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Question Number 104220 Answers: 3 Comments: 1

∫_0 ^π ((x^2 cos x)/((1+sin x)^2 )) dx ?

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{x}^{\mathrm{2}} \mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:{dx}\:?\: \\ $$

Question Number 104218 Answers: 1 Comments: 1

lim_(x→1) [((ln (1+x)+Σ_(n = 1) ^∞ [(((1+x^2^n ))/((1−x^2^n )))]^2^(−n) )/(ln ((1/(1−x)))))]

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)+\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right)}{\left(\mathrm{1}−{x}^{\mathrm{2}^{{n}} } \right)}\right]^{\mathrm{2}^{−{n}} } }{\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)}\right] \\ $$$$ \\ $$

Question Number 104217 Answers: 2 Comments: 0

lim_(x→0) ((tan (x+3)^2 −tan (9))/x) = ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{tan}\:\left(\mathrm{9}\right)}{{x}}\:=\:? \\ $$

Question Number 104215 Answers: 1 Comments: 0

lim_(x→1) ((9^3^(2ln x) −9^2^(ln x) )/(ln x)) =?

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{9}^{\mathrm{3}^{\mathrm{2ln}\:{x}} } −\mathrm{9}^{\mathrm{2}^{\mathrm{ln}\:{x}} } }{\mathrm{ln}\:{x}}\:=? \\ $$

Question Number 104964 Answers: 0 Comments: 0

Question Number 104207 Answers: 2 Comments: 0

Question Number 104199 Answers: 2 Comments: 0

solve for real values of x the equation 4(3^(2x+1) )+17(3^x )=7. if m and n are positive real numbers other than 1, show that the log_n m+log_(1/m) n=0

$${solve}\:{for}\:{real}\:{values}\:{of}\:{x}\:{the}\:{equation} \\ $$$$\mathrm{4}\left(\mathrm{3}^{\mathrm{2}{x}+\mathrm{1}} \right)+\mathrm{17}\left(\mathrm{3}^{{x}} \right)=\mathrm{7}. \\ $$$${if}\:{m}\:{and}\:{n}\:{are}\:{positive}\:{real}\:{numbers}\:{other} \\ $$$${than}\:\mathrm{1},\:{show}\:{that}\:{the}\:\mathrm{log}_{{n}} {m}+\mathrm{log}_{\frac{\mathrm{1}}{{m}}} {n}=\mathrm{0} \\ $$

Question Number 104197 Answers: 2 Comments: 0

calculate ∫_5 ^(+∞) (dx/((x^2 −9)^4 ))

$$\mathrm{calculate}\:\int_{\mathrm{5}} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{4}} } \\ $$

Question Number 104201 Answers: 0 Comments: 0

If a^2 + b^2 + c^2 = 1 and b + ic = (1 + a)z, then show that ((a + ib)/(i + c)) = ((1 + iz)/(1 − iz))

$$\mathrm{If}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:\:=\:\:\mathrm{1}\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\mathrm{b}\:\:+\:\:\mathrm{ic}\:\:=\:\:\left(\mathrm{1}\:\:+\:\:\mathrm{a}\right)\mathrm{z}, \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\:\:\:\:\:\frac{\mathrm{a}\:\:+\:\:\mathrm{ib}}{\mathrm{i}\:\:+\:\:\mathrm{c}}\:\:\:=\:\:\:\frac{\mathrm{1}\:\:+\:\:\mathrm{iz}}{\mathrm{1}\:\:−\:\:\mathrm{iz}} \\ $$

Question Number 104191 Answers: 1 Comments: 0

Given (E):x^4 −10x^2 +q=0 U=x^2 so (E_((U)) )=U^2 −40U+q=0 . We suppose that E_((U)) has two roots such as r_1 <r_(2 ) . We give also that r_1 +r_2 =40 and r_1 ×r_2 =q. 1)The equation E has four positive solution. Determinate them such that these solutions form an arithmetical progression.

$${Given}\:\left({E}\right):{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +{q}=\mathrm{0} \\ $$$${U}={x}^{\mathrm{2}} \:{so}\:\left({E}_{\left({U}\right)} \right)={U}^{\mathrm{2}} −\mathrm{40}{U}+{q}=\mathrm{0}\:. \\ $$$${We}\:{suppose}\:{that}\:{E}_{\left({U}\right)} {has}\:{two}\:{roots} \\ $$$${such}\:{as}\:{r}_{\mathrm{1}} <{r}_{\mathrm{2}\:} . \\ $$$${We}\:{give}\:{also}\:{that}\:{r}_{\mathrm{1}} +{r}_{\mathrm{2}} =\mathrm{40}\:{and} \\ $$$${r}_{\mathrm{1}} ×{r}_{\mathrm{2}} ={q}. \\ $$$$\left.\mathrm{1}\right){The}\:{equation}\:{E}\:{has}\:{four}\:{positive} \\ $$$${solution}.\:{Determinate}\:{them}\:{such} \\ $$$${that}\:{these}\:{solutions}\:{form}\:{an}\: \\ $$$${arithmetical}\:{progression}. \\ $$$$ \\ $$

Question Number 104190 Answers: 3 Comments: 1

integers x,y satisfy 2x+15y=2019. find the minimum of ∣y−x∣.

$${integers}\:{x},{y}\:{satisfy}\:\mathrm{2}{x}+\mathrm{15}{y}=\mathrm{2019}. \\ $$$${find}\:{the}\:{minimum}\:{of}\:\mid{y}−{x}\mid. \\ $$

Question Number 104187 Answers: 2 Comments: 0

Given: (E): (m+1)x^2 +(m−2)x+1=0 We suppose that it has two roots x_1 and x_2 . Determinate m such that: x_1 =1+x_2

$${Given}: \\ $$$$\left({E}\right):\:\left({m}+\mathrm{1}\right){x}^{\mathrm{2}} +\left({m}−\mathrm{2}\right){x}+\mathrm{1}=\mathrm{0} \\ $$$${We}\:{suppose}\:{that}\:{it}\:{has}\:{two}\:{roots}\:{x}_{\mathrm{1}} \\ $$$${and}\:{x}_{\mathrm{2}} . \\ $$$${Determinate}\:{m}\:{such}\:{that}: \\ $$$${x}_{\mathrm{1}} =\mathrm{1}+{x}_{\mathrm{2}} \\ $$

Question Number 104181 Answers: 1 Comments: 0

Given P(x)=x^4 +2x^3 −41x^2 +42x+360 Determinate Q(x) a quadratic poly− nom such that: P(x)=(Q(x))^2 −42(Q(x))+360

$${Given} \\ $$$${P}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{42}{x}+\mathrm{360} \\ $$$${Determinate}\:{Q}\left({x}\right)\:{a}\:{quadratic}\:{poly}− \\ $$$${nom}\:{such}\:{that}: \\ $$$${P}\left({x}\right)=\left({Q}\left({x}\right)\right)^{\mathrm{2}} −\mathrm{42}\left({Q}\left({x}\right)\right)+\mathrm{360} \\ $$

Question Number 104180 Answers: 0 Comments: 0

Question Number 104178 Answers: 3 Comments: 0

Solve in R a) 3∣x−(√3)∣−8(√((∣x−(√3)∣)+4))=0 b)(√((∣x^2 −x−6))∣)=x+1 c)(√((x^3 −27))+6<x+3

$${Solve}\:{in}\:\mathbb{R} \\ $$$$\left.{a}\right)\:\mathrm{3}\mid{x}−\sqrt{\mathrm{3}}\mid−\mathrm{8}\sqrt{\left(\mid{x}−\sqrt{\mathrm{3}}\mid\right)+\mathrm{4}}=\mathrm{0} \\ $$$$\left.{b}\left.\right)\sqrt{\left(\mid{x}^{\mathrm{2}} −{x}−\mathrm{6}\right.}\mid\right)={x}+\mathrm{1} \\ $$$$\left.{c}\right)\sqrt{\left({x}^{\mathrm{3}} −\mathrm{27}\right.}+\mathrm{6}<{x}+\mathrm{3} \\ $$

Question Number 104174 Answers: 2 Comments: 0

Π_(n=1) ^∞ ((n/(n+1)))^2

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$

Question Number 104173 Answers: 0 Comments: 0

For fun! S_n =1+2+3+4+5+6+7+... 2S_n = 2 + 4 + 6 +.. −S_n =1+3+5+7+9+11+..... −(−(1/(12)))=1+3+5+7+9+11+... 1+3+5+7+9+....=(1/(12)) S_n =1+2+3+4+5+6+7+.... S_n =1+(1+1)+(1+2)+(1+3)+... S_n =(1+1+1+....)+S_n 1+1+1+1+1+1+....=0

$$\:\mathrm{For}\:\mathrm{fun}! \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+... \\ $$$$\mathrm{2S}_{\mathrm{n}} =\:\:\:\:\mathrm{2}\:\:+\:\:\:\mathrm{4}\:\:\:\:\:\:+\:\:\mathrm{6}\:\:\:\:\:\:+.. \\ $$$$−\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+\mathrm{11}+..... \\ $$$$−\left(−\frac{\mathrm{1}}{\mathrm{12}}\right)=\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+\mathrm{11}+... \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+....=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$ \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+.... \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\left(\mathrm{1}+\mathrm{1}\right)+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{3}\right)+... \\ $$$$\mathrm{S}_{\mathrm{n}} =\left(\mathrm{1}+\mathrm{1}+\mathrm{1}+....\right)+\mathrm{S}_{\mathrm{n}} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+....=\mathrm{0} \\ $$

Question Number 104171 Answers: 0 Comments: 1

App Updates: v2.110 is available now on www.tinkutara.com and will be available in playstore in a day. This version improves drawing: − sidebar buttons are added − Multiple shape selection for alignment can be done from sidebar. − A new polyline option is added so a free handline can be drawn by tapping point. While in polyline mode arrow key can be used to adjust last point.

$$\mathrm{App}\:\mathrm{Updates}: \\ $$$$\mathrm{v2}.\mathrm{110}\:\mathrm{is}\:\mathrm{available}\:\mathrm{now}\:\mathrm{on} \\ $$$$\mathrm{www}.\mathrm{tinkutara}.\mathrm{com}\:\mathrm{and}\:\mathrm{will}\:\mathrm{be} \\ $$$$\mathrm{available}\:\mathrm{in}\:\mathrm{playstore}\:\mathrm{in}\:\mathrm{a}\:\mathrm{day}. \\ $$$$\mathrm{This}\:\mathrm{version}\:\mathrm{improves}\:\mathrm{drawing}: \\ $$$$−\:\mathrm{sidebar}\:\mathrm{buttons}\:\mathrm{are}\:\mathrm{added} \\ $$$$−\:\mathrm{Multiple}\:\mathrm{shape}\:\mathrm{selection}\:\mathrm{for} \\ $$$$\:\:\:\:\:\mathrm{alignment}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{from} \\ $$$$\mathrm{sidebar}. \\ $$$$−\:\mathrm{A}\:\mathrm{new}\:\mathrm{polyline}\:\mathrm{option}\:\mathrm{is}\:\mathrm{added} \\ $$$$\:\:\:\:\:\mathrm{so}\:\mathrm{a}\:\mathrm{free}\:\mathrm{handline}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn} \\ $$$$\:\:\:\:\:\mathrm{by}\:\mathrm{tapping}\:\mathrm{point}.\:\mathrm{While}\:\mathrm{in} \\ $$$$\:\:\:\:\:\mathrm{polyline}\:\mathrm{mode}\:\mathrm{arrow}\:\mathrm{key}\:\mathrm{can}\:\mathrm{be} \\ $$$$\:\:\:\:\:\mathrm{used}\:\mathrm{to}\:\mathrm{adjust}\:\mathrm{last}\:\mathrm{point}. \\ $$

Question Number 104165 Answers: 0 Comments: 1