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Question Number 72343    Answers: 0   Comments: 3

given that y = ln ( 1 + cos^2 x) find (dy/(dx )) at the point x = ((3π)/4) and if y =ln(x^2 + 4) find (dy/dx) at x = 1

$${given}\:{that}\:{y}\:=\:{ln}\:\left(\:\mathrm{1}\:+\:{cos}^{\mathrm{2}} {x}\right)\:{find}\:\frac{{dy}}{{dx}\:\:}\:{at}\:{the}\:{point}\:\:{x}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$${and}\:\:{if}\:\:{y}\:={ln}\left({x}^{\mathrm{2}} \:+\:\mathrm{4}\right)\:{find}\:\:\frac{{dy}}{{dx}}\:{at}\:{x}\:=\:\mathrm{1} \\ $$

Question Number 72232    Answers: 0   Comments: 6

to all those who deleted their posts after they had been answered: I will not answer you anymore this forum had been great but lately it has been filling with unpolite people I′m not a freebie solver for anybody′s homework

$$\mathrm{to}\:\mathrm{all}\:\mathrm{those}\:\mathrm{who}\:\mathrm{deleted}\:\mathrm{their}\:\mathrm{posts}\:\mathrm{after}\:\mathrm{they} \\ $$$$\mathrm{had}\:\mathrm{been}\:\mathrm{answered}:\:\mathrm{I}\:\mathrm{will}\:\mathrm{not}\:\mathrm{answer}\:\mathrm{you} \\ $$$$\mathrm{anymore} \\ $$$$\mathrm{this}\:\mathrm{forum}\:\mathrm{had}\:\mathrm{been}\:\mathrm{great}\:\mathrm{but}\:\mathrm{lately}\:\mathrm{it}\:\mathrm{has} \\ $$$$\mathrm{been}\:\mathrm{filling}\:\mathrm{with}\:\mathrm{unpolite}\:\mathrm{people} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{a}\:\mathrm{freebie}\:\mathrm{solver}\:\mathrm{for}\:\mathrm{anybody}'\mathrm{s}\:\mathrm{homework} \\ $$

Question Number 72153    Answers: 1   Comments: 3

∫((sin^3 x)/(√(cos x)))dx ∫sin(lnx)dx

$$\int\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}}{\sqrt{\mathrm{cos}\:\mathrm{x}}}\mathrm{dx} \\ $$$$\int\mathrm{sin}\left(\mathrm{lnx}\right)\mathrm{dx} \\ $$

Question Number 71852    Answers: 1   Comments: 0

show that 5^(22) + 17^(22) ≡ 6 (mod 11)

$${show}\:{that}\:\mathrm{5}^{\mathrm{22}} \:+\:\mathrm{17}^{\mathrm{22}} \:\equiv\:\mathrm{6}\:\left({mod}\:\mathrm{11}\right) \\ $$

Question Number 71838    Answers: 1   Comments: 3

solve the system of linear congruences x ≡ 2 (mod 3) x ≡ 4(mod 5) x ≡ 7 (mod 9) x≡ 11( mod 13) using the Brute force method

$${solve}\:{the}\:{system}\:{of}\:{linear}\:{congruences}\: \\ $$$$\:{x}\:\equiv\:\mathrm{2}\:\left({mod}\:\mathrm{3}\right) \\ $$$${x}\:\equiv\:\mathrm{4}\left({mod}\:\mathrm{5}\right) \\ $$$${x}\:\equiv\:\mathrm{7}\:\left({mod}\:\mathrm{9}\right) \\ $$$${x}\equiv\:\mathrm{11}\left(\:{mod}\:\mathrm{13}\right) \\ $$$${using}\:{the}\:{Brute}\:{force}\:{method} \\ $$

Question Number 71564    Answers: 1   Comments: 2

(z−i)^4 =−7+24i

$$\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{4}} =−\mathrm{7}+\mathrm{24i} \\ $$

Question Number 71432    Answers: 0   Comments: 2

is 1(5/(3 )) an example of a mixed fraction

$${is}\:\mathrm{1}\frac{\mathrm{5}}{\mathrm{3}\:}\:{an}\:{example}\:{of}\:{a}\:{mixed}\:{fraction} \\ $$

Question Number 71326    Answers: 2   Comments: 0

(−64)^(1/6) =?(Is there any short cut for mcq)

$$\left(−\mathrm{64}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} =?\left(\boldsymbol{\mathrm{I}}\mathrm{s}\:\mathrm{there}\:\mathrm{any}\:\mathrm{short}\:\mathrm{cut}\:\mathrm{for}\:\mathrm{mcq}\right) \\ $$

Question Number 71235    Answers: 2   Comments: 1

sinh[ln (x + (√(1 + x^2 ))) ] ≡ A. 2x B. (1/x) C. x^2 D. x

$${sinh}\left[{ln}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\:\right]\:\equiv\: \\ $$$$ \\ $$$${A}.\:\:\mathrm{2}{x} \\ $$$${B}.\:\:\frac{\mathrm{1}}{{x}} \\ $$$${C}.\:\:{x}^{\mathrm{2}} \\ $$$${D}.\:\:{x} \\ $$

Question Number 70914    Answers: 2   Comments: 3

1+(z+2i)+(z+2i)^2 +(z+2i)^3 +(z+2i)^4 =0 find z , z∈C

$$\mathrm{1}+\left(\mathrm{z}+\mathrm{2i}\right)+\left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{2}} +\left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{3}} +\left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{4}} =\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{z}\:,\:\mathrm{z}\in\mathrm{C} \\ $$

Question Number 70397    Answers: 0   Comments: 1

Question Number 70310    Answers: 3   Comments: 1

please help me find the term independent of x in the expansion of (x + (3/x))^(−12 )

$${please}\:{help}\:{me}\:{find}\:{the}\:{term}\:{independent}\:{of}\:{x} \\ $$$${in}\:{the}\:{expansion}\:{of}\: \\ $$$$\:\:\:\:\:\:\left({x}\:+\:\frac{\mathrm{3}}{{x}}\right)^{−\mathrm{12}\:} \\ $$

Question Number 70132    Answers: 1   Comments: 1

Σ_(n=1) ^(3050) i^n

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{3050}} {\sum}}\:{i}^{{n}} \\ $$

Question Number 70069    Answers: 1   Comments: 2

Π_(n=1) ^5 (((12n−2)^4 +18^2 )/((12n−8)^4 +18^2 )) =(((10^4 +324)(22^4 +324)(34^4 +324)(46^4 +324)(58^4 +324))/((4^4 +324)(16^4 +324)(28^4 +324)(40^4 +324)(52^4 +324)))

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{5}} {\prod}}\frac{\left(\mathrm{12}{n}−\mathrm{2}\right)^{\mathrm{4}} +\mathrm{18}^{\mathrm{2}} }{\left(\mathrm{12}{n}−\mathrm{8}\right)^{\mathrm{4}} +\mathrm{18}^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{10}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{22}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{34}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{46}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{58}^{\mathrm{4}} +\mathrm{324}\right)}{\left(\mathrm{4}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{16}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{28}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{40}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{52}^{\mathrm{4}} +\mathrm{324}\right)} \\ $$

Question Number 70035    Answers: 0   Comments: 4

Question Number 70030    Answers: 1   Comments: 0

Find the convergence of Σ_(n=1) ^∞ (((1/n) + 1)/(−n^2 ))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{convergence}\:\mathrm{of} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\frac{\mathrm{1}}{{n}}\:+\:\mathrm{1}}{−{n}^{\mathrm{2}} } \\ $$

Question Number 70051    Answers: 2   Comments: 0

((a+b)/c)=((cos(((a−b)/2)))/(cos(c/2)))

$$\frac{{a}+{b}}{{c}}=\frac{{cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)}{{cos}\frac{{c}}{\mathrm{2}}} \\ $$

Question Number 70022    Answers: 0   Comments: 1

Find all pairs of (p, q) integer(s) such that p^3 − q^5 = (p + q)^2

$${Find}\:\:\:{all}\:\:{pairs}\:\:{of}\:\:\:\left({p},\:{q}\right)\:\:{integer}\left({s}\right)\:\:{such}\:\:{that} \\ $$$${p}^{\mathrm{3}} \:−\:{q}^{\mathrm{5}} \:\:=\:\:\left({p}\:+\:{q}\right)^{\mathrm{2}} \\ $$

Question Number 70017    Answers: 0   Comments: 2

Solution- log_8 x+log_4 x+log_2 x=11 ⇒(1/(log_x 8))+(1/(log_x 4))+(1/(log_x 2))=11 ⇒(1/(log_x 2^3 ))+(1/(log_x 2^2 ))+(1/(log_x 2))=11 ⇒(1/(3log_x 2))+(1/(2log_x 2))+(1/(log_x 2))=11 ⇒((1/3)+(1/2)+1)(1/(log_x 2))=11 ⇒((11)/6)×(1/(log_x 2))=11 ⇒(1/(log_x 2))=11×(6/(11)) ⇒log_2 x=6 ⇒x=2^6 ∴x=64 is this rule correct????

$$\mathrm{Solution}- \\ $$$$\mathrm{log}_{\mathrm{8}} \mathrm{x}+\mathrm{log}_{\mathrm{4}} \mathrm{x}+\mathrm{log}_{\mathrm{2}} \mathrm{x}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{8}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{4}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3log}_{\mathrm{x}} \mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2log}_{\mathrm{x}} \mathrm{2}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{11}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11}×\frac{\mathrm{6}}{\mathrm{11}} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{2}} \mathrm{x}=\mathrm{6} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}^{\mathrm{6}} \\ $$$$\therefore\mathrm{x}=\mathrm{64} \\ $$$$ \\ $$$$\mathrm{is}\:\mathrm{this}\:\mathrm{rule}\:\mathrm{correct}???? \\ $$

Question Number 69894    Answers: 0   Comments: 3

∫ ((2x^5 −x)/(x^3 −2))dx

$$\int\:\frac{\mathrm{2}{x}^{\mathrm{5}} −{x}}{{x}^{\mathrm{3}} −\mathrm{2}}{dx} \\ $$

Question Number 69829    Answers: 2   Comments: 2

Question Number 69778    Answers: 2   Comments: 5

prove that the equation (b^2 −4ac)x^2 + 4(a + c)x −4 = 0 is always real.

$${prove}\:{that}\:{the}\:{equation}\: \\ $$$$\:\:\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right){x}^{\mathrm{2}} \:+\:\mathrm{4}\left({a}\:+\:{c}\right){x}\:−\mathrm{4}\:=\:\mathrm{0}\:{is}\:{always}\:{real}. \\ $$

Question Number 69766    Answers: 0   Comments: 4

find (dy/dx) at the point (0,3) when 2x^2 y + y + 4xy^2 = 2x + 3

$${find}\:\:\frac{{dy}}{{dx}}\:\:{at}\:{the}\:{point}\:\:\left(\mathrm{0},\mathrm{3}\right)\:\:{when}\:\:\mathrm{2}{x}^{\mathrm{2}} {y}\:+\:{y}\:+\:\mathrm{4}{xy}^{\mathrm{2}} \:=\:\mathrm{2}{x}\:+\:\mathrm{3}\: \\ $$

Question Number 69765    Answers: 1   Comments: 0

Given that y = (√(5x^2 + 3)) , show that when x^2 = (6/5) , (d^2 y/dx^(2 ) ) = ((125)/8)

$${Given}\:{that}\:\:{y}\:=\:\sqrt{\mathrm{5}{x}^{\mathrm{2}} \:+\:\mathrm{3}}\:,\:{show}\:{that}\:\:{when}\:{x}^{\mathrm{2}} \:=\:\frac{\mathrm{6}}{\mathrm{5}}\:,\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}\:} }\:=\:\frac{\mathrm{125}}{\mathrm{8}} \\ $$

Question Number 69764    Answers: 1   Comments: 1

find (dy/dx) if x = sin^2 t and y= tan t at t = (π/4)

$${find}\:\:\:\frac{{dy}}{{dx}}\:\:{if}\:\:{x}\:=\:{sin}^{\mathrm{2}} {t}\:\:{and}\:\:{y}=\:{tan}\:{t}\:{at}\:\:{t}\:=\:\frac{\pi}{\mathrm{4}} \\ $$

Question Number 69763    Answers: 1   Comments: 2

find (dy/dx) if y = 3^x e^(2x + 1) , at x =1

$${find}\:\frac{{dy}}{{dx}}\:\:{if}\:\:{y}\:=\:\mathrm{3}^{{x}} {e}^{\mathrm{2}{x}\:+\:\mathrm{1}} ,\:{at}\:{x}\:=\mathrm{1} \\ $$

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