Question Number 139618 by I want to learn more last updated on 29/Apr/21 Answered by mr W last updated on 29/Apr/21 $$\frac{\mathrm{0}.\mathrm{00015}×\mathrm{0}.\mathrm{4}×\mathrm{1000}}{\mathrm{0}.\mathrm{03}}=\mathrm{2}\:{mm} \\ $$$${a}\:{degree}\:{is}\:\mathrm{2}{mm}\:{length}\:{on}\:{the}\:{scale}. \\…
Question Number 139612 by 676597498 last updated on 29/Apr/21 $${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\sqrt{{x}}\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}} −\mathrm{1}}{dx}\:=\:\mathrm{1}−\frac{{e}}{\left({e}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 139609 by bemath last updated on 29/Apr/21 $$\mathrm{If}\:\mathrm{tan}\:\mathrm{14}°\:=\:\mathrm{x}\:\mathrm{then}\:\mathrm{tan}\:\mathrm{18}°\:=? \\ $$ Answered by qaz last updated on 29/Apr/21 $$\mathrm{18}°=\frac{\mathrm{18}×\mathrm{14}°}{\mathrm{14}}=\frac{\mathrm{9}}{\mathrm{7}}\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{18}°=\mathrm{tan}\:\left(\frac{\mathrm{9}}{\mathrm{7}}\mathrm{tan}^{−\mathrm{1}} {x}\right) \\…
Question Number 139608 by mohammad17 last updated on 29/Apr/21 $${find}\:{the}\:{first}\:{root}\:\left(−\mathrm{8}{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ Answered by floor(10²Eta[1]) last updated on 29/Apr/21 $$\sqrt{−\mathrm{8i}}=\mathrm{a}+\mathrm{bi} \\ $$$$−\mathrm{8i}=\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} +\mathrm{2abi} \\…
Question Number 74075 by ajfour last updated on 18/Nov/19 Commented by ajfour last updated on 18/Nov/19 $${To}\:{Tinkutara}\: \\ $$$${I}\:{was}\:{going}\:{to}\:{delete}\:{it}\:{on}\:{second} \\ $$$${thought}\:{but}\:{i}\:{touched}\:{upvote} \\ $$$${and}\:{cannot}\:{delete}\:{it}\:{now}, \\ $$$$\left({please}\:{help}\:{delete}\:{it}.\right)…
Question Number 74068 by FCB last updated on 18/Nov/19 Commented by FCB last updated on 19/Nov/19 $$\mathrm{thank}\:\mathrm{you} \\ $$ Commented by FCB last updated on…
Question Number 139601 by LUFFY last updated on 29/Apr/21 Commented by LUFFY last updated on 29/Apr/21 $$\mathrm{which}\:\mathrm{one}\:\mathrm{Black}\:\mathrm{or}\:\mathrm{White}\:\mathrm{is}\:\mathrm{correct}?????? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 139600 by rexford last updated on 29/Apr/21 Answered by mr W last updated on 29/Apr/21 $$\mathrm{2}^{{x}} =\mathrm{4}{x} \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{4}{x} \\ $$$${xe}^{−{x}\mathrm{ln}\:\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}…
Question Number 8528 by suci last updated on 14/Oct/16 $${A}\left(\mathrm{2},{b}\right)\:{translation}\:{T}_{\mathrm{1}} =\begin{pmatrix}{−\mathrm{3}}\\{\:\:\:{b}}\end{pmatrix} \\ $$$${followed}\:{by}\:{translation}\:{T}_{\mathrm{2}} =\begin{pmatrix}{{a}}\\{\mathrm{4}}\end{pmatrix} \\ $$$${A}'=\left({b}−\mathrm{1},{a}−\mathrm{3}\right) \\ $$$${determine}\:{the}\:{value}\:{of}\:{a}+{b}=…? \\ $$ Answered by sandy_suhendra last updated…
Question Number 8527 by suci last updated on 14/Oct/16 $$\sqrt{\mathrm{6}−\sqrt{\mathrm{32}}}\:=…? \\ $$ Answered by prakash jain last updated on 14/Oct/16 $$\sqrt{\mathrm{6}−\sqrt{\mathrm{32}}}=\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}=\sqrt{\mathrm{a}}−\sqrt{\mathrm{b}} \\ $$$$\mathrm{squaring}…