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Question Number 155992 Answers: 0 Comments: 0

(1+log _3 x).(√(log _(3x) ((x/3))^(1/3) )) ≤ 2

$$\:\left(\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \:\mathrm{x}\right).\sqrt{\mathrm{log}\:_{\mathrm{3x}} \:\sqrt[{\mathrm{3}}]{\frac{\mathrm{x}}{\mathrm{3}}}}\:\leqslant\:\mathrm{2} \\ $$

Question Number 155991 Answers: 2 Comments: 1

{ ((a(x+2)+y=3a)),((a+2x^3 =y^3 +(a+2)x^3 )) :} solve for x &y in term a

$$\:\begin{cases}{\mathrm{a}\left(\mathrm{x}+\mathrm{2}\right)+\mathrm{y}=\mathrm{3a}}\\{\mathrm{a}+\mathrm{2x}^{\mathrm{3}} =\mathrm{y}^{\mathrm{3}} +\left(\mathrm{a}+\mathrm{2}\right)\mathrm{x}^{\mathrm{3}} }\end{cases} \\ $$$$\:\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\&\mathrm{y}\:\mathrm{in}\:\mathrm{term}\:\mathrm{a} \\ $$

Question Number 155989 Answers: 0 Comments: 1

5sin^2 2x + 8cos^3 x = 8cos x ((3π)/2)≤x≤2π

$$\:\:\:\:\mathrm{5sin}\:^{\mathrm{2}} \mathrm{2x}\:+\:\mathrm{8cos}\:^{\mathrm{3}} \mathrm{x}\:=\:\mathrm{8cos}\:\mathrm{x} \\ $$$$\:\:\:\:\frac{\mathrm{3}\pi}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi \\ $$

Question Number 155988 Answers: 2 Comments: 0

2^2^

$$\mathrm{2}^{\overset{} {\mathrm{2}}} \\ $$

Question Number 155986 Answers: 0 Comments: 0

Question Number 155984 Answers: 0 Comments: 0

Question Number 155983 Answers: 1 Comments: 2

given y=(√((1−x)/(1+x))) , then (1−x^2 )(dy/dx)+ky=0. find k

$$\:{given}\:\:{y}=\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:,\:{then} \\ $$$$\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}+\boldsymbol{{k}}{y}=\mathrm{0}.\:{find}\:\boldsymbol{{k}} \\ $$

Question Number 156105 Answers: 1 Comments: 0

Find The derivatif of this function: 1). y^4 +3y−4x^3 =5x+1 2). 4xy^3 −x^2 y+x^3 −5x+6=0 3). 3y^4 +4x−x^2 sin y−4=0 4). y=x^2 sin y 5). sin^2 3y=x+y−1

$$\mathrm{Find}\:\mathrm{The}\:\mathrm{derivatif}\:\mathrm{of}\:\mathrm{this}\:\mathrm{function}: \\ $$$$\left.\mathrm{1}\right).\:\:\mathrm{y}^{\mathrm{4}} +\mathrm{3y}−\mathrm{4x}^{\mathrm{3}} =\mathrm{5x}+\mathrm{1} \\ $$$$\left.\mathrm{2}\right).\:\:\mathrm{4xy}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{x}^{\mathrm{3}} −\mathrm{5x}+\mathrm{6}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right).\:\:\mathrm{3y}^{\mathrm{4}} +\mathrm{4x}−\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\mathrm{y}−\mathrm{4}=\mathrm{0} \\ $$$$\left.\mathrm{4}\right).\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\mathrm{y} \\ $$$$\left.\mathrm{5}\right).\:\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{3y}=\mathrm{x}+\mathrm{y}−\mathrm{1} \\ $$$$\: \\ $$

Question Number 155977 Answers: 0 Comments: 0

Question Number 155972 Answers: 0 Comments: 0

Question Number 155973 Answers: 1 Comments: 2

f(x)=x^3 −3x^2 +4x−1 find a=? whenever f(a)=f^(−1) (a)

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{a}=? \\ $$$$\mathrm{whenever}\:\:\:\:\mathrm{f}\left(\mathrm{a}\right)=\mathrm{f}^{−\mathrm{1}} \left(\mathrm{a}\right) \\ $$$$ \\ $$

Question Number 155969 Answers: 1 Comments: 0

Question Number 155965 Answers: 1 Comments: 0

(x sin (y/x)−y cos (y/x))dx+x cos (y/x) dy=0

$$\left({x}\:\mathrm{sin}\:\frac{{y}}{{x}}−{y}\:\mathrm{cos}\:\frac{{y}}{{x}}\right){dx}+{x}\:\mathrm{cos}\:\frac{{y}}{{x}}\:{dy}=\mathrm{0} \\ $$

Question Number 155959 Answers: 1 Comments: 0

quel est le changement de variable qui permet de passer de l′equation differentielle : x^2 y′′−3xy′+4y=0 a une equation lineaire d′ordre 2 coefficient comstant en z

$${quel}\:{est}\:{le}\:{changement}\:{de}\:{variable}\:{qui}\:{permet}\:{de}\:{passer} \\ $$$${de}\:{l}'{equation}\:{differentielle}\:: \\ $$$${x}^{\mathrm{2}} {y}''−\mathrm{3}{xy}'+\mathrm{4}{y}=\mathrm{0} \\ $$$${a}\:{une}\:{equation}\:{lineaire}\:{d}'{ordre}\:\mathrm{2}\:\:{coefficient}\:{comstant}\:\:{en}\:{z} \\ $$

Question Number 159716 Answers: 1 Comments: 0

Show that ▽r^n =nr^(n−2) r

$${Show}\:{that}\:\bigtriangledown{r}^{{n}} ={nr}^{{n}−\mathrm{2}} {r} \\ $$

Question Number 155951 Answers: 1 Comments: 0

solve : ⌊ (1/x) ⌋ + ⌊ (3/x) ⌋= 4

$$ \\ $$$$\:\:\:\:\:\:\:{solve}\:\:: \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:\frac{\mathrm{1}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{3}}{{x}}\:\rfloor=\:\mathrm{4}\: \\ $$$$ \\ $$

Question Number 155945 Answers: 1 Comments: 1