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p-0-n-1-p-C-p-1-2-1-n-p-C-n-p-1-2-Please-Help-me-Aidez-moi-

Question Number 162234 by EvaNelle00 last updated on 27/Dec/21 $$ \\ $$$$\:\:\:\:\:\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{p}} {C}_{{p}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(−\mathrm{1}\right)^{{n}−{p}} {C}_{{n}−{p}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\:\:???? \\ $$$$\:\:\:\:\:\:\:{Please}\:{Help}\:{me}\left({Aidez}\:{moi}\right) \\ $$$$ \\…

Question-96682

Question Number 96682 by liki last updated on 03/Jun/20 Commented by liki last updated on 03/Jun/20 $$…\mathrm{am}\:\mathrm{i}\:\mathrm{right}\:\mathrm{if}\:\mathrm{use}\:\mathrm{concept}\:\mathrm{of}\:\mathrm{arthmetic}\:\mathrm{progression}?,\mathrm{please}\:\mathrm{if}\:\mathrm{i}\:\mathrm{do}\:\mathrm{mistake}\:\mathrm{need}\:\mathrm{correcton}! \\ $$ Commented by Rio Michael last updated…

3-2-1-3-2-1-3-3-1-3-3-1-3-4-1-3-4-1-3-2017-1-3-2017-1-

Question Number 162190 by naka3546 last updated on 27/Dec/21 $$\lfloor\:\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{1}}{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}\:+\:\frac{\mathrm{3}^{\mathrm{3}} +\mathrm{1}}{\mathrm{3}^{\mathrm{3}} −\mathrm{1}}\:+\:\frac{\mathrm{3}^{\mathrm{4}} +\mathrm{1}}{\mathrm{3}^{\mathrm{4}} −\mathrm{1}}\:+\:\ldots+\:\frac{\mathrm{3}^{\mathrm{2017}} +\mathrm{1}}{\mathrm{3}^{\mathrm{2017}} −\mathrm{1}}\:\rfloor\:=\:\:? \\ $$ Answered by mr W last…

a-4-b-4-13-is-a-possible-largest-prime-number-a-and-b-are-prime-numbers-Find-a-and-b-

Question Number 31109 by naka3546 last updated on 02/Mar/18 $$\boldsymbol{{a}}^{\mathrm{4}} \:+\:\boldsymbol{{b}}^{\mathrm{4}} \:+\:\mathrm{13}\:\:\:{is}\:\:{a}\:\:{possible}\:\:{largest}\:\:{prime}\:\:{number}\:. \\ $$$$\boldsymbol{{a}}\:\:{and}\:\:\boldsymbol{{b}}\:\:{are}\:\:{prime}\:\:{numbers}\:. \\ $$$${Find}\:\:\boldsymbol{{a}}\:\:{and}\:\:\boldsymbol{{b}}\:. \\ $$ Commented by rahul 19 last updated on…

Question-96616

Question Number 96616 by mathocean1 last updated on 03/Jun/20 Answered by Rio Michael last updated on 03/Jun/20 $${A}\:=\:\begin{pmatrix}{{a}}&{{b}}\\{{b}}&{{c}}\end{pmatrix}\:\mathrm{and}\:{B}\:=\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\ $$$${AB}\:=\:\begin{pmatrix}{{a}}&{{b}}\\{{b}}&{{c}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\begin{pmatrix}{{ax}\:+\:{by}}\\{{bx}\:+\:{cy}}\end{pmatrix} \\ $$$${BA}\:=\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\begin{pmatrix}{{a}}&{{b}}\\{{b}}&{{c}}\end{pmatrix}\:\mathrm{impossible}\:\mathrm{since}\: \\ $$$$\mathrm{order}\:\left(\mathrm{A}\right)\:=\:\mathrm{2}\:×\mathrm{2}\:\:\mathrm{and}\:\mathrm{order}\:\left(\mathrm{B}\right)\:=\:\mathrm{2}×\:\mathrm{1} \\…

Question-96606

Question Number 96606 by Hamida last updated on 03/Jun/20 Commented by Tony Lin last updated on 03/Jun/20 $${y}'=\frac{\mathrm{5}}{\mathrm{13}}{e}^{\frac{\mathrm{5}}{\mathrm{13}}{x}} +\mathrm{13}{sec}\left(\mathrm{13}{x}\right){tan}\left(\mathrm{13}{x}\right) \\ $$ Terms of Service Privacy…

Question-162139

Question Number 162139 by LEKOUMA last updated on 27/Dec/21 Answered by alephzero last updated on 27/Dec/21 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\sqrt{{x}}}{\mathrm{1}−\mathrm{ln}\:\left({e}−{x}\right)}\:=\: \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\left(\sqrt{{x}}\right)}{\frac{{d}}{{dx}}\left(\mathrm{1}−\mathrm{ln}\:\left({e}−{x}\right)\right)}\:=\: \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{+}…