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Question Number 160701 Answers: 1 Comments: 0

Question Number 160695 Answers: 0 Comments: 0

Question Number 160694 Answers: 1 Comments: 0

Prove that 1 + (1/( (√2))) + (1/( (√3))) + …+ (1/( (√n))) < 2(√n)

$${Prove}\:\:{that}\:\: \\ $$$$\:\:\:\:\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\:\ldots+\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\:<\:\mathrm{2}\sqrt{{n}} \\ $$

Question Number 160689 Answers: 1 Comments: 0

Question Number 160682 Answers: 0 Comments: 3

Question Number 160677 Answers: 1 Comments: 1

{ ((((x+y)/(xyz)) = −(1/4))),((((y+z)/(xyz)) = −(1/(24)))),((((x+z)/(xyz)) = (1/(24)))) :}

$$\:\:\:\begin{cases}{\frac{\mathrm{x}+\mathrm{y}}{\mathrm{xyz}}\:=\:−\frac{\mathrm{1}}{\mathrm{4}}}\\{\frac{\mathrm{y}+\mathrm{z}}{\mathrm{xyz}}\:=\:−\frac{\mathrm{1}}{\mathrm{24}}}\\{\frac{\mathrm{x}+\mathrm{z}}{\mathrm{xyz}}\:=\:\frac{\mathrm{1}}{\mathrm{24}}}\end{cases}\: \\ $$$$\:\: \\ $$

Question Number 160672 Answers: 3 Comments: 1

Calculate lim_(x→0) ((tgx^m )/((sin x)^n )), lim_(x→0) ((xcos x−x)/((e^x −1)ln (1+3x^2 )))

$${Calculate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tgx}^{{m}} }{\left(\mathrm{sin}\:{x}\right)^{{n}} },\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\mathrm{cos}\:{x}−{x}}{\left({e}^{{x}} −\mathrm{1}\right)\mathrm{ln}\:\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)} \\ $$

Question Number 160671 Answers: 0 Comments: 0

please ∫((lnx)/(x+lnx))dx calculate.

$${please}\:\:\int\frac{{lnx}}{{x}+{lnx}}{dx}\:\:\:{calculate}. \\ $$

Question Number 160665 Answers: 0 Comments: 2

In how many ways can you divide 20 students into 5 groups with 4 students in each group such that any two students don′t meet each other in more than one group.

$${In}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{divide}\:\mathrm{20} \\ $$$${students}\:{into}\:\mathrm{5}\:{groups}\:{with}\:\mathrm{4}\:{students} \\ $$$${in}\:{each}\:{group}\:{such}\:{that}\:{any}\:{two} \\ $$$${students}\:{don}'{t}\:{meet}\:{each}\:{other}\:{in} \\ $$$${more}\:{than}\:{one}\:{group}. \\ $$

Question Number 160661 Answers: 0 Comments: 0

Question Number 160659 Answers: 1 Comments: 0

Given that y′′−4y′+3y=0 and y(0)=0, y′(0)=2, find y(ln 2).

$$\mathrm{Given}\:\mathrm{that}\:{y}''−\mathrm{4}{y}'+\mathrm{3}{y}=\mathrm{0}\:\mathrm{and}\:{y}\left(\mathrm{0}\right)=\mathrm{0}, \\ $$$${y}'\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{find}\:{y}\left(\mathrm{ln}\:\mathrm{2}\right). \\ $$

Question Number 160658 Answers: 1 Comments: 3

Find: lim_(x→0) 4x [ (1/(4x)) ] = ?

$$\boldsymbol{\mathrm{Find}}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{4}\boldsymbol{\mathrm{x}}\:\left[\:\frac{\mathrm{1}}{\mathrm{4}\boldsymbol{\mathrm{x}}}\:\right]\:=\:? \\ $$$$ \\ $$

Question Number 160654 Answers: 1 Comments: 0

Question Number 160649 Answers: 0 Comments: 0

les digonales d′un quadrile^ re ABCD (inscriptible) se coupent en O on note AB=a ; BC=b ; CD=c ; AD=d ; OA=e OB=f ; Oc=g ; OD=h montrer que: (a/c)+(c/a)+(b/d)+(d/b)≤(e/g)+(g/e)+(f/h)+(h/f)

$${les}\:{digonales}\:{d}'{un}\:{quadril}\grave {{e}re}\:{ABCD} \\ $$$$\left({inscriptible}\right)\:{se}\:{coupent}\:{en}\:{O}\:{on}\:{note} \\ $$$${AB}={a}\:;\:{BC}={b}\:;\:{CD}={c}\:;\:{AD}={d}\:;\:{OA}={e} \\ $$$${OB}={f}\:;\:{Oc}={g}\:;\:{OD}={h} \\ $$$${montrer}\:{que}: \\ $$$$\frac{{a}}{{c}}+\frac{{c}}{{a}}+\frac{{b}}{{d}}+\frac{{d}}{{b}}\leqslant\frac{{e}}{{g}}+\frac{{g}}{{e}}+\frac{{f}}{{h}}+\frac{{h}}{{f}} \\ $$

Question Number 160645 Answers: 2 Comments: 1

∫ sin^8 x dx =?

$$\:\int\:\mathrm{sin}\:^{\mathrm{8}} \mathrm{x}\:\mathrm{dx}\:=? \\ $$

Question Number 160644 Answers: 0 Comments: 0

King′s inequality if a≤b ; n≤m then: ((a^n + b^m )/2) ≥ (((1 + b)/2))^(m-n) ∙ (((a + b)/2))^n

$$\mathrm{King}'\mathrm{s}\:\mathrm{inequality} \\ $$$$\mathrm{if}\:\:\:\mathrm{a}\leqslant\mathrm{b}\:\:\:;\:\:\:\mathrm{n}\leqslant\mathrm{m}\:\:\:\mathrm{then}: \\ $$$$\frac{\mathrm{a}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{b}^{\boldsymbol{\mathrm{m}}} }{\mathrm{2}}\:\geqslant\:\left(\frac{\mathrm{1}\:+\:\mathrm{b}}{\mathrm{2}}\right)^{\boldsymbol{\mathrm{m}}-\boldsymbol{\mathrm{n}}} \:\centerdot\:\left(\frac{\mathrm{a}\:+\:\mathrm{b}}{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$

Question Number 160638 Answers: 0 Comments: 0