Question Number 174807 by Eulerian last updated on 11/Aug/22 $$\:\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{vertex}\:\mathrm{form}\:\mathrm{of}\:\mathrm{parabola}\:\mathrm{is}\:\mathrm{given}\:\mathrm{as} \\ $$$$\:\mathrm{y}\:=\:\mathrm{a}\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} +\mathrm{k} \\ $$$$\: \\ $$$$\:\mathrm{From}\:\mathrm{the}\:\mathrm{given}\:\mathrm{diagram}\:\mathrm{of}\:\mathrm{bridge}\:\mathrm{that}\:\mathrm{resembles}\:\mathrm{a}\:\mathrm{parabola}, \\ $$$$\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{vertex}\:\mathrm{points}\:\mathrm{of}\:\left(\mathrm{0},\:\mathrm{30}\right)\:\mathrm{and}\:\mathrm{other}\:\mathrm{points}\:\mathrm{due}\:\mathrm{to}\:\mathrm{towers} \\ $$$$\:\mathrm{that}\:\mathrm{supports}\:\mathrm{the}\:\mathrm{parabolic}−\mathrm{shape}\:\mathrm{cable},\:\mathrm{which}\:\mathrm{is}\:\left(\mathrm{200},\:\mathrm{150}\right). \\ $$$$\: \\ $$$$\:\therefore\:\:\mathrm{y}\:=\:\mathrm{a}\left(\mathrm{x}−\mathrm{0}\right)^{\mathrm{2}}…
Question Number 108104 by bemath last updated on 14/Aug/20 Answered by bobhans last updated on 14/Aug/20 $$\:\:\:\:\:\frac{\mathbb{B}\mathrm{ob}\mathbb{H}\mathrm{ans}}{\bigstar} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{let}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}\:=\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{gradient}\:=\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\mathrm{where}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{k}\:\left(\frac{\mathrm{y}−\mathrm{2}}{\mathrm{x}−\mathrm{1}}\right)\:\mathrm{or}\:\frac{\mathrm{dy}}{\mathrm{y}−\mathrm{2}}\:=\:\mathrm{k}\:\frac{\mathrm{dx}}{\mathrm{x}−\mathrm{1}} \\ $$$$\int\:\frac{\mathrm{dy}}{\mathrm{y}−\mathrm{2}}\:=\:\int\:\frac{\mathrm{k}\:\mathrm{dx}}{\mathrm{x}−\mathrm{1}}\:\Rightarrow\mathrm{ln}\:\left(\mathrm{y}−\mathrm{2}\right)=\:\mathrm{k}\:\mathrm{ln}\:\left(\mathrm{x}−\mathrm{1}\right)\:+\:\mathrm{c} \\ $$$$\mathrm{substitute}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},\mathrm{5}\right)\:\&\:\left(\mathrm{9},\mathrm{8}\right)…
Question Number 40826 by behi83417@gmail.com last updated on 28/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18 $${A}\left({a},{a}\right)\:\:{B}\left(\alpha,\beta\right) \\ $$$${mid}\:{point}\:{AB}\left(\frac{{a}+\alpha}{\mathrm{2}},\frac{{a}+\beta}{\mathrm{2}}\right) \\ $$$${h}=\frac{{a}+\alpha}{\mathrm{2}}\:\:\:\:{k}=\frac{{a}+\beta}{\mathrm{2}} \\ $$$$\alpha=\mathrm{2}{h}−{a}\:\:\:\:\beta==\mathrm{2}{k}−{a} \\ $$$$…
Question Number 104060 by bramlex last updated on 19/Jul/20 $${evaluate}\:\int\int_{{s}} \left({xz}+{y}^{\mathrm{2}} \right){dS}\:{where} \\ $$$${S}\:{is}\:{the}\:{surface}\:{described}\:{by}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{16} \\ $$$$,\:\mathrm{0}\leqslant{z}\leqslant\mathrm{3} \\ $$ Answered by john santu last…
Question Number 38475 by Rio Mike last updated on 26/Jun/18 $${A}\:{committee}\:{of}\:\mathrm{2}\:{girls}\:{and}\:\mathrm{3}{boys} \\ $$$${is}\:{to}\:{be}\:{form}\:{from}\:\mathrm{6}{girls}\:{and}\:\mathrm{8}{boys} \\ $$$${how}\:{many}\:{different}\:{committee}\:{can} \\ $$$${be}\:{formed} \\ $$$$? \\ $$ Answered by MrW3 last…
Question Number 102474 by pticantor last updated on 09/Jul/20 $$\boldsymbol{{U}}{n}=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{{n}} \\ $$$${show}\:{that}\:{we}\:{have}\:\boldsymbol{{p}}_{\boldsymbol{{n}}} \in\mathbb{N}\:/ \\ $$$$\boldsymbol{{U}}_{\boldsymbol{{n}}} =\sqrt{{p}_{{n}} }+\sqrt{{p}_{{n}} +\mathrm{1}} \\ $$ Answered by ~blr237~ last updated…
Question Number 167221 by rexford last updated on 09/Mar/22 $${The}\:{plane}\:{y}=\mathrm{1}\:{slices}\:{the}\:{surface}\: \\ $$$${z}={arctan}\left(\frac{{x}+{y}}{\mathrm{1}−{xy}}\right) \\ $$$${in}\:{a}\:{curve}\:{C}. \\ $$$${Find}\:{the}\:{slope}\:{of}\:{the}\:{tangent}\:{line}\:{to} \\ $$$${C}\:{at}\:{x}=\mathrm{2} \\ $$ Answered by TheSupreme last updated…
Question Number 100320 by bemath last updated on 26/Jun/20 $$\mathrm{Evaluate}\:\int\underset{\mathrm{s}} {\int}\:\overset{\rightarrow} {\mathrm{F}}.\hat {\mathrm{n}}\:\mathrm{dS}\:\mathrm{where}\:\overset{\rightarrow} {\mathrm{F}}=\mathrm{4x}\hat {\mathrm{i}}\:−\mathrm{2y}^{\mathrm{2}} \hat {\mathrm{j}}\:+\mathrm{z}^{\mathrm{2}} \hat {\mathrm{k}}\: \\ $$$$\mathrm{and}\:\mathrm{S}\:\mathrm{is}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cylinder} \\ $$$$\mathrm{bounded}\:\mathrm{by}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{4}\:,\mathrm{z}\:=\:\mathrm{0}\:\mathrm{and}\:\mathrm{z}=\mathrm{3}\:.…
Question Number 100074 by kungmikami last updated on 24/Jun/20 Answered by smridha last updated on 25/Jun/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\int_{\mathrm{0}} ^{\sqrt{\boldsymbol{{y}}}} \left(\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}+\boldsymbol{{xy}}^{\mathrm{2}} \right)\boldsymbol{{dx}}\right]\boldsymbol{{dy}} \\ $$$$=\int_{\mathrm{0}}…
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