Menu Close

Category: Algebra

If-y-f-x-d-2-x-dy-2-e-y-1-and-the-tangent-line-to-the-curve-of-the-function-f-x-on-the-point-x-1-1-is-paralel-to-the-straight-line-g-x-x-3-then-find-f-x-

Question Number 207442 by York12 last updated on 15/May/24 $$\mathrm{If}\:{y}={f}\left({x}\right),\:\frac{{d}^{\mathrm{2}} {x}}{{dy}^{\mathrm{2}} }={e}^{{y}+\mathrm{1}} ,\:\mathrm{and}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:\mathrm{on}\:\mathrm{the}\:\mathrm{point} \\ $$$$\left({x}_{\mathrm{1}} ,−\mathrm{1}\right)\:\mathrm{is}\:\mathrm{paralel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{line}\:{g}\left({x}\right)={x}−\mathrm{3},\:\mathrm{then}\:\mathrm{find}\:{f}'\left({x}\right). \\ $$ Commented by York12 last updated on 15/May/24…

Find-2-5-5-2-1-5-5-3-1-2-1-3-2-sin-7pi-4-

Question Number 207385 by hardmath last updated on 13/May/24 $$\mathrm{Find}: \\ $$$$\sqrt{\left(\mathrm{2},\mathrm{5}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }\:−\:\sqrt[{\mathrm{3}}]{\left.\left(\mathrm{1},\mathrm{5}−\sqrt{\mathrm{5}}\right)^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:−\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$ Commented by Frix last updated on 13/May/24 $$=\frac{\mathrm{7}}{\mathrm{2}}−\sqrt{\mathrm{5}}−\frac{\sqrt{\mathrm{6}\left(−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{4}}−\frac{\sqrt{\mathrm{2}\left(−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}\right.}}{\mathrm{4}}\mathrm{i}…

z-3-2-1-2-i-find-z-11-

Question Number 207408 by hardmath last updated on 13/May/24 $$\boldsymbol{\mathrm{z}}\:\:=\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}}\:\:\:\:\:\mathrm{find}:\:\:\boldsymbol{\mathrm{z}}^{\mathrm{11}} \:=\:? \\ $$ Answered by Frix last updated on 13/May/24 $${z}=\mathrm{e}^{−\frac{\pi}{\mathrm{6}}\mathrm{i}} \\ $$$${z}^{\mathrm{11}} =\mathrm{e}^{−\frac{\mathrm{11}\pi}{\mathrm{6}}\mathrm{i}} =\mathrm{e}^{\left(\mathrm{2}\pi−\frac{\mathrm{11}\pi}{\mathrm{6}}\right)\mathrm{i}}…

Geometric-series-b-4-b-7-b-10-b-1-b-3-b-5-2-12-find-b-5-b-2-

Question Number 207395 by hardmath last updated on 13/May/24 $$\mathrm{Geometric}\:\mathrm{series}: \\ $$$$\frac{\mathrm{b}_{\mathrm{4}} \:\centerdot\:\mathrm{b}_{\mathrm{7}} \:\centerdot\:\mathrm{b}_{\mathrm{10}} }{\mathrm{b}_{\mathrm{1}} \:\centerdot\:\mathrm{b}_{\mathrm{3}} \:\centerdot\:\mathrm{b}_{\mathrm{5}} }\:\:=\:\:\mathrm{2}^{\mathrm{12}} \:\:\:\:\:\mathrm{find}:\:\:\:\frac{\mathrm{b}_{\mathrm{5}} }{\mathrm{b}_{\mathrm{2}} }\:\:=\:\:? \\ $$ Answered by…

a-b-a-how-is-the-solution-

Question Number 207332 by mustafazaheen last updated on 12/May/24 $$\left(\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{b}}\right)×\left(\overset{\rightarrow} {\mathrm{a}}\right)=? \\ $$$$\mathrm{how}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution} \\ $$ Commented by Ghisom last updated on 12/May/24 $$\left(\begin{pmatrix}{{a}_{\mathrm{1}}…

y-tgx-ctgx-8-0-pi-2-Find-min-y-

Question Number 207361 by hardmath last updated on 12/May/24 $$\mathrm{y}\:=\:\frac{\mathrm{tg}\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{ctg}\boldsymbol{\mathrm{x}}}{\mathrm{8}}\:\:\:\:\:,\:\:\:\:\:\left(\mathrm{0}\:;\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{min}\left(\mathrm{y}\right)\:=\:? \\ $$ Answered by Berbere last updated on 12/May/24 $${ctg}\left({x}\right)=\frac{\mathrm{1}}{{y}};{y}={tan}\left({x}\right) \\ $$$$\Leftrightarrow{Min}\left(\frac{\mathrm{1}}{\mathrm{8}}\left({y}+\frac{\mathrm{1}}{{y}}\right);{y}\in\right]\mathrm{0},\infty\left[\right) \\…

x-3-x-2-x-2-0-Find-x-

Question Number 207327 by hardmath last updated on 11/May/24 $$\left(\mathrm{x}−\mathrm{3}\right)\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{2}}\:\:=\:\:\mathrm{0} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by A5T last updated on 11/May/24 $${x}=\mathrm{3}\:{or}\:\mathrm{2}\:{or}\:−\mathrm{1} \\ $$…

Find-1-4-cos-2-40-1-cos-20-2-4-cos-2-40-1-cos-20-

Question Number 207326 by hardmath last updated on 11/May/24 $$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}°\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:=\:? \\ $$$$\mathrm{2}.\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}°\:+\:\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com