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AllQuestion and Answers: Page 1039

Question Number 107621 Answers: 0 Comments: 1

Question Number 107618 Answers: 1 Comments: 0

Question Number 107617 Answers: 1 Comments: 0

Question Number 107609 Answers: 2 Comments: 0

Question Number 107598 Answers: 0 Comments: 3

Calculate: (√(1 + (√(2 + (√(3 + (√(4 + (√(5 + ...))))))))))

$$\:\:\:\:\mathrm{Calculate}:\:\:\sqrt{\mathrm{1}\:+\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{3}\:+\:\sqrt{\mathrm{4}\:+\:\sqrt{\mathrm{5}\:+\:...}}}}} \\ $$

Question Number 107596 Answers: 2 Comments: 0

Given I_(m,n) = ∫_1 ^e x^m (ln x)^n dx where m,n ∈ N^∗ Show that (1 + m)I_(m,n) = e^(m+1) −nI_(m,n−1) for m >0 and n>0 also, evaluate I_(2,3)

$$\mathrm{Given}\: \\ $$$$\:{I}_{{m},{n}} \:=\:\underset{\mathrm{1}} {\overset{{e}} {\int}}{x}^{{m}} \:\left(\mathrm{ln}\:{x}\right)^{{n}} \:{dx}\:\mathrm{where}\:{m},{n}\:\in\:\mathbb{N}^{\ast} \\ $$$$\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{1}\:+\:{m}\right){I}_{{m},{n}} \:=\:{e}^{{m}+\mathrm{1}} −{nI}_{{m},{n}−\mathrm{1}} \:\mathrm{for}\:{m}\:>\mathrm{0}\:\mathrm{and}\:{n}>\mathrm{0} \\ $$$$\mathrm{also},\:\mathrm{evaluate}\:{I}_{\mathrm{2},\mathrm{3}} \\ $$

Question Number 107594 Answers: 2 Comments: 0

Find the greatest common divisor of 1122 and 1001 and express the greatest common divisor d in the form. d = 1122x + 1001y Using the above result solve the congruence equation 37x ≡ 11 (mod 33)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{common}\:\mathrm{divisor}\:\mathrm{of}\:\mathrm{1122}\:\mathrm{and}\:\mathrm{1001}\:\mathrm{and}\: \\ $$$$\mathrm{express}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{common}\:\mathrm{divisor}\:{d}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}. \\ $$$$\:\:{d}\:=\:\mathrm{1122}{x}\:+\:\mathrm{1001}{y} \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{above}\:\mathrm{result}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{congruence}\:\mathrm{equation} \\ $$$$\:\mathrm{37}{x}\:\equiv\:\mathrm{11}\:\left(\mathrm{mod}\:\mathrm{33}\right) \\ $$

Question Number 107591 Answers: 2 Comments: 2

Question Number 107589 Answers: 2 Comments: 0

Question Number 107586 Answers: 0 Comments: 2

App Updates: v2.135 • fix for background color problems • new drawing tools added in build and edit menu add equality marker to line etc • A new drawling tool to draw smooth curves.

$$\mathrm{App}\:\mathrm{Updates}:\:\mathrm{v2}.\mathrm{135} \\ $$$$\bullet\:\mathrm{fix}\:\mathrm{for}\:\mathrm{background}\:\mathrm{color}\:\mathrm{problems} \\ $$$$\bullet\:\mathrm{new}\:\mathrm{drawing}\:\mathrm{tools}\:\mathrm{added}\:\mathrm{in} \\ $$$$\:\:\:\mathrm{build}\:\mathrm{and}\:\mathrm{edit}\:\mathrm{menu} \\ $$$$\:\:\:\mathrm{add}\:\mathrm{equality}\:\mathrm{marker}\:\mathrm{to}\:\mathrm{line}\:\mathrm{etc} \\ $$$$\bullet\:\mathrm{A}\:\mathrm{new}\:\mathrm{drawling}\:\mathrm{tool}\:\mathrm{to}\:\mathrm{draw} \\ $$$$\:\:\:\:\mathrm{smooth}\:\mathrm{curves}. \\ $$

Question Number 107572 Answers: 1 Comments: 1

Question Number 107571 Answers: 0 Comments: 0

Given 0<x≤(π/2), 0<y≤(π/2), Let z_1 =((cos x)/(sin y))+((cos y)/(sin x)) i ,and ∣z_1 ∣=2 ; If z_2 =(√x)+(√(y ))i ,then what is the maximum value of ∣z_1 −z_2 ∣.

$${G}\mathrm{iven}\:\mathrm{0}<\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}},\:\mathrm{0}<\mathrm{y}\leqslant\frac{\pi}{\mathrm{2}}, \\ $$$$\mathrm{Let}\:{z}_{\mathrm{1}} =\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{y}}+\frac{\mathrm{cos}\:{y}}{\mathrm{sin}\:{x}}\:{i}\:,\mathrm{and}\:\mid{z}_{\mathrm{1}} \mid=\mathrm{2}\:; \\ $$$$\mathrm{If}\:{z}_{\mathrm{2}} =\sqrt{{x}}+\sqrt{{y}\:}{i}\:,\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid. \\ $$

Question Number 107567 Answers: 3 Comments: 0

∫(√(3x^2 −2x)) dx

$$\int\sqrt{\mathrm{3x}^{\mathrm{2}} −\mathrm{2x}}\:\mathrm{dx} \\ $$

Question Number 107555 Answers: 2 Comments: 0

Question Number 107550 Answers: 5 Comments: 0

L=lim_(x→0) (2^(x−1) +(1/2))^(1/x) = ?

$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{2}^{{x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}/{x}} \:\:=\:? \\ $$

Question Number 107547 Answers: 1 Comments: 6

Question Number 107537 Answers: 0 Comments: 3

Question Number 107536 Answers: 2 Comments: 1

♠BeMath★ Without L′Hopital and series find lim_(x→0) ((tan x−x)/x^3 )

$$\:\spadesuit\mathcal{B}{e}\mathcal{M}{ath}\bigstar \\ $$$${Without}\:{L}'{Hopital}\:{and}\:{series}\: \\ $$$${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−{x}}{{x}^{\mathrm{3}} } \\ $$

Question Number 107534 Answers: 5 Comments: 2

⊨BeMath⊨ sin^6 x + cos^6 x = (7/(16)) ; x ∈ (0,(π/2))

$$\:\:\:\vDash\mathcal{B}{e}\mathcal{M}{ath}\vDash \\ $$$$\:\mathrm{sin}\:^{\mathrm{6}} {x}\:+\:\mathrm{cos}\:^{\mathrm{6}} {x}\:=\:\frac{\mathrm{7}}{\mathrm{16}}\:;\:{x}\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$

Question Number 107533 Answers: 2 Comments: 0

The position as a function of time x(t) for a particle in motion is given as x(t) = (3 m/s^2 )t^2 . Find the velocity of this particle as a function of time.

$$\mathrm{The}\:\mathrm{position}\:\:\mathrm{as}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:{x}\left({t}\right)\:\mathrm{for}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{in} \\ $$$$\mathrm{motion}\:\mathrm{is}\:\mathrm{given}\:\mathrm{as}\:\:{x}\left({t}\right)\:=\:\left(\mathrm{3}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right){t}^{\mathrm{2}} \:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{velocity} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{particle}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}. \\ $$

Question Number 107521 Answers: 1 Comments: 0

Question Number 107515 Answers: 2 Comments: 0

∫(x^4 /(1+x^8 ))dx

$$\int\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{8}} }{dx} \\ $$

Question Number 107512 Answers: 0 Comments: 3

I seen that the function insert separator (right and left)don′t work,ask Titurkara fix this problem!

$$\mathrm{I}\:\mathrm{seen}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{insert}\:\mathrm{separator}\:\left(\mathrm{right}\:\mathrm{and}\:\mathrm{left}\right)\mathrm{don}'\mathrm{t}\: \\ $$$$\mathrm{work},\mathrm{ask}\:\mathrm{Titurkara}\:\mathrm{fix}\:\mathrm{this}\:\mathrm{problem}! \\ $$

Question Number 107516 Answers: 3 Comments: 0

⊚BeMath⊚ lim_(t→0) ((1−(√(cos 2t)))/(sin ((π/2)−t)−cos 2t)) ?

$$\:\:\:\:\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc \\ $$$$\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{2}{t}}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{t}\right)−\mathrm{cos}\:\mathrm{2}{t}}\:? \\ $$

Question Number 107500 Answers: 1 Comments: 0

Given a ∈R−{±1} 1. Show that ∀x∈R 1−2acos(x)+a^2 >0 2. Show that; Π_(k=1) ^n (1−2acos(((2kπ)/n))+a^2 )=Π_(k=1) ^n (a−e^(2ikπ/n) )(a−e^(−2ikπ/n) ) 3. Deduce that; Π_(k=1) ^n (1−2acos(((2kπ)/n))+a^2 )=(a^n −1)^2 4. Using Reimann′s sum, calculate I=∫_0 ^(2π) ln(1−2acos(x)+a^2 )dx

$$\mathrm{Given}\:\mathrm{a}\:\in\mathbb{R}−\left\{\pm\mathrm{1}\right\} \\ $$$$\mathrm{1}.\:\mathrm{Show}\:\mathrm{that}\:\forall\mathrm{x}\in\mathbb{R}\:\mathrm{1}−\mathrm{2acos}\left(\mathrm{x}\right)+\mathrm{a}^{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{2}.\:\mathrm{Show}\:\mathrm{that}; \\ $$$$\:\:\:\:\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\mathrm{2acos}\left(\frac{\mathrm{2k}\pi}{\mathrm{n}}\right)+\mathrm{a}^{\mathrm{2}} \right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{a}−\mathrm{e}^{\mathrm{2ik}\pi/\mathrm{n}} \right)\left(\mathrm{a}−\mathrm{e}^{−\mathrm{2ik}\pi/\mathrm{n}} \right) \\ $$$$\mathrm{3}.\:\mathrm{Deduce}\:\mathrm{that}; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\mathrm{2acos}\left(\frac{\mathrm{2k}\pi}{\mathrm{n}}\right)+\mathrm{a}^{\mathrm{2}} \right)=\left(\mathrm{a}^{\mathrm{n}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}.\:\:\mathrm{Using}\:\mathrm{Reimann}'\mathrm{s}\:\mathrm{sum},\:\mathrm{calculate} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{ln}\left(\mathrm{1}−\mathrm{2acos}\left(\mathrm{x}\right)+\mathrm{a}^{\mathrm{2}} \right)\mathrm{dx} \\ $$

Question Number 107498 Answers: 1 Comments: 0

Show that Π_(k=1) ^n (a−e^((2ikπ)/n) )(a−e^(−((2ikπ)/n)) )=(a^n −1)^2

$$\mathrm{Show}\:\mathrm{that}\: \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{a}−\mathrm{e}^{\frac{\mathrm{2}{i}\mathrm{k}\pi}{\mathrm{n}}} \right)\left(\mathrm{a}−\mathrm{e}^{−\frac{\mathrm{2}{i}\mathrm{k}\pi}{\mathrm{n}}} \right)=\left(\mathrm{a}^{\mathrm{n}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$

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