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Category: Coordinate Geometry

Question-12902

Question Number 12902 by tawa last updated on 06/May/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 $$\left.\mathrm{1}\right)\:\boldsymbol{{c}}=\mathrm{0}\:\:\:\:\:\:\:\left(\boldsymbol{{x}}=\mathrm{0}\Rightarrow\boldsymbol{{y}}=\mathrm{0}\right) \\ $$$$\boldsymbol{{y}}^{'} =\mathrm{2}\boldsymbol{{ax}}+\boldsymbol{{b}}\:\Downarrow \\ $$$$\left.\mathrm{2}\right)\begin{cases}{\mathrm{2}\boldsymbol{{a}}+\boldsymbol{{b}}=\mathrm{4}}\\{\mathrm{4}\boldsymbol{{a}}+\boldsymbol{{b}}=\mathrm{5}}\end{cases}\Rightarrow\mathrm{2}\boldsymbol{{a}}=\mathrm{1}\Rightarrow\boldsymbol{{a}}=\frac{\mathrm{1}}{\mathrm{2}},\boldsymbol{{b}}=\mathrm{3} \\ $$$$\boldsymbol{{y}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}\:\:\:.\:\blacksquare…

given-the-regular-pyramid-T-ABCD-with-a-square-base-length-AB-8-TC-6-point-P-is-mid-BC-if-x-is-the-angle-between-TP-and-BD-determine-the-value-of-cos-x-

Question Number 78330 by john santu last updated on 16/Jan/20 $${given}\:{the}\:{regular}\:{pyramid} \\ $$$${T}.{ABCD}\:{with}\:{a}\:{square}\:{base} \\ $$$$.\:{length}\:{AB}\:=\:\mathrm{8}\:,\:{TC}\:=\:\mathrm{6}.\:{point} \\ $$$${P}\:{is}\:{mid}\:{BC}.\:{if}\:{x}\:{is}\:{the}\:{angle}\: \\ $$$${between}\:{TP}\:{and}\:{BD}.\:{determine} \\ $$$${the}\:{value}\:{of}\:\mathrm{cos}\:{x}. \\ $$ Terms of…

Given-two-functions-f-x-and-g-x-with-f-1-7-g-2-1-f-1-204-and-g-x-22-What-is-the-derivative-of-f-g-x-at-x-2-

Question Number 12576 by tawa last updated on 26/Apr/17 $$\mathrm{Given}\:\mathrm{two}\:\mathrm{functions}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{f}\left(\mathrm{1}\right)\:=\:\mathrm{7},\:\mathrm{g}\left(\mathrm{2}\right)\:=\:\mathrm{1}\:,\:\mathrm{f}'\left(\mathrm{1}\right)\:=\:\mathrm{204} \\ $$$$\mathrm{and}\:\mathrm{g}'\left(\mathrm{x}\right)\:=\:\mathrm{22}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of}\:\:\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\:\mathrm{at}\:\mathrm{x}\:=\:\mathrm{2} \\ $$ Answered by mrW1 last updated on 26/Apr/17 $$\frac{{df}}{{dx}}=\frac{{df}\left({g}\right)}{{dg}}×\frac{{dg}\left({x}\right)}{{dx}}={f}'\left({g}\left({x}\right)\right)×{g}'\left({x}\right) \\ $$$$={f}'\left({g}\left(\mathrm{2}\right)\right)×{g}'\left(\mathrm{2}\right)={f}'\left(\mathrm{1}\right)×{g}'\left(\mathrm{2}\right)=\mathrm{204}×\mathrm{22} \\…

a-circle-offends-the-y-axis-at-point-0-b-and-through-the-intersection-of-the-curve-y-x-2-x-1-4-value-of-b-

Question Number 77854 by jagoll last updated on 11/Jan/20 $${a}\:{circle}\: \\ $$$${offends}\:{the}\:{y}\:{axis}\:{at}\:{point}\left(\mathrm{0},{b}\right)\: \\ $$$${and}\:{through}\:{the}\:{intersection} \\ $$$${of}\:{the}\:{curve}\:{y}\:=\:{x}\:−\mathrm{2}\sqrt{{x}}+\frac{\mathrm{1}}{\mathrm{4}}.\: \\ $$$${value}\:{of}\:{b}\:=\:? \\ $$ Commented by john santu last…

Let-g-x-be-an-ininitely-differentiable-function-such-that-g-2x-6-g-3x-1-for-all-x-given-that-g-44-3-33-find-g-8-

Question Number 12246 by tawa last updated on 16/Apr/17 $$\mathrm{Let}\:\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{be}\:\mathrm{an}\:\mathrm{ininitely}\:\mathrm{differentiable}\:\mathrm{function}\:,\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{g}\left(\mathrm{2x}\:+\:\mathrm{6}\right)\:=\:\mathrm{g}^{'} \left(\mathrm{3x}\:−\:\mathrm{1}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}. \\ $$$$\mathrm{given}\:\mathrm{that}\:\:\mathrm{g}\left(\frac{\mathrm{44}}{\mathrm{3}}\right)\:=\:\mathrm{33}\:.\:\:\mathrm{find}\:\:\:\mathrm{g}''\left(\mathrm{8}\right) \\ $$ Answered by mrW1 last updated on 16/Apr/17 $${u}=\mathrm{3}{x}−\mathrm{1}\Rightarrow{x}=\frac{{u}+\mathrm{1}}{\mathrm{3}}\Rightarrow\mathrm{2}{x}+\mathrm{6}=\mathrm{2}×\frac{{u}+\mathrm{1}}{\mathrm{3}}+\mathrm{6}=\frac{\mathrm{2}{u}+\mathrm{20}}{\mathrm{3}}…

Question-12127

Question Number 12127 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Apr/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17 $${in}\:{triangle}:{ABC}\:{we}\:{have}: \\ $$$${AB}={m},{AC}={n},{AD}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}, \\ $$$${BD}\bot{DC},\measuredangle{BDA}=\measuredangle{ADC}. \\ $$$${find}:{BD}\:{and}\:{DC}\:{in}\:{term}\:{of}:\:{m},{n}. \\ $$…