Question Number 135816 by ShiimiGee last updated on 16/Mar/21 Commented by ShiimiGee last updated on 16/Mar/21 $$ \\ $$ Commented by liberty last updated on…
Question Number 135800 by benjo_mathlover last updated on 16/Mar/21 Answered by Olaf last updated on 16/Mar/21 $$\mathrm{A}\begin{pmatrix}{−\mathrm{1}}\\{−\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix}\:\mathrm{B}\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\:\mathrm{C}\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix}\:\mathrm{D}\begin{pmatrix}{−\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{AB}\:\mathrm{and}\:\mathrm{CD}\:\mathrm{are}\:\mathrm{two}\:\mathrm{diagonals} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{cube} \\ $$$$\overset{\rightarrow} {\mathrm{AB}}\:=\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:\overset{\rightarrow} {\mathrm{CD}}\:=\:\begin{pmatrix}{−\mathrm{2}}\\{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}…
Question Number 4317 by shiv009 last updated on 10/Jan/16 Commented by prakash jain last updated on 10/Jan/16 $$\mathrm{Vertices}\:\mathrm{of}\:\mathrm{triangle}\:\left(\mathrm{1},\mathrm{3}\right),\:\left(\mathrm{4},−\mathrm{1}\right)\: \\ $$$$\mathrm{area}\:\mathrm{5}.\:\mathrm{Find}\:\mathrm{position}\:\mathrm{of}\:\mathrm{3}^{\mathrm{rd}} \mathrm{vertex}. \\ $$$$\left(\mathrm{0},\:\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{4}}\right) \\…
Question Number 135243 by benjo_mathlover last updated on 11/Mar/21 Commented by benjo_mathlover last updated on 11/Mar/21 Answered by liberty last updated on 13/Mar/21 $${since}\:{AB}\mid\mid{CD}\:{triangles}\:{ABP}\:{and} \\…
Question Number 134457 by bramlexs22 last updated on 03/Mar/21 $$ \\ $$There's a given right triangle ABC and it has a height CD (it has…
Question Number 3291 by Rasheed Soomro last updated on 09/Dec/15 $$\mathcal{A}\:{curve}\:{is}\:{defined}\:\:{by}\:{f}\left({x}\right).\mathcal{W}{hat}\:{is}\:{the}\:{length} \\ $$$${of}\:{curve}\:{from}\:{x}={a}\:\:{to}\:\:{x}={b}\:? \\ $$ Answered by 123456 last updated on 09/Dec/15 $$\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{differentiable} \\ $$$$\underset{{a}}…
Question Number 134287 by EDWIN88 last updated on 02/Mar/21 $$\mathrm{The}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{x}+\mathrm{y}−\mathrm{1}=\mathrm{0}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{cicle} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{6x}−\mathrm{8y}\:=\:\mathrm{0}\:\mathrm{at}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:.\:\mathrm{Then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{which}\:\mathrm{AB}\:\mathrm{is}\:\mathrm{diameter} \\ $$ Answered by bramlexs22 last updated on 02/Mar/21…
Question Number 134074 by EDWIN88 last updated on 27/Feb/21 Commented by EDWIN88 last updated on 27/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{r}\:=\:\mathrm{10}\:\mathrm{cm} \\ $$ Answered by bramlexs22 last updated on…
Question Number 133892 by EDWIN88 last updated on 25/Feb/21 $$\mathrm{If}\:{u}\:\mathrm{and}\:{v}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation}\: \\ $$$$\mathrm{6x}^{\mathrm{2}} −\mathrm{6px}\:+\mathrm{14p}−\mathrm{2}=\mathrm{0},\:\mathrm{where}\:{u}\:;\:{v}\:\mathrm{non}\:\mathrm{integer} \\ $$$$\mathrm{and}\:{u},\:{v}\:\geqslant\:\mathrm{1}\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mid{u}−{v}\mid\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{14}\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{15}\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{16}\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{17}\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{18} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 133771 by bramlexs22 last updated on 24/Feb/21 $$\mathrm{A}\:\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{centered} \\ $$$$\mathrm{at}\:\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{is}\:\mathrm{conjoined}\:\mathrm{to}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{radius}\:\mathrm{2}\:\mathrm{centered}\:\mathrm{at}\:\left(\mathrm{2},\mathrm{0}\right).\: \\ $$$$\mathrm{Forming}\:\mathrm{a}\:\mathrm{single}\:\mathrm{shape}\:.\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{shape}'\mathrm{s}\:\mathrm{area}? \\ $$ Commented by bramlexs22 last updated…