Question Number 73223 by malwaan last updated on 08/Nov/19 $$\boldsymbol{{lim}}\:\frac{\mathrm{0}}{\infty}\:\overset{?} {=}\:\mathrm{0}\:? \\ $$$$\boldsymbol{{lim}}\:\frac{\infty}{\mathrm{0}}\:\overset{?} {=}\:\infty\:? \\ $$ Commented by MJS last updated on 08/Nov/19 $$\mathrm{these}\:\mathrm{are}\:\mathrm{not}\:\mathrm{of}\:\mathrm{correct}\:\mathrm{syntax}\:\Rightarrow\:\mathrm{we}\:\mathrm{cannot} \\…
Question Number 7686 by Alimudin last updated on 09/Sep/16 $${f}\:×=\left(\mathrm{81}\right)\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}\left(\mathrm{9}\right)\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{4} \\ $$$$\left(\mathrm{2}+\frac{\mathrm{1}}{×}\right)^{\mathrm{3}} =… \\ $$$$ \\ $$ Commented by Rasheed Soomro last updated on 09/Sep/16…
Question Number 138758 by EnterUsername last updated on 17/Apr/21 Answered by mr W last updated on 18/Apr/21 Commented by mr W last updated on 18/Apr/21…
Question Number 73221 by aliesam last updated on 08/Nov/19 Answered by MJS last updated on 08/Nov/19 $${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{a}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{a}}\\{{a}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{E}=\begin{pmatrix}{{b}}\\{\mathrm{0}}\end{pmatrix}\:\:{F}=\begin{pmatrix}{{b}}\\{{b}}\end{pmatrix}\:\:{G}=\begin{pmatrix}{\mathrm{0}}\\{{b}}\end{pmatrix} \\ $$$${T}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\:{I}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\frac{{a}−{b}}{\mathrm{2}}}\end{pmatrix} \\ $$$${M}=\begin{pmatrix}{{b}}\\{\frac{{a}−{b}}{\mathrm{2}}}\end{pmatrix}\:\:{N}=\begin{pmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({a}+{b}\right)+{b}}\\{\frac{\mathrm{3}{a}−{b}}{\mathrm{4}}}\end{pmatrix}\:\:{L}=\begin{pmatrix}{{b}}\\{{a}}\end{pmatrix} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{incircle}\:\mathrm{of}\:{ETIM}=\frac{{r}}{\mathrm{2}}=\frac{{a}−{b}}{\mathrm{4}}…
Question Number 7683 by new phone last updated on 08/Sep/16 $$\mathrm{If}\:\mathrm{sin}\:\theta+\mathrm{2cos}\:\theta=\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{2sin}\:\theta−\mathrm{cos}\:\theta=\mathrm{2} \\ $$ Commented by sandy_suhendra last updated on 08/Sep/16 $${sin}\theta+^{} \mathrm{2}{cos}\theta=\mathrm{1} \\…
Question Number 138748 by physicstutes last updated on 17/Apr/21 $$\mathrm{A}\:\mathrm{committee}\:\mathrm{of}\:\mathrm{3}\:\mathrm{members}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{8}\:\mathrm{members}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{committees}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{if}\:\mathrm{two}\:\mathrm{particular} \\ $$$$\mathrm{club}\:\mathrm{members}\:\mathrm{cannot}\:\mathrm{both}\:\mathrm{be}\:\mathrm{in}\:\mathrm{a}\:\mathrm{committee} \\ $$ Answered by mr W last updated on 17/Apr/21 $${C}_{\mathrm{3}}…
Question Number 73210 by TawaTawa last updated on 08/Nov/19 Commented by mr W last updated on 09/Nov/19 $$\left(\mathrm{2}\right)\:{is}\:{not}\:{defined}.\:{no}\:{unique}\:{solution}. \\ $$ Commented by TawaTawa last updated…
Question Number 73211 by vaibhav_0 last updated on 08/Nov/19 $${y}=\left({c}+\mathrm{3}\right)\sqrt{{x}}\:+\frac{\mathrm{3}+{d}}{{x}}−\frac{{a}+\mathrm{4}}{{x}^{{a}+\mathrm{3}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 138742 by mnjuly1970 last updated on 17/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:……..\:{nice}\:\:\:….\:\:\:{calculus}… \\ $$$$\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{{e}^{\pi} −\mathrm{1}} −{x}^{{e}^{\gamma} −\mathrm{1}} }{{ln}\left(\sqrt[{\mathrm{3}}]{{x}}\:\right)}{dx}\overset{?} {=}\mathrm{3}\left(\pi−\gamma\right) \\ $$ Answered by…
Question Number 7668 by 314159 last updated on 08/Sep/16 Commented by Yozzia last updated on 08/Sep/16 $${Suppose}\:{a}+{b}={c},\:{with}\:{a},{b},{c}>\mathrm{0}. \\ $$$${Let}\:{n}=\frac{{a}}{\mathrm{1}+{a}}+\frac{{b}}{\mathrm{1}+{b}}−\frac{{c}}{\mathrm{1}+{c}}. \\ $$$${n}=\frac{{a}}{\mathrm{1}+{a}}+\frac{{b}}{\mathrm{1}+{b}}−\frac{{a}+{b}}{\mathrm{1}+{a}+{b}} \\ $$$${n}=\frac{{a}}{\mathrm{1}+{a}}−\frac{{a}}{\mathrm{1}+{a}+{b}}+\frac{{b}}{\mathrm{1}+{b}}−\frac{{b}}{\mathrm{1}+{a}+{b}} \\ $$$${n}={a}\left(\frac{\mathrm{1}}{\mathrm{1}+{a}}−\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}\right)+{b}\left(\frac{\mathrm{1}}{\mathrm{1}+{b}}−\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}\right)…