Question Number 29490 by Victor31926 last updated on 09/Feb/18 $$\mathrm{A}\:\mathrm{gas}\:\mathrm{expands}\:\mathrm{according}\:\mathrm{to}\:\mathrm{the}\:\mathrm{law}\:\mathrm{PV}=\mathrm{K}\:\left(\mathrm{constant}\right). \\ $$$$\mathrm{Initialy},\:\boldsymbol{\mathrm{v}}=\mathrm{1000cubic}\:\mathrm{metres}\:\mathrm{and}\:\boldsymbol{\mathrm{p}}=\mathrm{40N}/\mathrm{m}^{\mathrm{2}} . \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{pressure}\:\mathrm{is}\:\mathrm{decreased}\:\mathrm{at}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{5N}/\mathrm{m}^{\mathrm{2}} /\mathrm{min}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{at}\:\mathrm{which} \\ $$$$\mathrm{the}\:\mathrm{gas}\:\mathrm{is}\:\mathrm{expanding}\:\mathrm{when}\:\mathrm{its}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{2000cubic}\:\mathrm{metres}. \\ $$$$ \\ $$ Answered by Rasheed.Sindhi…
Question Number 160558 by mnjuly1970 last updated on 01/Dec/21 $$ \\ $$$$\:\:\:\:{solve} \\ $$$$ \\ $$$$\lfloor\:{x}−\:\sqrt{\mathrm{1}−{x}^{\:\mathrm{2}} }\:\rfloor+\lfloor\:{x}+\:\sqrt{\mathrm{1}−{x}^{\:\mathrm{2}} }\:\rfloor=\mathrm{0} \\ $$$$ \\ $$ Answered by mr…
Question Number 160545 by alcohol last updated on 01/Dec/21 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}}{{n}^{\mathrm{2}} +{k}}\: \\ $$$$\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}\:} {\overset{{n}} {\sum}}{cos}\left(\frac{\mathrm{1}}{\:\sqrt{{n}+{k}}}\right) \\ $$$${convergente}? \\ $$$$ \\ $$ Terms of…
Question Number 160528 by amin96 last updated on 01/Dec/21 $$\left(\mathrm{2}\boldsymbol{\mathrm{cosh}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{y}}\right)\right)\boldsymbol{\mathrm{dx}}+\left(\boldsymbol{\mathrm{sinh}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{y}}\right)\right)\boldsymbol{\mathrm{dy}}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160530 by alcohol last updated on 01/Dec/21 Answered by Rasheed.Sindhi last updated on 01/Dec/21 $$\mathrm{I}\left(\mathrm{1}\right) \\ $$$$\mathrm{v}=\frac{\mathrm{s}}{\mathrm{t}};\:\mathrm{s}:\:\mathrm{distance},\mathrm{v}:\:\mathrm{average}\:\mathrm{speed},\:\mathrm{t}:\mathrm{time} \\ $$$$\underline{{a}.\:\:\mathrm{10km}/\mathrm{h}} \\ $$$$\mathrm{t}=\frac{\mathrm{s}}{\mathrm{v}}=\frac{\mathrm{40km}}{\mathrm{10km}/\mathrm{h}}=\mathrm{4}\:\mathrm{hours} \\ $$$$\mathrm{First}\:\mathrm{20km}\:\mathrm{with}\:\mathrm{20km}/\mathrm{h}…
Question Number 94984 by bobhans last updated on 22/May/20 $$\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\frac{\mathrm{1}}{\mathrm{x}}\:+\sqrt{\mathrm{x}}\:\right]\:? \\ $$ Commented by MJS last updated on 22/May/20 $$\mathrm{sorry}\:\mathrm{but}\:\mathrm{how}\:\mathrm{can}\:\mathrm{someone}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{a}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{time} \\ $$$$\mathrm{be}\:\mathrm{unable}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{basic}\:\mathrm{derivations}? \\…
Question Number 29442 by prof Abdo imad last updated on 08/Feb/18 $${splve}\:{the}\:{d}.{e}\:\:\:{xy}^{'} =\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:}\:+{y}\:\:{with}\:{x}>\mathrm{0} \\ $$ Answered by mrW2 last updated on 09/Feb/18 $${xy}^{'}…
Question Number 94861 by Ar Brandon last updated on 21/May/20 $$\mathrm{Show}\:\mathrm{that} \\ $$$$\left(\mathrm{g}\circ\mathrm{f}\right)'\left(\mathrm{x}\right)=\mathrm{g}'\left(\mathrm{f}\left(\mathrm{x}\right)\right)\centerdot\mathrm{f}'\left(\mathrm{x}\right) \\ $$ Commented by john santu last updated on 21/May/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{u}\:\Rightarrow\mathrm{y}=\mathrm{g}\left(\mathrm{u}\right) \\…
Question Number 160363 by mnjuly1970 last updated on 28/Nov/21 $$ \\ $$$$\:\:\:\:\:{s}>\mathrm{0} \\ $$$$\:\:\:\:{lim}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\left(\:−\mathrm{1}\right)^{\:{k}} .\left(\frac{\:{k}}{\mathrm{2}{n}}\:\right)^{\:{s}} =\:\:? \\ $$ Terms of Service Privacy Policy…
Question Number 94765 by Hamida last updated on 20/May/20 Terms of Service Privacy Policy Contact: info@tinkutara.com