Question Number 25706 by tawa tawa last updated on 13/Dec/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 156548 by mnjuly1970 last updated on 12/Oct/21 Commented by amin96 last updated on 12/Oct/21 $$ \\ $$ Commented by amin96 last updated on…
Question Number 90989 by ajfour last updated on 27/Apr/20 Commented by ajfour last updated on 27/Apr/20 $${If}\:{eq}.\:{of}\:{parabola}\:{is}\:\:\mathrm{9}{y}=\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$${and}\:{AB}={BC}={CD},\:{find}\:{eq}.\:{of} \\ $$$${the}\:{line}\:{AD}. \\ $$ Answered…
Question Number 25441 by rather ishfaq last updated on 10/Dec/17 $${in}\:{what}\:{ratio}\:{in}\:{which}\:{y}−{x}+\mathrm{2}=\mathrm{0}\:{divides}\:{the}\:{line}\:{joining}\:\left(\mathrm{3},−\mathrm{1}\right)\:{and}\:\left(\mathrm{8},\mathrm{9}\right). \\ $$$$ \\ $$ Answered by mrW1 last updated on 10/Dec/17 $${Line}\:{from}\:{A}\left(\mathrm{3},−\mathrm{1}\right)\:{to}\:{B}\left(\mathrm{8},\mathrm{9}\right): \\ $$$$\left({x},{y}\right)=\left(\mathrm{3},−\mathrm{1}\right)+\lambda\left(\mathrm{5},\mathrm{10}\right)…
Question Number 25367 by ajfour last updated on 08/Dec/17 Commented by ajfour last updated on 08/Dec/17 $${Find}\:{area}\:{of}\:\bigtriangleup{PQR}\:. \\ $$ Answered by jota+ last updated on…
Question Number 25236 by Mr eaay last updated on 06/Dec/17 Commented by prakash jain last updated on 07/Dec/17 $$\mid\frac{\alpha{x}_{\mathrm{1}} +\beta{y}_{\mathrm{1}} +{c}}{\:\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }}\mid \\ $$…
Question Number 90773 by ajfour last updated on 26/Apr/20 Commented by ajfour last updated on 26/Apr/20 $${If}\:{equation}\:{of}\:{parabola}\:{is}\:{y}={x}^{\mathrm{2}} , \\ $$$$\:{find}\:{equation}\:{of}\:{AC}. \\ $$ Answered by mr…
Question Number 156296 by cortano last updated on 10/Oct/21 Commented by MJS_new last updated on 10/Oct/21 $$\mathrm{20} \\ $$ Commented by cortano last updated on…
Question Number 25201 by tawa tawa last updated on 06/Dec/17 Answered by ajfour last updated on 06/Dec/17 $${y}=\frac{\mathrm{6}}{\mathrm{3}{x}−\mathrm{2}}\:\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}\mid_{{x}=\mathrm{2}} =−\frac{\mathrm{18}}{\left(\mathrm{3}{x}−\mathrm{2}\right)^{\mathrm{2}} }\mid_{{x}=\mathrm{2}} \\ $$$$\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\mathrm{2}} =−\frac{\mathrm{9}}{\mathrm{8}}\:. \\ $$$${V}=\int_{\mathrm{1}}…
Question Number 25200 by tawa tawa last updated on 06/Dec/17 $$\mathrm{The}\:\mathrm{point}\:\mathrm{A}\:\mathrm{has}\:\mathrm{coordinate}\:\left(−\:\mathrm{1},\:\:−\:\mathrm{5}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{point}\:\mathrm{B}\:\mathrm{has}\:\mathrm{coordinates}\:\left(\mathrm{7},\:\mathrm{1}\right). \\ $$$$\mathrm{The}\:\mathrm{perpendicular}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{AB}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{x}\:−\:\mathrm{axis}\:\mathrm{at}\:\mathrm{C}\:\mathrm{and}\:\mathrm{the}\:\mathrm{y}\:−\:\mathrm{axis}\:\mathrm{at} \\ $$$$\mathrm{D}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{CD} \\ $$ Answered by mrW1 last updated on 06/Dec/17 $${Middle}\:{of}\:{AB}={M}\left(\frac{−\mathrm{1}+\mathrm{7}}{\mathrm{2}},\frac{−\mathrm{5}+\mathrm{1}}{\mathrm{2}}\right)={M}\left(\mathrm{3},−\mathrm{2}\right)…