}+{sin}^{\mathrm{2}} \left(\mathrm{72}°\right){cos}\left(\mathrm{36}°\right)=\frac{\mathrm{45}+\mathrm{11}\sqrt{\mathrm{5}}}{\mathrm{16}} \\ $$

Question Number 63579    Answers: 0   Comments: 4

lim_(x→0) ((2^(sec(x)) − 2^(cos(x)) )/x^2 )

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{2}^{\mathrm{sec}\left(\mathrm{x}\right)} \:−\:\mathrm{2}^{\mathrm{cos}\left(\mathrm{x}\right)} }{\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 63410    Answers: 2   Comments: 2

Question Number 63405    Answers: 0   Comments: 0

find ∫ (√((x^2 −4x+1)/(x+2)))dx

$${find}\:\int\:\sqrt{\frac{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}{{x}+\mathrm{2}}}{dx} \\ $$

Question Number 63404    Answers: 1   Comments: 2

1) find ∫ ((x+1)/(x^3 −3x −2))dx 2) calculate ∫_4 ^(+∞) ((x+1)/(x^3 −3x +2))dx

$$\left.\mathrm{1}\right)\:{find}\:\:\int\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}\:−\mathrm{2}}{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{4}} ^{+\infty} \:\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}\:+\mathrm{2}}{dx} \\ $$

Question Number 63407    Answers: 0   Comments: 0

find a basics of solution of the ode (x^2 −x)y′′ −xy′+y=0

$${find}\:{a}\:{basics}\:{of}\:{solution}\:{of}\:{the}\:{ode} \\ $$$$ \\ $$$$\left({x}^{\mathrm{2}} −{x}\right){y}''\:−{xy}'+{y}=\mathrm{0} \\ $$

Question Number 63395    Answers: 0   Comments: 3

let f(t) =∫_0 ^∞ ((ln(1+tx))/(1+x^2 ))dx with ∣t∣<1 1) determine a explicit form of f(t) 2) find the value of ∫_0 ^∞ ((ln(1+x))/(1+x^2 ))dx

$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{1}+{tx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:{with}\:\:\mid{t}\mid<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 63614    Answers: 0   Comments: 0

solve the equation with using (FPI_Fixed Point Iteration) x−tan(x)=0 , in[4,5]

$${solve}\:{the}\:{equation}\:{with}\:{using}\:\left({FPI\_Fixed}\:{Point}\:{Iteration}\right) \\ $$$$ \\ $$$${x}−{tan}\left({x}\right)=\mathrm{0}\:,\:{in}\left[\mathrm{4},\mathrm{5}\right] \\ $$

Question Number 63389    Answers: 0   Comments: 2

f function integrable on [a,b] is max ∫_a ^b f(x)dx =∫_a ^b maxf(x)dx? if not give a opposite example .

$${f}\:{function}\:{integrable}\:{on}\:\left[{a},{b}\right] \\ $$$${is}\:{max}\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\:=\int_{{a}} ^{{b}} \:{maxf}\left({x}\right){dx}?\:\:{if}\:{not}\:{give}\:{a}\:{opposite}\:{example}\:. \\ $$

Question Number 63383    Answers: 1   Comments: 1

solve this equation in all part of complex number: (√((x^9 −3x^2 +1)(x−6)+4))=(x^9 −3x^2 +1)(x−6)−16

$${solve}\:{this}\:{equation}\:{in}\:{all}\: \\ $$$${part}\:{of}\:{complex}\:{number}: \\ $$$$\sqrt{\left({x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{6}\right)+\mathrm{4}}=\left({x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{6}\right)−\mathrm{16} \\ $$

Question Number 63381    Answers: 0   Comments: 2

solve for both x and n in equation: x^n =216 in all part of integer A {_(n=3) ^(x=6) B{_(n=4) ^(x=5) C{_(n=5) ^(x=4) D {_(n=6) ^(x=3)

$${solve}\:{for}\:{both}\:{x}\:{and}\:{n} \\ $$$${in}\:{equation}:\:{x}^{{n}} =\mathrm{216}\:{in}\:{all} \\ $$$${part}\:{of}\:{integer} \\ $$$$\mathscr{A}\:\underset{{n}=\mathrm{3}} {\overset{{x}=\mathrm{6}} {\left\{}}\right. \\ $$$$\mathscr{B}\underset{{n}=\mathrm{4}} {\overset{{x}=\mathrm{5}} {\left\{}}\right. \\ $$$$\mathscr{C}\underset{{n}=\mathrm{5}} {\overset{{x}=\mathrm{4}} {\left\{}}\right. \\ $$$$\mathscr{D}\:\underset{{n}=\mathrm{6}} {\overset{{x}=\mathrm{3}} {\left\{}}\right. \\ $$

Question Number 63378    Answers: 1   Comments: 0

∫_2 ^x (1/x)dx=2ln(3)−ln(2) please help me to solve for x

$$\underset{\mathrm{2}} {\overset{{x}} {\int}}\frac{\mathrm{1}}{{x}}{dx}=\mathrm{2}{ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right) \\ $$$${please}\:{help}\:{me}\:{to}\:{solve}\:{for}\:{x} \\ $$

Question Number 63363    Answers: 1   Comments: 3

Question Number 63361    Answers: 0   Comments: 3

Question Number 63351    Answers: 3   Comments: 2

Question Number 63372    Answers: 1   Comments: 2

For what values of a and b will the integral ∫_a ^b (√(10−x−x^2 ))dx be at maximum

$${For}\:{what}\:{values}\:{of}\:{a}\:{and}\:{b}\:{will}\:{the} \\ $$$${integral}\:\int_{{a}} ^{{b}} \sqrt{\mathrm{10}−{x}−{x}^{\mathrm{2}} }{dx}\:{be}\:{at} \\ $$$${maximum} \\ $$

Question Number 63373    Answers: 0   Comments: 2

just found this on the web I thought it might help in some cases where quartics appear i.e. Sir Aifour′s geometric questions. sometimes we know the nature of the roots, but how to use this information? ax^4 +bx^3 +cx^2 +dx+e=0 1. divide by a 2. x=z−(b/(4a)) this leads to the reduced z^4 +pz^2 +qz+r=0 now we find the nature of the roots: T_1 =16p^4 r−4p^3 q^2 −128p^2 r^2 +144pq^2 r−27q^4 +256r^3 T_2 =p^2 +12r T_3 =−p^2 +4r T_1 <0 ⇒ 2 distinct real and 2 conjugated complex roots T_1 >0∧(p<0∧T_3 <0) ⇒ 4 distinct real roots T_1 >0∧(p>0∨T_3 >0) ⇒ 2 pairs of conjugated complex roots T_1 =0∧(p<0∧T_3 <0∧T_2 ≠0) ⇒ 1 real double and 2 real simple roots T_1 =0∧(T_3 >0∨(p>0∧(T_3 ≠0∨q≠0))) ⇒ 1 real double and 2 conjugated complex roots T_1 =0∧(T_2 =0∧T_3 ≠0) ⇒ 1 real triple and 1 real simple roots T_1 =0∧(T_3 =0∧p<0) ⇒ 2 real double roots T_1 =0∧(T_3 =0∧p>0∧q=0) ⇒ 2 conjugated complex double roots T_1 =0∧T_2 =0 ⇒ all roots are equal

$$\mathrm{just}\:\mathrm{found}\:\mathrm{this}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web} \\ $$$$\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{might}\:\mathrm{help}\:\mathrm{in}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{where} \\ $$$$\mathrm{quartics}\:\mathrm{appear}\:\mathrm{i}.\mathrm{e}.\:\mathrm{Sir}\:\mathrm{Aifour}'\mathrm{s}\:\mathrm{geometric} \\ $$$$\mathrm{questions}.\:\mathrm{sometimes}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{roots},\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{this}\:\mathrm{information}? \\ $$$$ \\ $$$${ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e}=\mathrm{0} \\ $$$$\mathrm{1}.\:\mathrm{divide}\:\mathrm{by}\:{a} \\ $$$$\mathrm{2}.\:{x}={z}−\frac{{b}}{\mathrm{4}{a}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{reduced} \\ $$$$ \\ $$$${z}^{\mathrm{4}} +{pz}^{\mathrm{2}} +{qz}+{r}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}: \\ $$$${T}_{\mathrm{1}} =\mathrm{16}{p}^{\mathrm{4}} {r}−\mathrm{4}{p}^{\mathrm{3}} {q}^{\mathrm{2}} −\mathrm{128}{p}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{144}{pq}^{\mathrm{2}} {r}−\mathrm{27}{q}^{\mathrm{4}} +\mathrm{256}{r}^{\mathrm{3}} \\ $$$${T}_{\mathrm{2}} ={p}^{\mathrm{2}} +\mathrm{12}{r} \\ $$$${T}_{\mathrm{3}} =−{p}^{\mathrm{2}} +\mathrm{4}{r} \\ $$$${T}_{\mathrm{1}} <\mathrm{0}\:\Rightarrow\:\mathrm{2}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} >\mathrm{0}\wedge\left({p}<\mathrm{0}\wedge{T}_{\mathrm{3}} <\mathrm{0}\right)\:\Rightarrow\:\mathrm{4}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} >\mathrm{0}\wedge\left({p}>\mathrm{0}\vee{T}_{\mathrm{3}} >\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({p}<\mathrm{0}\wedge{T}_{\mathrm{3}} <\mathrm{0}\wedge{T}_{\mathrm{2}} \neq\mathrm{0}\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{double}\:\mathrm{and}\:\mathrm{2}\:\mathrm{real}\:\mathrm{simple}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} >\mathrm{0}\vee\left({p}>\mathrm{0}\wedge\left({T}_{\mathrm{3}} \neq\mathrm{0}\vee{q}\neq\mathrm{0}\right)\right)\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{double}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{2}} =\mathrm{0}\wedge{T}_{\mathrm{3}} \neq\mathrm{0}\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{triple}\:\mathrm{and}\:\mathrm{1}\:\mathrm{real}\:\mathrm{simple}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} =\mathrm{0}\wedge{p}<\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{real}\:\mathrm{double}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} =\mathrm{0}\wedge{p}>\mathrm{0}\wedge{q}=\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{double}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge{T}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\mathrm{all}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{equal} \\ $$

Question Number 63336    Answers: 2   Comments: 1

Question Number 63324    Answers: 1   Comments: 0

Question Number 63399    Answers: 1   Comments: 1

if α and β are the roots of 4x^(2 ) −6x+1===00====================== =0. find α^3 −β^3 .

$${if}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{roots}\:{of}\:\mathrm{4}{x}^{\mathrm{2}\:} −\mathrm{6}{x}+\mathrm{1}===\mathrm{00}====================== \\ $$$$=\mathrm{0}.\:{find}\:\alpha^{\mathrm{3}} −\beta^{\mathrm{3}} . \\ $$

Question Number 63301    Answers: 1   Comments: 9

find (dy/dx) if x(x +y) = y^2

$${find}\:\frac{{dy}}{{dx}}\:{if}\:\:{x}\left({x}\:+{y}\right)\:=\:{y}^{\mathrm{2}} \\ $$

Question Number 63300    Answers: 0   Comments: 2

show that a) 1 + tan ((π/4) + A) = (2/(1−tanA)) b) 2cos2θsinθ + 9sinθ + 3 ≡ 11sinθ − 4sin^3 θ + 3

$${show}\:{that}\:\: \\ $$$$\left.{a}\right)\:\mathrm{1}\:+\:{tan}\:\left(\frac{\pi}{\mathrm{4}}\:+\:{A}\right)\:=\:\frac{\mathrm{2}}{\mathrm{1}−{tanA}} \\ $$$$\left.{b}\right)\:\mathrm{2}{cos}\mathrm{2}\theta{sin}\theta\:+\:\mathrm{9}{sin}\theta\:+\:\mathrm{3}\:\equiv\:\mathrm{11}{sin}\theta\:−\:\mathrm{4}{sin}^{\mathrm{3}} \theta\:+\:\mathrm{3} \\ $$

Question Number 63298    Answers: 0   Comments: 2

A particle P, moves on the curve with polar equation r = e^(kθ) , where (r,θ) are polar coordinates referred to a fixed pole and k is a positive constant. Given that the radial velocity of P is (k/r) show that the transverse acceleration of th particle is zero.

$${A}\:{particle}\:{P},\:{moves}\:{on}\:{the}\:{curve}\:{with}\:{polar}\:{equation}\:\: \\ $$$${r}\:=\:{e}^{{k}\theta} \:,\:{where}\:\left({r},\theta\right)\:{are}\:{polar}\:{coordinates}\:{referred}\:{to}\:{a}\:{fixed} \\ $$$${pole}\:{and}\:{k}\:{is}\:{a}\:{positive}\:{constant}.\:{Given}\:{that}\:{the}\:{radial}\:{velocity} \\ $$$${of}\:{P}\:{is}\:\frac{{k}}{{r}}\:\:{show}\:{that}\:{the}\:{transverse}\:{acceleration}\:{of}\:{th}\:{particle} \\ $$$${is}\:{zero}. \\ $$$$ \\ $$

Question Number 63296    Answers: 1   Comments: 1

A random Variable Y has probability function P, defined by P(y) = { (((y^2 /k) , y= 1,2,3)),(((((y−7)^2 )/k) , y= 4,5,6)),((0 , otherwise.)) :} Find (i) The value of the constant k. (ii) the mean and varriance of Y. (iii) The variance R, where R= 2Y −3.

$${A}\:{random}\:{Variable}\:{Y}\:{has}\:{probability}\:{function}\:{P},\:{defined}\:{by} \\ $$$$\:{P}\left({y}\right)\:=\:\begin{cases}{\frac{{y}^{\mathrm{2}} }{{k}}\:,\:{y}=\:\mathrm{1},\mathrm{2},\mathrm{3}}\\{\frac{\left({y}−\mathrm{7}\right)^{\mathrm{2}} }{{k}}\:,\:{y}=\:\mathrm{4},\mathrm{5},\mathrm{6}}\\{\mathrm{0}\:\:\:\:,\:{otherwise}.}\end{cases} \\ $$$${Find}\: \\ $$$$\left({i}\right)\:{The}\:{value}\:{of}\:{the}\:{constant}\:{k}. \\ $$$$\left({ii}\right)\:{the}\:{mean}\:{and}\:{varriance}\:{of}\:{Y}. \\ $$$$\left({iii}\right)\:{The}\:{variance}\:{R},\:{where}\:{R}=\:\mathrm{2}{Y}\:−\mathrm{3}. \\ $$

Question Number 63273    Answers: 0   Comments: 1

let F(x) =∫_x^2 ^x^3 ((sin(t))/(t+x)) dt 1) calculate lim_(x→0) F(x) and lim_(x→+∞) F(x) 2)calculste lim_(x→0) F^′ (x) and lim_(x→+∞) F^′ (x)

$${let}\:{F}\left({x}\right)\:=\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\:\:\:\:\frac{{sin}\left({t}\right)}{{t}+{x}}\:{dt} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:{F}\left({x}\right)\:{and}\:{lim}_{{x}\rightarrow+\infty} {F}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculste}\:{lim}_{{x}\rightarrow\mathrm{0}} \:{F}^{'} \left({x}\right)\:{and}\:{lim}_{{x}\rightarrow+\infty} \:{F}^{'} \left({x}\right) \\ $$

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