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Three-towns-X-Y-and-Z-are-on-a-straight-road-and-Y-is-the-mid-way-between-X-and-Z-A-motor-cyclist-moving-with-uniform-acceleration-passes-X-Y-and-Z-The-speed-with-which-the-motocyclist-passes-X-a

Question Number 26328 by tawa tawa last updated on 24/Dec/17 $$\mathrm{Three}\:\mathrm{towns}\:\mathrm{X},\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}\:\mathrm{are}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}\:\mathrm{and}\:\mathrm{Y}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mid}−\mathrm{way}\:\mathrm{between} \\ $$$$\mathrm{X}\:\mathrm{and}\:\mathrm{Z}.\:\mathrm{A}\:\mathrm{motor}\:\mathrm{cyclist}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{uniform}\:\mathrm{acceleration}\:\mathrm{passes}\:\mathrm{X},\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}.\: \\ $$$$\mathrm{The}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the}\:\mathrm{motocyclist}\:\mathrm{passes}\:\mathrm{X}\:\mathrm{and}\:\mathrm{Z}\:\mathrm{are}\:\mathrm{20m}/\mathrm{s}\:\mathrm{and}\:\mathrm{40m}/\mathrm{s}\: \\ $$$$\mathrm{respectively}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the}\:\mathrm{motorcyclist}\:\mathrm{passes}\:\mathrm{Y}. \\ $$ Commented by tawa tawa last updated…

x-5-x-2-17-x-find-the-value-of-x-

Question Number 26262 by soyebshaikh41@gmail.com last updated on 23/Dec/17 $$\left(\mathrm{x}+\mathrm{5}\right)\:\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{17}−\mathrm{x}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Dec/17 $$\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{10}−\mathrm{17}+\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{27}=\mathrm{0} \\…

given-these-bellow-sequenses-find-fifth-term-1-1-5-1-4-3-11-2-7-2-3-5-9-11-25-19-57-35-3-1-6-7-11-13-16-19-21-4

Question Number 91785 by zainal tanjung last updated on 03/May/20 $$\mathrm{given}\:\:\mathrm{these}\:\mathrm{bellow}\:\mathrm{sequenses}\:.\: \\ $$$$\mathrm{find}\:\mathrm{fifth}\:\mathrm{term} \\ $$$$ \\ $$$$\left.\mathrm{1}\right).\:\:\frac{\mathrm{1}}{\mathrm{5}},\:\frac{\mathrm{1}}{\mathrm{4}}\:,\:\frac{\mathrm{3}}{\mathrm{11}},\:\frac{\mathrm{2}}{\mathrm{7}},… \\ $$$$\left.\mathrm{2}\right).\:\:\frac{\mathrm{3}}{\mathrm{5}},\:\frac{−\mathrm{9}}{\mathrm{11}}\:,\:\frac{−\mathrm{25}}{\mathrm{19}},\:\frac{\mathrm{57}}{\mathrm{35}},… \\ $$$$\left.\mathrm{3}\right).\:\:\:\frac{\mathrm{1}}{\mathrm{6}},\:\frac{\mathrm{7}}{\mathrm{11}}\:,\:\frac{\mathrm{13}}{\mathrm{16}},\:\frac{\mathrm{19}}{\mathrm{21}},… \\ $$$$\left.\mathrm{4}\right).\:\:\:\mathrm{4},−\:\frac{\mathrm{2}}{\mathrm{3}}\:,\:−\frac{\mathrm{4}}{\mathrm{13}},−\:\frac{\mathrm{1}}{\mathrm{5}},… \\ $$$$\left.\mathrm{5}\right).\:\:\mathrm{2},\:\mathrm{9}\:,\:\mathrm{28}\:,\:\mathrm{65},……

1-1-2-2-3-3-4-4-5-2-4-6-10-18-34-3-5-7-11-19-35-4-4-6-10-18-34-5-4-11-30-85-248-

Question Number 91783 by zainal tanjung last updated on 03/May/20 $$\left.\mathrm{1}\right).\:\:\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{3}}{\mathrm{4}}\:,\:\frac{\mathrm{4}}{\mathrm{5}},\:…,\:…. \\ $$$$\left.\mathrm{2}\right).\:\:\mathrm{4},\mathrm{6},\mathrm{10},\mathrm{18},\mathrm{34},…,…. \\ $$$$\left.\mathrm{3}\right).\:\:\mathrm{5},\mathrm{7},\mathrm{11},\mathrm{19},\mathrm{35},…,…. \\ $$$$\left.\mathrm{4}\right).\:\:\mathrm{4},\mathrm{6},\mathrm{10},\mathrm{18},\mathrm{34},…,…. \\ $$$$\left.\mathrm{5}\right).\:\:\mathrm{4},\mathrm{11},\mathrm{30},\mathrm{85},\mathrm{248},…,… \\ $$ Commented by Tony Lin…

If-normal-plasma-is-7-4-and-normal-CO-2-is-1-2mm-what-is-the-normal-H-2-CO-3-

Question Number 26246 by tawa tawa last updated on 22/Dec/17 $$\mathrm{If}\:\:\mathrm{normal}\:\mathrm{plasma}\:\mathrm{is}\:\mathrm{7}.\mathrm{4}\:\mathrm{and}\:\mathrm{normal}\:\:\mathrm{CO}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{1}.\mathrm{2mm},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{normal}\:\left(\mathrm{H}_{\mathrm{2}} \mathrm{CO}_{\mathrm{3}} ^{−} \right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-91715

Question Number 91715 by 174 last updated on 02/May/20 Commented by mathmax by abdo last updated on 02/May/20 $${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{{n}}}} \:\:\:\:\frac{{n}\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{dx}\:{changement}\:{nx}\:={t}\:{give}\:…