Question Number 96317 by M±th+et+s last updated on 31/May/20 $${find}\:{the}\:{sum} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}×\mathrm{3}}{\mathrm{4}×\mathrm{6}}+\frac{\mathrm{1}×\mathrm{3}×\mathrm{5}}{\mathrm{4}×\mathrm{6}×\mathrm{8}}…..=? \\ $$$${find} \\ $$$$\int\frac{\sqrt{{x}}}{\:\sqrt{{x}}+\sqrt{\mathrm{3}−{x}}}{dx} \\ $$$$ \\ $$$$ \\ $$ Commented by bemath…
Question Number 96314 by Ar Brandon last updated on 31/May/20 $$\mathfrak{a}\backslash\:\mathcal{L}\mathfrak{et}\:\boldsymbol{\mathrm{E}}\left(\mathfrak{x}\right)\:\boldsymbol{\mathrm{d}}\mathfrak{enote}\:\mathfrak{the}\:\mathfrak{whole}\:\mathfrak{number}\:\mathfrak{part}\:\mathfrak{of}\:\mathfrak{the}\:\mathfrak{real} \\ $$$$\left.\mathfrak{number}\:\mathfrak{x},\:\boldsymbol{\mathrm{d}}\mathfrak{etermine}\:\boldsymbol{\mathrm{E}}\left(\mathfrak{x}^{\mathfrak{x}} \right)\:\mathfrak{an}\boldsymbol{\mathrm{d}}\:\boldsymbol{\mathrm{E}}\left(\mathfrak{x}^{\mathfrak{x}^{\mathfrak{x}} } \right)\:\mathfrak{for}\:\mathfrak{x}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\mathfrak{b}\backslash\:\mathcal{C}\mathfrak{alculate}\:\underset{\mathfrak{x}\rightarrow\mathrm{0}} {\mathfrak{lim}}\boldsymbol{\mathrm{E}}\left(\mathfrak{x}^{\mathfrak{x}^{\mathfrak{x}} } \right) \\ $$ Terms of…
Question Number 30780 by mondodotto@gmail.com last updated on 25/Feb/18 Answered by Rasheed.Sindhi last updated on 25/Feb/18 $$\left(\mathrm{4x},\mathrm{4y},\mathrm{4z}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\Rightarrow\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\right. \\ $$$$\left.\mathrm{because}\:\frac{\mathrm{4y}}{\mathrm{4x}}=\frac{\mathrm{4z}}{\mathrm{4y}}\Rightarrow\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{z}}{\mathrm{y}}\right) \\ $$$$\mathrm{y}=\mathrm{x}+\mathrm{d}\:,\mathrm{z}=\mathrm{x}+\mathrm{2d} \\ $$$$\mathrm{x}+\left(\mathrm{x}+\mathrm{d}\right)+\left(\mathrm{x}+\mathrm{2d}\right)=\mathrm{70} \\ $$$$\mathrm{3x}+\mathrm{3d}=\mathrm{70}\Rightarrow\mathrm{x}+\mathrm{d}=\frac{\mathrm{70}}{\mathrm{3}}\Rightarrow\mathrm{x}=\frac{\mathrm{70}−\mathrm{3d}}{\mathrm{3}}…
Question Number 96310 by pticantor last updated on 31/May/20 Commented by pticantor last updated on 31/May/20 $${please}\:{i}\:{need}\:{help}\:{to}\:{solve}\:{that}\:{exercise} \\ $$ Commented by prakash jain last updated…
Question Number 96293 by 06122004 last updated on 31/May/20 Answered by mathmax by abdo last updated on 31/May/20 $$\mathrm{let}\:\mathrm{P}\left(\mathrm{x}\right)\:=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cosa}}\right)\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{2a}\right)}\right)….\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{na}\right)}\right)=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{ka}\right)}\right) \\ $$$$\mathrm{so}\:\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{ka}\right)}\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{this}\:\mathrm{polynom} \\ $$$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{P}^{'}…
Question Number 96287 by Ar Brandon last updated on 31/May/20 $$\mathcal{C}\mathrm{onsider}\:\mathrm{the}\:\mathrm{system}\:\mathrm{in}\:\mathbb{N}^{\mathrm{3}} \\ $$$$\left(\mathrm{S}\right):\:\begin{cases}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} }\\{\mathrm{q}+\mathrm{p}+\mathrm{r}=\mathrm{24}}\\{\mathrm{r}<\mathrm{p}+\mathrm{q}}\end{cases} \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{triplet}\:\left(\mathrm{p}:\mathrm{q}:\mathrm{r}\right)\:\mathrm{is}\:\mathrm{solution}\:\mathrm{to}\:\left(\mathrm{S}\right)\:\mathrm{if} \\ $$$$\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\mathrm{r}<\mathrm{12}.\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation}; \\ $$$$\mathrm{n}^{\mathrm{2}} −\left(\mathrm{24}−\mathrm{r}\right)\mathrm{n}+\mathrm{24}\left(\mathrm{12}−\mathrm{r}\right)=\mathrm{0}\:\mathrm{where}\:\mathrm{n}\:\mathrm{is}\:\mathrm{an}\:\mathrm{unknown}.\mathrm{p} \\ $$…
Question Number 161823 by MathsFan last updated on 22/Dec/21 $$\:{Given}\:{that}\:−\mathrm{1}<{x}<\mathrm{1},\:{find}\:{the} \\ $$$$\:{expansion}\:{of}\:\:\frac{\mathrm{3}−\mathrm{2}{x}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{4}+{x}^{\mathrm{2}} \right)}\:{in} \\ $$$$\:{ascending}\:{power}\:{of}\:{x},\:{up}\:{to}\:{and} \\ $$$$\:{including}\:{the}\:{term}\:{in}\:{x}^{\mathrm{3}} \\ $$ Answered by mr W last updated…
Question Number 96279 by Ar Brandon last updated on 31/May/20 $$\mathcal{F}\mathfrak{ind}\:\mathcal{P}\left(\mathfrak{x}\right)=\prod_{\mathrm{2}} \left(\mathfrak{x}\right)×\prod_{\mathrm{2}\alpha} \left(\mathfrak{x}\right)\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161811 by mnjuly1970 last updated on 22/Dec/21 Answered by mnjuly1970 last updated on 22/Dec/21 $$\mathrm{D}{E}=\sqrt{\mathrm{68}} \\ $$$$\:\:\:\mathrm{68}=\:\mathrm{64}\:+\mathrm{100}−\mathrm{160}{cos}\:\left(\mathrm{E}\overset{\wedge} {\mathrm{A}D}=\alpha\:\right) \\ $$$$\:\:\:\:{cos}\:\left(\alpha\right)=\:\frac{\mathrm{96}}{\mathrm{160}}\:=\:\frac{\mathrm{3}}{\mathrm{5}}\: \\ $$$$\:\:\:\:\:\:{cos}\:\left(\frac{\alpha}{\mathrm{2}}\:\right)=\frac{\mathrm{8}}{{x}}\:\:\Rightarrow\:\frac{\:\mathrm{1}+\frac{\mathrm{3}}{\mathrm{5}}}{\mathrm{2}}\:=\:\frac{\mathrm{64}}{{x}^{\:\mathrm{2}} }…
Question Number 30711 by mondodotto@gmail.com last updated on 24/Feb/18 Answered by mrW2 last updated on 24/Feb/18 $$\mathrm{2}{x}_{\mathrm{1}} −{x}_{\mathrm{2}} ={h} \\ $$$$\Leftrightarrow−\mathrm{6}{x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} =−\mathrm{3}{h} \\ $$$$\left({a}\right)…