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Question Number 189533 Answers: 1 Comments: 0
$${when}\:\:{a}+{b}=\mathrm{60}^{°} \:\:\:\:\:{find}\:\:\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)}=? \\ $$
Question Number 189531 Answers: 1 Comments: 0
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{natural} \\ $$$$\mathrm{numbers}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{satisfy} \\ $$$$\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{gcd}\left(\mathrm{x},\mathrm{y}^{\mathrm{2}} \right)=\mathrm{2023} \\ $$
Question Number 189523 Answers: 0 Comments: 0
Question Number 189522 Answers: 0 Comments: 2
Question Number 189521 Answers: 0 Comments: 0
Question Number 189517 Answers: 1 Comments: 1
Question Number 189514 Answers: 0 Comments: 0
Question Number 189509 Answers: 1 Comments: 0
Question Number 189507 Answers: 0 Comments: 0
Question Number 189506 Answers: 1 Comments: 0
Question Number 189496 Answers: 2 Comments: 0
$$\int\:\mathrm{xe}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:\mathrm{dx} \\ $$
Question Number 189495 Answers: 0 Comments: 7
Question Number 189489 Answers: 1 Comments: 1
$$ \\ $$$$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}} {cos}\left({x}\right){ln}\left({x}\right){dx}=? \\ $$$$\:\:\:\:\:−−− \\ $$$$\:\:\:\:\:\:{f}\:\left({a}\:\right)=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}} {cos}\left({x}\right){x}^{\:{a}} \:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{Re}\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}} .{e}^{\:−{ix}} .{x}^{\:{a}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{Re}\:\int_{\mathrm{0}} ^{\:\infty} \:{e}^{\:−{x}\:\left(\mathrm{1}+{i}\right)} .{x}^{\:{a}} \:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{Re}\left(\mathscr{L}\:\:\left\{\:{x}^{\:{a}} \:\right\}\mid_{\:{s}=\:{i}+\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:{Re}\left(\:\frac{\Gamma\:\left(\mathrm{1}+{a}\right)}{{s}^{\:{a}+\mathrm{1}} }\:\mid_{\:\mathrm{1}+{i}} =\:\frac{\Gamma\:\left(\mathrm{1}+{a}\right)}{\left(\mathrm{1}+{i}\right)^{\:{a}+\mathrm{1}} }\:\right) \\ $$$$\:\:\:\:\:\:\:\:{Re}\:\left(\Gamma\left(\mathrm{1}+{a}\right).\mathrm{2}^{\:\frac{\mathrm{1}+{a}}{\mathrm{2}}} .\:{e}^{\:−\frac{{i}\pi}{\mathrm{4}}\:\left(\mathrm{1}+{a}\right)} \right) \\ $$$$\:\:\:\:\Omega=\:{f}\:'\left(\mathrm{0}\right)=....... \\ $$$$\:\: \\ $$
Question Number 189484 Answers: 1 Comments: 1
Question Number 189501 Answers: 3 Comments: 0
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}+\mathrm{2}{x}+\mathrm{3}{x}+\mathrm{4}{x}+.....+{nx}} −\mathrm{1}}{{x}}=? \\ $$
Question Number 189482 Answers: 2 Comments: 0
$$\mathrm{If}\:\mathrm{tan}\:\mathrm{11}°\:=\:{x}\:\mathrm{then}\:\mathrm{tan}\:\mathrm{1}°\:=\:?\: \\ $$
Question Number 189473 Answers: 1 Comments: 0
Question Number 189461 Answers: 0 Comments: 0
Question Number 189455 Answers: 0 Comments: 0
Question Number 189449 Answers: 1 Comments: 0
Question Number 189446 Answers: 1 Comments: 0
$${x}^{\mathrm{2}} =\sqrt[{\mathrm{5}}]{\mathrm{2}}+{y} \\ $$$${y}^{\mathrm{2}} =\sqrt[{\mathrm{5}}]{\mathrm{2}}+{x} \\ $$$${x}\centerdot{y}=?\:\:\:\:\:\:\:\:\:\:\:{x}\neq{y} \\ $$
Question Number 189445 Answers: 2 Comments: 0
Question Number 189429 Answers: 1 Comments: 1
Question Number 189428 Answers: 1 Comments: 0
Question Number 189426 Answers: 0 Comments: 0
Question Number 189421 Answers: 1 Comments: 1
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