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AllQuestion and Answers: Page 1188

Question Number 91183 Answers: 0 Comments: 0

Question Number 91182 Answers: 1 Comments: 0

lim_(x→0) ((2x^6 +3x^2 −3tan^2 x)/(3x^6 ))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3tan}\:^{\mathrm{2}} {x}}{\mathrm{3}{x}^{\mathrm{6}} } \\ $$

Question Number 91178 Answers: 0 Comments: 2

Question Number 91176 Answers: 0 Comments: 1

My Post Filter Issue Dear Mr W, can you send a few screenshots on what is seen when using filter myposts. Thank You

$$\mathrm{My}\:\mathrm{Post}\:\mathrm{Filter}\:\mathrm{Issue} \\ $$$$\mathrm{Dear}\:\mathrm{Mr}\:\mathrm{W},\:\mathrm{can}\:\mathrm{you}\:\mathrm{send}\:\mathrm{a}\:\mathrm{few}\:\mathrm{screenshots} \\ $$$$\mathrm{on}\:\mathrm{what}\:\mathrm{is}\:\mathrm{seen}\:\mathrm{when}\:\mathrm{using}\:\mathrm{filter} \\ $$$$\mathrm{myposts}.\:\mathrm{Thank}\:\mathrm{You} \\ $$

Question Number 91177 Answers: 1 Comments: 1

6^((log_2 x)^2 ) + x^((log_2 x)) = 12

$$\mathrm{6}^{\left(\mathrm{log}_{\mathrm{2}} \:{x}\right)^{\mathrm{2}} } \:+\:{x}^{\left(\mathrm{log}_{\mathrm{2}} \:{x}\right)} \:=\:\mathrm{12}\: \\ $$

Question Number 91167 Answers: 1 Comments: 4

Question Number 91166 Answers: 0 Comments: 7

Question Number 91158 Answers: 0 Comments: 3

lim_(x→∞) (√(4x^2 +2x)) − ((8x^3 +4x^2 ))^(1/(3 ))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}}\:−\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} } \\ $$

Question Number 91157 Answers: 0 Comments: 1

lim_(x→0) ((3tan 4x−4tan 3x)/(3sin 4x−4sin 3x)) = ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4}{x}−\mathrm{4tan}\:\mathrm{3}{x}}{\mathrm{3sin}\:\mathrm{4}{x}−\mathrm{4sin}\:\mathrm{3}{x}}\:=\:? \\ $$

Question Number 91154 Answers: 1 Comments: 1

lim_(x→0) ((12−6x^2 −12cos x)/x^4 )

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{12}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{12cos}\:{x}}{{x}^{\mathrm{4}} } \\ $$

Question Number 91147 Answers: 2 Comments: 1

(D^2 +1)^2 y = x^2 cos x

$$\left({D}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {y}\:=\:{x}^{\mathrm{2}} \mathrm{cos}\:{x}\: \\ $$

Question Number 91139 Answers: 0 Comments: 1

(dy/dx) = ((y−x+1)/(y−x+5))

$$\frac{{dy}}{{dx}}\:=\:\frac{{y}−{x}+\mathrm{1}}{{y}−{x}+\mathrm{5}} \\ $$

Question Number 91149 Answers: 1 Comments: 1

A particle starts from rest and moves in a straight line on a smooth horizontal surface. Its acceleration at time t seconds is given by k(4v + 1) ms^(−2) where k is a positve constant and v ms^(−1) is the speed of the particle. Given that v = ((e^2 −1)/4) when t = 1. show that v = (1/4)(e^(2t) −1)

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{and}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\: \\ $$$$\mathrm{horizontal}\:\mathrm{surface}.\:\mathrm{Its}\:\mathrm{acceleration}\:\mathrm{at}\:\mathrm{time}\:{t}\:\mathrm{seconds}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}\left(\mathrm{4}{v}\:+\:\mathrm{1}\right)\:\mathrm{ms}^{−\mathrm{2}} \\ $$$$\mathrm{where}\:{k}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positve}\:\mathrm{constant}\:\mathrm{and}\:{v}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}. \\ $$$$\mathrm{Given}\:\mathrm{that}\:{v}\:=\:\frac{{e}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}}\:\mathrm{when}\:{t}\:=\:\mathrm{1}.\:\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{v}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left({e}^{\mathrm{2}{t}} −\mathrm{1}\right) \\ $$

Question Number 91133 Answers: 1 Comments: 2

lim_(x→1) (((x−1)+((1−x))^(1/(3 )) )/((1−x^2 ))^(1/(3 )) ) =

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left({x}−\mathrm{1}\right)+\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}−{x}}}{\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\: \\ $$

Question Number 91119 Answers: 0 Comments: 0

∫_0 ^( ∫_0 ^( k) (1 + (1/x))^x dx) sin (x^e ) dx = (π/e) k = ?

$$\: \\ $$$$\:\int_{\mathrm{0}} ^{\:\int_{\mathrm{0}} ^{\:{k}} \:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\right)^{{x}} {dx}} \:\mathrm{sin}\:\left({x}^{{e}} \right)\:{dx}\:=\:\frac{\pi}{{e}} \\ $$$$\:{k}\:=\:? \\ $$

Question Number 91099 Answers: 1 Comments: 12

Question Number 91088 Answers: 0 Comments: 0

let U_n =∫_0 ^(1/2) (dx/(√(1−x^n ))) calculate lim_(n→+∞) U_n

$${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{{n}} }}\:\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \\ $$

Question Number 91086 Answers: 1 Comments: 0

∫ ((2−x^2 )/(1+x(√(1−x^2 )))) dx

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

Question Number 91085 Answers: 0 Comments: 1

f(x) = ⌈2x−1⌉ f(3) , f(4) = ?

$${f}\left({x}\right)\:=\:\lceil\mathrm{2}{x}−\mathrm{1}\rceil\: \\ $$$${f}\left(\mathrm{3}\right)\:,\:{f}\left(\mathrm{4}\right)\:=\:? \\ $$

Question Number 91074 Answers: 1 Comments: 1

this trig integral has quite a few insights on trig integrals andd u subs as well as on the properties of logarithms. try it out it′s a nice one ∫tan(x)dx

$${this}\:{trig}\:{integral}\:{has}\:{quite}\:{a}\:{few}\: \\ $$$${insights}\:{on}\:{trig}\:{integrals}\:{andd}\: \\ $$$${u}\:{subs}\:{as}\:{well}\:{as}\:{on}\:{the}\:{properties} \\ $$$${of}\:{logarithms}.\:{try}\:{it}\:{out}\:{it}'{s}\:{a}\:{nice} \\ $$$${one} \\ $$$$\int{tan}\left({x}\right){dx} \\ $$

Question Number 91054 Answers: 1 Comments: 2

Question Number 91047 Answers: 1 Comments: 6

Find the square root of: (√7) + (√5)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root}\:\mathrm{of}:\:\:\:\:\sqrt{\mathrm{7}}\:\:+\:\:\sqrt{\mathrm{5}} \\ $$

Question Number 91038 Answers: 1 Comments: 1

if sin((α/2))=(4/5) and cos((β/2))=(3/5) prove sin(α)=cos(β)

$${if}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right)=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${and}\:{cos}\left(\frac{\beta}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${prove} \\ $$$${sin}\left(\alpha\right)={cos}\left(\beta\right) \\ $$

Question Number 91037 Answers: 0 Comments: 0

hi every one what is the scientific reason for using trigonometric compensation in integration? and what is the rule that we rely on in other compensation? (√(a^2 −x^2 ))→→x=a sin(θ) OR a cos(θ) (√(a^2 +x^2 ))→→x=a tan(θ) OR a cot(θ) (√(x^2 −a^2 ))→→x=a sec(θ) or a csc(θ)

$${hi}\:{every}\:{one} \\ $$$${what}\:{is}\:{the}\:{scientific}\:{reason}\:{for}\:{using} \\ $$$${trigonometric}\:{compensation}\: \\ $$$${in}\:{integration}? \\ $$$${and}\:{what}\:{is}\:{the}\:{rule}\:{that}\:{we}\:{rely}\:{on} \\ $$$${in}\:{other}\:{compensation}? \\ $$$$ \\ $$$$\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\rightarrow\rightarrow{x}={a}\:{sin}\left(\theta\right)\:{OR}\:{a}\:{cos}\left(\theta\right) \\ $$$$\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\rightarrow\rightarrow{x}={a}\:{tan}\left(\theta\right)\:{OR}\:{a}\:{cot}\left(\theta\right) \\ $$$$\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\rightarrow\rightarrow{x}={a}\:{sec}\left(\theta\right)\:{or}\:{a}\:{csc}\left(\theta\right) \\ $$

Question Number 91036 Answers: 1 Comments: 0

Solve: (x+a)^2 (d^2 y/dx^2 )−4(x+a)(dy/dx)+6y= x

$$\:\mathrm{Solve}: \\ $$$$\:\:\left(\mathrm{x}+\mathrm{a}\right)^{\mathrm{2}} \frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }−\mathrm{4}\left(\mathrm{x}+\mathrm{a}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{6y}=\:\mathrm{x} \\ $$

Question Number 91024 Answers: 1 Comments: 5

x^2 y′′+3xy′ +2y = 4x^2

$${x}^{\mathrm{2}} {y}''+\mathrm{3}{xy}'\:+\mathrm{2}{y}\:=\:\mathrm{4}{x}^{\mathrm{2}} \\ $$

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