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AllQuestion and Answers: Page 1405

Question Number 70429    Answers: 3   Comments: 6

Question Number 70425    Answers: 0   Comments: 0

Question Number 70719    Answers: 1   Comments: 0

∫sin (101x)sin^(99) x dx

$$\int\mathrm{sin}\:\left(\mathrm{101x}\right)\mathrm{sin}\:^{\mathrm{99}} \mathrm{x}\:\mathrm{dx} \\ $$

Question Number 70418    Answers: 0   Comments: 0

Question Number 70417    Answers: 1   Comments: 0

Question Number 70395    Answers: 1   Comments: 1

Question Number 70394    Answers: 1   Comments: 0

montrer que sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)

$$\mathrm{montrer}\:\mathrm{que} \\ $$$$\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}=\mathrm{4cos}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{C}}{\mathrm{2}} \\ $$

Question Number 70397    Answers: 0   Comments: 1

Question Number 70390    Answers: 0   Comments: 0

Question Number 70423    Answers: 2   Comments: 1

Question Number 70385    Answers: 1   Comments: 0

The value of determinant determinant (((x+2),( x+3),(x+5)),((x+4),( x+6),(x+9)),((x+8),(x+11),(x+15))) is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{determinant} \\ $$$$\begin{vmatrix}{{x}+\mathrm{2}}&{\:{x}+\mathrm{3}}&{{x}+\mathrm{5}}\\{{x}+\mathrm{4}}&{\:{x}+\mathrm{6}}&{{x}+\mathrm{9}}\\{{x}+\mathrm{8}}&{{x}+\mathrm{11}}&{{x}+\mathrm{15}}\end{vmatrix}\:\mathrm{is} \\ $$

Question Number 70383    Answers: 0   Comments: 1

If a matrix A is such that 3A^3 +2A^2 +5A+I=0, then A^(−1) is equal to

$$\mathrm{If}\:\mathrm{a}\:\mathrm{matrix}\:{A}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that}\:\mathrm{3}{A}^{\mathrm{3}} +\mathrm{2}{A}^{\mathrm{2}} +\mathrm{5}{A}+{I}=\mathrm{0}, \\ $$$$\mathrm{then}\:{A}^{−\mathrm{1}} \mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 70399    Answers: 1   Comments: 3

Solve x^4 + x^3 −2ax^2 −ax + a^2 = 0, a ∈ R

$$\boldsymbol{{Solve}}\:\:\boldsymbol{{x}}^{\mathrm{4}} \:+\:\boldsymbol{{x}}^{\mathrm{3}} \:−\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} \:−\boldsymbol{{ax}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} =\:\mathrm{0},\:\:\boldsymbol{{a}}\:\in\:\mathbb{R} \\ $$

Question Number 70370    Answers: 1   Comments: 4

If gcd(p , q)=1,prove that gcd(p(p+q) , q(p+q) , pq)=1 Related to Q#69939

$${If}\:\:{gcd}\left({p}\:,\:{q}\right)=\mathrm{1},{prove}\:{that} \\ $$$$\:\:\:\:\:{gcd}\left({p}\left({p}+{q}\right)\:,\:{q}\left({p}+{q}\right)\:,\:{pq}\right)=\mathrm{1} \\ $$$$\mathrm{R}\boldsymbol{\mathrm{elated}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{Q}}#\mathrm{69939} \\ $$

Question Number 70369    Answers: 0   Comments: 1

Question Number 70361    Answers: 0   Comments: 0

Hello si(x)=−∫_x ^∞ ((sin(x))/x)dx show ∫_0 ^(+∞) x^(a−1) si(x)dx=−((Γ(a)sin(((πa)/2)))/a) hint ipp +complex Analysis

$${Hello}\: \\ $$$${si}\left({x}\right)=−\int_{{x}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$$${show}\:\int_{\mathrm{0}} ^{+\infty} {x}^{{a}−\mathrm{1}} {si}\left({x}\right){dx}=−\frac{\Gamma\left({a}\right){sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}{{a}} \\ $$$${hint}\:{ipp}\:+{complex}\:{Analysis} \\ $$

Question Number 70312    Answers: 1   Comments: 1

If, (1/a^2 )+(1/b^2 )+(1/c^2 ) = (1/(ab))+(1/(bc))+(1/(ca)) then prove that, a=b=c.

$$\mathrm{If},\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{ca}}\:\mathrm{then}\:\mathrm{prove}\: \\ $$$$\mathrm{that},\:\mathrm{a}=\mathrm{b}=\mathrm{c}. \\ $$

Question Number 70364    Answers: 0   Comments: 3

solve L=lim_(x→0) ((e^x −(1/(1−x)))/x^2 )

$$\mathrm{solve}\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} −\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} } \\ $$

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 70277    Answers: 2   Comments: 0

Question Number 70270    Answers: 1   Comments: 0

If log_x y = 6 & log_(14x) 8y = 3 then find the value of x & y.

$$\mathrm{If}\:\mathrm{log}_{\mathrm{x}} \mathrm{y}\:=\:\mathrm{6}\:\&\:\mathrm{log}_{\mathrm{14x}} \mathrm{8y}\:=\:\mathrm{3}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\&\:\mathrm{y}. \\ $$

Question Number 70262    Answers: 0   Comments: 1

calculate ∫_0 ^(π/4) ln(cosx)dx and ∫_0 ^(π/4) ln(sinx)dx

$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cosx}\right){dx}\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sinx}\right){dx} \\ $$

Question Number 70256    Answers: 0   Comments: 1

Question Number 70253    Answers: 0   Comments: 3

Question Number 70252    Answers: 1   Comments: 0

If, log x^y = 6 and log 14x^(8y) = 3 then find the value of x, y.

$$\mathrm{If},\:\mathrm{log}\:\mathrm{x}^{\mathrm{y}} \:=\:\mathrm{6}\:\mathrm{and}\:\mathrm{log}\:\mathrm{14x}^{\mathrm{8y}} \:=\:\mathrm{3}\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x},\:\mathrm{y}. \\ $$

Question Number 70287    Answers: 0   Comments: 2

I wanted to say this earlier... I love mathematics and I also love people. But I′m not here to solve the same old boring problems copied from facebook or whatsapp or other platforms. They are not interesting at all. They have been coming in as a kind of competition, or simply to brag, they′ve been traded from one non−mathematician to the other. No use to copy−paste problems that you found somewhere on the web and that you might not even be able to fully understand. If you need an explanation, you′re welcome, but I won′t do your homework and I won′t answer impolite posts. Last but not least: I won′t try to solve the famous unsolved problems much better people were not able to cope with.

$$\mathrm{I}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{say}\:\mathrm{this}\:\mathrm{earlier}... \\ $$$$\mathrm{I}\:\mathrm{love}\:\mathrm{mathematics}\:\mathrm{and}\:\mathrm{I}\:\mathrm{also}\:\mathrm{love}\:\mathrm{people}. \\ $$$$\mathrm{But}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{here}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{same}\:\mathrm{old}\:\mathrm{boring} \\ $$$$\mathrm{problems}\:\mathrm{copied}\:\mathrm{from}\:\mathrm{facebook}\:\mathrm{or}\:\mathrm{whatsapp} \\ $$$$\mathrm{or}\:\mathrm{other}\:\mathrm{platforms}.\:\mathrm{They}\:\mathrm{are}\:\mathrm{not}\:\mathrm{interesting} \\ $$$$\mathrm{at}\:\mathrm{all}.\:\mathrm{They}\:\mathrm{have}\:\mathrm{been}\:\mathrm{coming}\:\mathrm{in}\:\mathrm{as}\:\mathrm{a}\:\mathrm{kind} \\ $$$$\mathrm{of}\:\mathrm{competition},\:\mathrm{or}\:\mathrm{simply}\:\mathrm{to}\:\mathrm{brag},\:\mathrm{they}'\mathrm{ve} \\ $$$$\mathrm{been}\:\mathrm{traded}\:\mathrm{from}\:\mathrm{one}\:\mathrm{non}−\mathrm{mathematician} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{other}. \\ $$$$ \\ $$$$\mathrm{No}\:\mathrm{use}\:\mathrm{to}\:\mathrm{copy}−\mathrm{paste}\:\mathrm{problems}\:\mathrm{that}\:\mathrm{you} \\ $$$$\mathrm{found}\:\mathrm{somewhere}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web}\:\mathrm{and}\:\mathrm{that}\:\mathrm{you} \\ $$$$\mathrm{might}\:\mathrm{not}\:\mathrm{even}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{fully}\:\mathrm{understand}. \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{need}\:\mathrm{an}\:\mathrm{explanation},\:\mathrm{you}'\mathrm{re}\:\mathrm{welcome}, \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{won}'\mathrm{t}\:\mathrm{do}\:\mathrm{your}\:\mathrm{homework}\:\mathrm{and}\:\mathrm{I}\:\mathrm{won}'\mathrm{t} \\ $$$$\mathrm{answer}\:\mathrm{impolite}\:\mathrm{posts}. \\ $$$$ \\ $$$$\mathrm{Last}\:\mathrm{but}\:\mathrm{not}\:\mathrm{least}:\:\mathrm{I}\:\mathrm{won}'\mathrm{t}\:\mathrm{try}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the} \\ $$$$\mathrm{famous}\:\mathrm{unsolved}\:\mathrm{problems}\:\mathrm{much}\:\mathrm{better} \\ $$$$\mathrm{people}\:\mathrm{were}\:\mathrm{not}\:\mathrm{able}\:\mathrm{to}\:\mathrm{cope}\:\mathrm{with}. \\ $$

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