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Question Number 149309    Answers: 3   Comments: 0

if t=tanx what is the value of sinx and cosx ?

$${if}\:\:\:\:{t}={tanx}\:{what}\:{is}\:{the}\:{value}\:{of} \\ $$$${sinx}\:{and}\:{cosx}\:? \\ $$

Question Number 149296    Answers: 2   Comments: 0

if a ; b > 0 find [ (((a^2 + 4)(b^2 + 9))/(ab)) ]_(min) = ?

$${if}\:\:\:{a}\:;\:{b}\:>\:\mathrm{0} \\ $$$${find}\:\:\:\left[\:\frac{\left({a}^{\mathrm{2}} \:+\:\mathrm{4}\right)\left({b}^{\mathrm{2}} \:+\:\mathrm{9}\right)}{{ab}}\:\right]_{\boldsymbol{{min}}} =\:? \\ $$

Question Number 149292    Answers: 1   Comments: 0

if sin2𝛂 = - (1/3) find 3tg^2 (((3𝛑)/4) - 𝛂) = ?

$${if}\:\:\:{sin}\mathrm{2}\boldsymbol{\alpha}\:=\:-\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${find}\:\:\:\mathrm{3}{tg}^{\mathrm{2}} \left(\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{4}}\:-\:\boldsymbol{\alpha}\right)\:=\:? \\ $$

Question Number 149291    Answers: 2   Comments: 0

f(x) = 2sinx^2 βˆ’cos^2 xβˆ’2 f^β€² (x) = ?

$${f}\left({x}\right)\:=\:\mathrm{2}{sinx}^{\mathrm{2}} βˆ’{cos}^{\mathrm{2}} {x}βˆ’\mathrm{2} \\ $$$${f}\:^{'} \left({x}\right)\:=\:? \\ $$

Question Number 152597    Answers: 0   Comments: 0

Question Number 149275    Answers: 0   Comments: 0

Question Number 149274    Answers: 1   Comments: 0

Question Number 149273    Answers: 3   Comments: 0

∫_0 ^(Ο€/4) ((e^(tan x) sin^2 x)/(cos^4 x)) dx =?

$$\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{e}^{\mathrm{tan}\:\mathrm{x}} \:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\:\mathrm{dx}\:=? \\ $$

Question Number 149268    Answers: 4   Comments: 0

Question Number 149266    Answers: 1   Comments: 0

Solve for natural numbers: x^2 + y^2 +x + y = 3xy

$${Solve}\:{for}\:{natural}\:{numbers}: \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+{x}\:+\:{y}\:=\:\mathrm{3}{xy} \\ $$

Question Number 149271    Answers: 2   Comments: 0

Question Number 149259    Answers: 1   Comments: 0

cos^(βˆ’1) ((√(sin (41Β°)sin (19Β°)+(3/4))) )=?

$$\:\mathrm{cos}^{βˆ’\mathrm{1}} \left(\sqrt{\mathrm{sin}\:\left(\mathrm{41}Β°\right)\mathrm{sin}\:\left(\mathrm{19}Β°\right)+\frac{\mathrm{3}}{\mathrm{4}}}\:\right)=? \\ $$

Question Number 149252    Answers: 4   Comments: 0

e^y =(sin x)(cos x) find (dy/dx)

$$ \\ $$$${e}^{{y}} =\left(\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{x}\right) \\ $$$${find}\:\frac{{dy}}{{dx}} \\ $$

Question Number 149251    Answers: 2   Comments: 0

lim_(xβ†’0^+ ) ((tan x (√(tan x))βˆ’sin x (√(sin x)))/(x^3 (√x))) =?

$$\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{tan}\:\mathrm{x}}βˆ’\mathrm{sin}\:\mathrm{x}\:\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{3}} \:\sqrt{\mathrm{x}}}\:=? \\ $$$$ \\ $$

Question Number 149247    Answers: 0   Comments: 3

Question Number 149244    Answers: 0   Comments: 0

Let a,b,c be positive real numbers with sum 3. Prove that (1/a)+(1/b)+(1/c) β‰₯ (3/(2a^2 +bc))+(3/(2b^2 +ac))+(3/(2c^2 +ab))

$$\:\:{Let}\:{a},{b},{c}\:{be}\:{positive}\:{real}\:{numbers} \\ $$$${with}\:{sum}\:\mathrm{3}.\:{Prove}\:{that}\: \\ $$$$\:\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}{a}^{\mathrm{2}} +{bc}}+\frac{\mathrm{3}}{\mathrm{2}{b}^{\mathrm{2}} +{ac}}+\frac{\mathrm{3}}{\mathrm{2}{c}^{\mathrm{2}} +{ab}} \\ $$

Question Number 149241    Answers: 2   Comments: 0

Question Number 149240    Answers: 1   Comments: 0

solve for real number x^4 βˆ’3x^2 +1 =(√((4/(4βˆ’x^2 )) βˆ’1))

$$\:\:\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{number}\: \\ $$$$\:\mathrm{x}^{\mathrm{4}} βˆ’\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\:=\sqrt{\frac{\mathrm{4}}{\mathrm{4}βˆ’\mathrm{x}^{\mathrm{2}} }\:βˆ’\mathrm{1}} \\ $$$$\:\: \\ $$

Question Number 149239    Answers: 1   Comments: 0

lim_(xβ†’a^+ ) (((√x)βˆ’(√a) βˆ’(√(xβˆ’a)))/( (√(x^2 βˆ’a^2 )))) =?

$$\:\:\:\underset{{x}\rightarrow\mathrm{a}^{+} } {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}}βˆ’\sqrt{\mathrm{a}}\:βˆ’\sqrt{\mathrm{x}βˆ’\mathrm{a}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} βˆ’\mathrm{a}^{\mathrm{2}} }}\:=?\: \\ $$

Question Number 149238    Answers: 0   Comments: 1

z^3 βˆ’z=c z^4 βˆ’z^2 =cz let z=x+q x^4 +4qx^3 +6q^2 x^2 +4q^3 x+q^4 βˆ’x^2 βˆ’2qxβˆ’q^2 βˆ’cxβˆ’cq=0 β‡’ x^4 +4qx^3 +(6q^2 βˆ’1)x^2 + (4q^3 βˆ’2qβˆ’c)x+q^4 βˆ’q^2 βˆ’cq=0 let q^2 =1/6 β‡’ x^4 +4qx^3 βˆ’(((4q)/3)+c)xβˆ’(5/(36))βˆ’cq=0 ≑ (x^2 +mx+k)(x^2 +sx+h)=0 β‡’ m+s=4q h+k+ms=0 mh+ks=βˆ’(((4q)/3)+c) kh=βˆ’(5/(36))βˆ’cq h, k= βˆ’((ms)/2)Β±(√(((m^2 s^2 )/4)+(5/(36))+cq)) m(βˆ’((ms)/2)+D)+s(βˆ’((ms)/2)βˆ’D) +((4q)/3)+c=0 (mβˆ’s)Dβˆ’((ms)/2)(4q)+((4q)/3)+c=0 now letβ€²s assume ms=p & since m+s=4q (mβˆ’s)^2 =(8/3)βˆ’4p β‡’ ((8/3)βˆ’4p)((p^2 /4)+(5/(36))+cq) =(2pqβˆ’((4q)/3)βˆ’c)^2 p^3 +((5/9)+4cqβˆ’(8/9)βˆ’4cq)p +(((4q)/3)+c)^2 βˆ’(8/3)((5/(36))+cq)=0 β‡’ p^3 βˆ’(p/3)+((14)/(27))=0 D_0 =((7/(27)))^2 βˆ’((1/9))^3 ....

$${z}^{\mathrm{3}} βˆ’{z}={c} \\ $$$${z}^{\mathrm{4}} βˆ’{z}^{\mathrm{2}} ={cz} \\ $$$${let}\:\:{z}={x}+{q} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} +\mathrm{6}{q}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{3}} {x}+{q}^{\mathrm{4}} \\ $$$$βˆ’{x}^{\mathrm{2}} βˆ’\mathrm{2}{qx}βˆ’{q}^{\mathrm{2}} βˆ’{cx}βˆ’{cq}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} +\left(\mathrm{6}{q}^{\mathrm{2}} βˆ’\mathrm{1}\right){x}^{\mathrm{2}} + \\ $$$$\:\:\:\:\left(\mathrm{4}{q}^{\mathrm{3}} βˆ’\mathrm{2}{q}βˆ’{c}\right){x}+{q}^{\mathrm{4}} βˆ’{q}^{\mathrm{2}} βˆ’{cq}=\mathrm{0} \\ $$$${let}\:\:{q}^{\mathrm{2}} =\mathrm{1}/\mathrm{6}\:\:\Rightarrow \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} βˆ’\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right){x}βˆ’\frac{\mathrm{5}}{\mathrm{36}}βˆ’{cq}=\mathrm{0} \\ $$$$\equiv\:\left({x}^{\mathrm{2}} +{mx}+{k}\right)\left({x}^{\mathrm{2}} +{sx}+{h}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{m}+{s}=\mathrm{4}{q} \\ $$$$\:\:\:\:\:\:{h}+{k}+{ms}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{mh}+{ks}=βˆ’\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right) \\ $$$$\:\:\:\:\:\:{kh}=βˆ’\frac{\mathrm{5}}{\mathrm{36}}βˆ’{cq} \\ $$$${h},\:{k}=\:βˆ’\frac{{ms}}{\mathrm{2}}\pm\sqrt{\frac{{m}^{\mathrm{2}} {s}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{36}}+{cq}} \\ $$$${m}\left(βˆ’\frac{{ms}}{\mathrm{2}}+{D}\right)+{s}\left(βˆ’\frac{{ms}}{\mathrm{2}}βˆ’{D}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}=\mathrm{0} \\ $$$$\left({m}βˆ’{s}\right){D}βˆ’\frac{{ms}}{\mathrm{2}}\left(\mathrm{4}{q}\right)+\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}=\mathrm{0} \\ $$$${now}\:{let}'{s}\:{assume}\:\:{ms}={p} \\ $$$$\&\:{since}\:\:{m}+{s}=\mathrm{4}{q} \\ $$$$\left({m}βˆ’{s}\right)^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}}βˆ’\mathrm{4}{p} \\ $$$$\Rightarrow\:\:\left(\frac{\mathrm{8}}{\mathrm{3}}βˆ’\mathrm{4}{p}\right)\left(\frac{{p}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{36}}+{cq}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{2}{pq}βˆ’\frac{\mathrm{4}{q}}{\mathrm{3}}βˆ’{c}\right)^{\mathrm{2}} \\ $$$$\:\:{p}^{\mathrm{3}} +\left(\frac{\mathrm{5}}{\mathrm{9}}+\mathrm{4}{cq}βˆ’\frac{\mathrm{8}}{\mathrm{9}}βˆ’\mathrm{4}{cq}\right){p} \\ $$$$\:\:\:\:\:\:\:+\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right)^{\mathrm{2}} βˆ’\frac{\mathrm{8}}{\mathrm{3}}\left(\frac{\mathrm{5}}{\mathrm{36}}+{cq}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{3}} βˆ’\frac{{p}}{\mathrm{3}}+\frac{\mathrm{14}}{\mathrm{27}}=\mathrm{0} \\ $$$${D}_{\mathrm{0}} =\left(\frac{\mathrm{7}}{\mathrm{27}}\right)^{\mathrm{2}} βˆ’\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{3}} \\ $$$$.... \\ $$

Question Number 149237    Answers: 1   Comments: 0

Question Number 149234    Answers: 1   Comments: 0

(x/2) + log_2 (x + 12) = 6 β‡’ x = ?

$$\frac{{x}}{\mathrm{2}}\:+\:{log}_{\mathrm{2}} \left({x}\:+\:\mathrm{12}\right)\:=\:\mathrm{6} \\ $$$$\Rightarrow\:\boldsymbol{{x}}\:=\:? \\ $$

Question Number 149233    Answers: 0   Comments: 0

Question Number 149219    Answers: 2   Comments: 0

Question Number 149212    Answers: 1   Comments: 0

if x;y;z>0 and xyz=1 prove that: (x^4 /(x+yz)) + (y^4 /(y+zx)) + (z^4 /(z+xy)) β‰₯ (3/2)

$${if}\:\:\:{x};{y};{z}>\mathrm{0}\:\:\:{and}\:\:\:{xyz}=\mathrm{1}\:\:\:{prove}\:{that}: \\ $$$$\frac{{x}^{\mathrm{4}} }{{x}+{yz}}\:+\:\frac{{y}^{\mathrm{4}} }{{y}+{zx}}\:+\:\frac{{z}^{\mathrm{4}} }{{z}+{xy}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Question Number 149192    Answers: 1   Comments: 0

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